INSTRUCTOR’S SOLUTIONS MANUAL CALCULUS AND ITS APPLICATIONS THIRD EDITION Marvin L. Bittinger David by welldoneassistant

INSTRUCTOR’S SOLUTIONS MANUAL

CALCULUS

AND ITS APPLICATIONS THIRD EDITION Marvin L. Bittinger David J. Ellenbogen Scott A. Surgent Gene F. Kramer

Table of Contents Chapter R: Functions, Graphs, and Models ............................................................1 Chapter 1: Differentiation ...................................................................................101 Chapter 2: Exponential and Logarithmic Functions ...........................................249 Chapter 3: Applications of Differentiation .........................................................340 Chapter 4: Integration .........................................................................................550 Chapter 5: Applications of Integration ...............................................................691 Chapter 6: Functions of Several Variables .........................................................809 Chapter 7: Trigonometric Functions ...................................................................945 Chapter 8: Differential Equations .......................................................................994 Chapter 9: Sequences and Series ......................................................................1069 Chapter 10: Probability Distributions ...............................................................1161 Chapter 11: Systems and Matrices.....................................................................1218 Chapter 12: Combinatorics and Probability.......................................................1325

iii

Chapter R Functions, Graphs, and Models Exercise Set R.1 1.

1

6

Graph y  x  4 . We choose some x-values and calculate the corresponding y-values to find some ordered pairs that are solutions of the equation. Then we plot the points and connect them with a smooth curve.

2

 x, y   2, 2   0, 4   3, 7 

Graph y  x  1 . 6.

1 x. 4

Graph y  

Graph y   3 x . We choose some x-values and calculate the corresponding y-values to find some ordered pairs that are solutions of the equation. Then we plot the points and connect them with a smooth curve.

 x, y   1, 3  0, 0   2, 6 

5 Graph y   x  3 . 3

2 x4. 3 We choose some x-values and calculate the corresponding y-values to find some ordered pairs that are solutions of the equation. Then we plot the points and connect them with a smooth curve. y x  x, y  Graph y 

3

6

4

2

 3, 6   0, 4   3, 2 

2 Chapter R: Functions, Graphs, and Models 7.

Graph x  y  4 .

Next, we choose some x-values and calculate the corresponding y-values to find some ordered pairs that are solutions of the equation. Then we plot the points and connect them with a smooth curve.

Graph x  y  5 . We solve for y first. x y 5 subtract x from both sides y  5 x commutative property y  x  5 Next, we choose some x-values and calculate the corresponding y-values to find some ordered pairs that are solutions of the equation. Then we plot the points and connect them with a smooth curve.

1

 x, y   1, 6   0, 5   2, 3 

2

 x, y   2, 0   2,1  6, 2 

11. Graph 2 x  5 y  10 .

12. Graph 5 x  6 y  12 . We solve for y first. 5 x  6 y  12 subtract 5 x from both sides 6 y  12  5 x 9.

1 divide both sides by  6 12  5 x  6 5 y  2  x 6 5 y  x2 6 We choose some x-values and calculate the corresponding y-values to find some ordered pairs that are solutions of the equation. Then we plot the points and connect them with a smooth curve. y

Graph 6 x  3 y  9 .

10. Graph 8 y  2 x  4 . We solve for y first. 8 y  2x  4 8 y  2x  4 2 4 y  x 8 8 1 1 y  x 4 2

add 2x to both sides

6

7

divide both sides by 8

2

 x, y   6, 7   0, 2   6, 3 

Exercise Set R.1 3 16. Graph x  2  y 2 . Since x is expressed in terms of y we first choose values for y and then compute x. Then we plot the points that are found and connect them with a smooth curve.

13. Graph y  x 2  3 .

14. Graph y  x 2  5 . We choose some x-values and calculate the corresponding y-values to find some ordered pairs that are solutions of the equation. Then we plot the points and connect them with a smooth curve.

 x, y 

2

1

4

5

4

1

 2, 1  1, 4   0, 5  1, 4   2, 1

 x, y 

2

1

2

 2, 2  1, 1  2, 0   1,1  2, 2 

17. Graph y  5 .

We choose some x-values and calculate the corresponding y-values to find some ordered pairs that are solutions of the equation. Then we plot the points and connect them with a smooth curve.

2

1

15. Graph x  y 2  2 .

 x, y   2, 5   1, 5   0, 5  1, 5   2, 5 

4 Chapter R: Functions, Graphs, and Models 18. Graph y  2 .

22. Graph y  1  x 3 . First we solve for y. y  1  x3

19. Graph y  7  x 2 . We choose some x-values and calculate the corresponding y-values to find some ordered pairs that are solutions of the equation. Then we plot the points and connect them with a smooth curve.

 x, y 

2

1

 2, 3   1, 6   0, 7  1, 6   2, 3 

20. Graph y  5  x 2 .

21. Graph y  7  x 3 .

y  x3  1 Next, we choose some x-values and calculate the corresponding y-values to find some ordered pairs that are solutions of the equation. Then we plot the points and connect them with a smooth curve.

 x, y 

2

9

1

2

1

 2, 9   1, 2   0, 1 1, 0   2, 7 

23. Consider the equation, R  10.55 x  40.6 . We substitute x  7 in R  10.55 x  40.6 . R  10.55 x  40.6 R  10.55  7   40.6

R  73.85  40.6 R  114.45 billion dollars. According to this model, the world record for the mile in 2025 is approximately 114.45 billion dollars. We substitute x  10 in R  10.55 x  40.6 . R  10.55 x  40.6 R  10.55 10  40.6

R  105.5  40.6 R  146.1 billion dollars. According to this model, the world record for the mile in 2028 is approximately 146.1 billion dollars. We substitute x  17 in R  10.55 x  40.6 . R  10.55 x  40.6 R  10.55 17   40.6 R  179.35  40.6 R  219.95 billion dollars.

Exercise Set R.1 5

Where, a  0.0467, b  1.6384, c  118.9231

According to this model, the world record for the mile in 2035 is approximately 219.95 billion dollars. 24.

t

A  0.5t 4  3.45t 3  96.65t 2  347.7t , 0t6 We substitute t  2

t

 344.4 Approximately 344.4 milligrams of ibuprofen will remain in the blood stream 2 hours after 400 mg have been swallowed.

25. a) S  0.0467 t 2  1.6384t  1.0769 We substitute t  10 . S  0.0467 10   1.6384 10   1.0769 2

S  0.0467 100   1.6384 10   1.0769 S  4.67  16.384  1.0769 S  22.13 mm . In 2003 , the sea had risen 22.13 mm over the 1993 level. b) S  0.0467 t 2  1.6384t  1.0769 We substitute t  32 . S  0.0467  32   1.6384  32   1.0769 2

S  0.0467 1024  1.6384  32  1.0769 S  47.8208  52.4288  1.0769 S  101.3 mm . In 2025 , the sea had risen 101.3 mm over the 1993 level. c) S  0.0467 t 2  1.6384t  1.0769 We substitute S  120 . 120  0.0467  t   1.6384  t   1.0769 2

0.0467 t 2  1.6384t  118.9231  0

t

b  b 2  4ac 2a

1.6384 

1.6384   4  0.0467  118.9231 2  0.0467  2

t  35.88339166, and t  70.96690343 Since t must be positive, t  1993  35.88  2028.88 In 2028 , the sea level will have risen more than 120 meter over the 1993 level.

A  0.5  2   3.45  2   96.65  2   347.7  2  4

b  b 2  4ac 2a

26. a. A  0.002 x 2  0.924 x  0.152 We substitute x  21.3 A  0.002(21.3) 2  0.924(21.3)  0.152  18.62 The optimum angle, in degrees, to tilt the solar panel in Honolulu is 18.62. b. A  0.002 x 2  0.924 x  0.152 We substitute x  39.1 A  0.002(39.1) 2  0.924(39.1)  0.152  32.919 The optimum angle, in degrees, to tilt the solar panel in Kansas City is 32.919. c.

A  0.002 x 2  0.924 x  0.152 We substitute x  53.5 A  0.002(53.5) 2  0.924(53.5)  0.152  43.558 The optimum angle, in degrees, to tilt the solar panel in Edmonton is 43.558.

27. a) N  0.018t  3.09 We substitute t  18 N  0.018 18  3.09

N  0.324  3.09 N  3.414 In 2025 , the number of girl athletes will be approximately 3.414 million. b) N  0.018t  3.09 We substitute N  3.6 . 3.6  0.018t  3.09 0.51  0.018t t  28.33 In the year 2007  28.33  2035.33 , the number of girl athletes will be approximately 3.6 million.

6 Chapter R: Functions, Graphs, and Models

N  328.851 thousand. According to this model, the number of homeless is 328.851 thousand.

28. a) N   0.0028t 2  0.0606t  3.008 We substitute t  18 N  0.0028 18   0.0606 18   3.008 2

N  0.0028  324  0.0606 18  3.008 N  0.9072  1.0908  3.008 N  3.19 million. According to this model, in 2025 there will be approximately 3.19 million girl high school athletes. This model predicts a slightly lower level than the model in exercise 27. b) N   0.0028t 2  0.0606t  3.008 We substitute t  30 . N  0.0028  30   0.0606  30   3.008 2

N  0.0028  900  0.0606  30  3.008 N  2.52  1.818  3.008 N  2.31 million. According to this model, in 2037 there will be approximately 2.31 million girl high athletes. c) Answers may vary. d) Answers may vary. 29. a) Consider the equation, N   0.080t 2  10.838t  549.855 .

We substitute t  8 . N   0.080t 2  10.838t  549.855

N  0.080  8   10.838  8   549.855 2

N  0.080  64  10.838  8  549.855 N  5.12  86.704  549.855 N  458.031 thousand. According to this model, the number of homeless is 458.031 thousand. We substitute t  13 . N   0.080t 2  10.838t  549.855

N  0.080 13  10.838 13  549.855 2

N  0.080 169  10.838 13  549.855 N  13.52  140.894  549.855 N  395.441 thousand. According to this model, the number of homeless is 395.441 billion dollars. We substitute t  18 . N   0.080t 2  10.838t  549.855

N  0.080 18   10.838 18   549.855 2

N  0.080  324  10.838 18  549.855 N  25.92  195.084  549.855

30.

s  t   16t 2

s  t   28 28  16t 2 28  t2 16 28  t2 16 1.3228  t Danny took approximately 1.32 seconds to hit the ramp.

31. a) A  P 1  i 

A  500 1  0.0425  A  500 1.0425 

A  500 1.132995515625 A  556.50 At the end of 3 year, the investment is worth $556.50 . i  b) A  P  1   n  

2 3

 0.0425  A  500  1   2  

 2  0.0425  A  500   2    2.0425  A  500    2 

A  500 1.02125 

A  500 1.1344684 A  567.23 At the end of 3 year, the investment is worth $567.23 . i  c) A  P  1   n  

4 3

 0.0425  A  500 1   4  

 4  0.0425  A  500   4   12

 4.0425  A  500    4 

Exercise Set R.1 7 A  500 1.010625 

i  b) A  P  1   n  

A  500 1.135221 A  567.61 At the end of 3 year, the investment is worth $567.61 . i  d) A  P  1   n  

3653

1095

24

 1000 1.050625 

i  c) A  P  1   n  

1095

 365  0.0425  A  500   365  

 0.05  A  1000  1   4  

1095

 365.0425  A  500    365 

44

 1000 1.0125 

A  500 1.0001164 

1095

A  500 1.13592873 A  567.99 At the end of 3 year, the investment is worth $567.99 . e) There are 24  365  8760 hours in one year. 87603 i  A  P 1    n  0.0425  A  500  1   8760  

26280

 8760  0.0425  A  500   8760    8, 760.0425  A  500    8760 

26280

 1219.89 At the end of 4 year, the investment is worth $1219.89. i  d) A  P  1   n 

26280

A  500 1.1359845 A  567.99 At the end of 3 year, the investment is worth $567.99 . t

 1215.51 At the end of 4 year, the investment is worth $1512.51.

 0.05  A  1000  1   365  

3654

 1000 1.000136986 

1460

 1221.39 At the end of 4 year, the investment is worth $1221.39. e) There are 24  365  8760 hours in one year. i  A  P 1   n 

A  500 1.00000485159817 

A  1000 1  0.05 

 0.05  A  1000  1   2  

 1218.40 At the end of 4 year, the investment is worth $1218.40.

 0.0425  A  500  1   365  

32. a) A  P 1  i 

0.05   A  1000  1    8760 

87604

 1000 1.000005708 

35,040

 1221.40 At the end of 4 year, the investment is worth $1221.40. 33. Principal Amount: $300, 000. Interest: 4.2% Time: 1 year. nt  r Formula Used: A  P  1    n a) Compounded Annually. 1 A  300, 000 1  0.042  A  $ 312, 600

8 Chapter R: Functions, Graphs, and Models

b) Compounded Semi Annually. 2  0.042  A  300, 000 1   2   A  $ 312, 732.30

c) Compounded Quarterly. 4  0.042  A  300, 000 1   4   A  $ 312, 799.84

d) Compounded Daily. 365  0.042  A  300, 000  1   365   A  $ 312, 867.59

e) Compounded Hourly. 365 24 0.042   A  300, 000  1    365  24  A  $ 312,868.31

34. Principal Amount: $100, 000 . Interest: 5.1% Time: 1 year. nt  r Formula Used: A  P  1    n a) Compounded Annually. 1 A  100, 000 1  0.051 A  $1 05,100 b) Compounded Semi Annually.

 0.051  A  100, 000  1   2  

n. Then we use a calculator to perform the computation. 36  0.046  0.046    1    12  12   M  18, 000  36    0.046   1   1  12      536.25 The monthly payment on the loan will be approximately $536.25. 36. Principal Amount: $100, 000 Interest: 3.9% Time: 30 Years Number of months: n  30 12  360 .  0.039  0.039 360   1    12  12    M  100, 000 360  0.039  1    1 12  

 0.00325 1.00325   100, 000 360

1.00325

A  $1 05,165.03

360

1

M  $471.67 The monthly payment on the loan will be approximately $471.67.

c) Compounded Quarterly. 4  0.051  A  100, 000  1   4  

 1  r  n  1   37. W  P  r   We substitute 3000 for P, 0.0305  3.05%  0.0305  for r, and 18 for n.

A  $1 05,198.37

d) Compounded Daily. 365  0.051  A  100, 000  1   365   A  $1 05, 231.91

e) Compounded Hourly. 365 24 0.051   A  100, 000  1    365  24  A  $105, 232.27

35. Using the formula: n  r  r    1     12 12   M  P  n   r   1    1    12   We substitute 18,000 for P , 0.046  4.6%  0.046  for r, and 36  3  12  36  for

 1  0.0305 18  1   W  3000  0.0305    70, 561.01 Rounded to the nearest cent, the annuity will be worth $70,561.01 after 18 years.

Exercise Set R.1 9 38. We substitute 50,000 for W, 0.0725  1   4 %  0.0425  in for r, and 20 for n. Then 4   we proceed to solve for P.

 1  0.0425  20  1   50, 000  P  0.0425   50, 000 P  1  0.0425  20  1    0.0425   1635.99  P You will need to invest $1635.99 annually to reach a goal of $50,000 after 20 years. 39. a) 5.8% or above Locate 5.8% on the vertical axis and then think of horizontal lines extending across the graph from this point. The years for which the graph lies above this line are the years for which the unemployment rate was at or above 5.8%. This time period was 2008 – 2014 and 2020. b) Below 7% Locate 7 on the vertical axis and then think of a horizontal line extending across the graph from this point. The years for which the graph lies below this line are the years for which unemployment was below 7% . Those time periods are 2007, 2008, 2014 – 2019 and 2021. c) Highest point Locate the highest point on the graph and extend a line vertically to the horizontal axis. The year in which the unemployment rate was the highest was 2010 at 9.6%. d) Lowest point Locate the lowest point on the graph and extend a line vertically to the horizontal axis. The year in which the unemployment rate was the highest was 2019 at 3.7%. 40. [Answer may vary] No, we can’t estimate the unemployment rate with the help of the graph provided. For first two years, the rate increases. Then, smoothly decreases. At the end, reach some peak value. At the right most, again decreases rapidly.

As the changes are not having any uniformity and regularity, it is hard to estimate the unemployment rate for any other years after 2019 or before 2010.  1  r  n  1   we 41. a) Using the formula W  P  r   substitute 1200 for P, 0.04  4%  0.04  for r and 35 for n.  1  0.04 35  1   W  1200  0.04    88, 382.67 Tig will have approximately $88,382.67 in her account when she retires. b) Tig invested $1200 per year for 35 years. Therefore, the total amount of her original payments is: $1200  35  $42, 000 . Since the total amount in the account was $88,382.67, the interest earned over the 35 years is: $88, 382.67  $42, 000  $46, 382.67

Therefore, $42,000 was the total amount of Sally’s payments and $46,382.67 was the total amount of her interest. 42. a) Using the formula: n  r  r    1    12 12   M  P  n   r   1    1    12   We substitute 88,382.67 for P, 0.04  4%  0.04  for r, and 180  (80  65)  12  180  for n. Then we

use a calculator to perform the computation. 180  0.04  0.04    1    12  12   M  206, 780.16  180    0.04   1   1  12      653.76 Tig should take a monthly payment of $653.76. b) Tig received 180 payments of $653.76, therefore she received a total of $653.76  180  $117, 676.80 during the 15 years. Of that 42,000 was what she originally contributed, leaving $75,676.80 in interest.

Chapter R: Functions, Graphs, and Models

43. Using the formula:

360

r  Y  1    1 n   We substitute 0.041  4.1%  0.041 for r, and

4 (quarterly) for n. Then we use a calculator to perform the computation. 4

 0.041  Y  1   1 4    0.0416  4.16% The annual yield of 4.1% compounded quarterly would be approximately 4.16%.

44. Using the formula:

0.04   Y  1   1 360    0.0408  4.08% The annual yield of 4% compounded daily would be approximately 4.08%.

47. a. Using the formula: n

r  Y  1    1 n  We substitute 0.03  3%  0.03  for r, and 1

(annually) for n. Then we use a calculator to perform the computation. 1

r  Y  1    1 n  We substitute 0.053  5.3%  0.053  for r, and

12 (monthly) for n. Then we use a calculator to perform the computation. 12

0.053   Y  1   1 12    0.0543  5.43% The annual yield of 5.3% compounded monthly would be approximately 5.43%.

0.03   Y  1   1 1    0.03  3% The annual yield for Sierra Savings will be 3%.

Using the formula: n

r  Y  1    1 n  We substitute 0.0297  2.97%  0.0297  for

r, and 52 (weekly) for n. Then we use a calculator to perform the computation. 52

45. Using the formula: n

r  Y  1    1 n   We substitute 0.0375  3.75%  0.0375  for r,

and 52 (weekly) for n. Then we use a calculator to perform the computation. 52

0.0375   Y  1   1 52    0.0382  3.82% The annual yield of 3.75% compounded weekly would be approximately 3.82%.

46. Using the formula:

0.0297   Y  1   1 52    0.0301  3.01% The annual yield for Foothill Bank will be approximately 3.01% b. Foothill Bank has a higher yield.

48. a. Using the formula: n

r  Y  1    1 n  We substitute 0.025  2.5%  0.025  for r,

and 1 (annually) for n. Then we use a calculator to perform the computation. 1

r  Y  1    1 n  We substitute 0.04  4%  0.04  for r, and 360

(daily) for n. Then we use a calculator to perform the computation.

0.025   Y  1   1 1    0.025  2.5% The annual yield for Western Bank will be 2.5%.

Using the formula: n

r  Y  1    1 n 

Exercise Set R.1 11 We substitute 0.0243  2.43%  0.0243  for r, and 12 (monthly) for n. Then we use a calculator to perform the computation.

52. Graph y  2 x The resulting graph is

0.0243   Y  1   1 12    0.0246  2.46% The annual yield for Commonwealth Savings will be approximately 2.46% b. Western Bank has a higher yield.

53. Graph of y  x  x . In standard window of DESMOS: 3

49. Using the formula: n

r  Y  1    1 n  We substitute 0.022  2.2%  0.022  for Y, and

12 (monthly) for n. Then we use a calculator to perform the computation. 12

r   0.022   1    1  12  r  0.02178  2.178% Mesalands would need to pay at least 2.178% to exceed an annual yield of 2.2%.

50. Using the formula: n

r  Y  1    1 n   We substitute 0.0375  3.75%  0.0375  for Y,

Graph of y  2 In standard window of DESMOS:

and 4 (quarterly) for n. Then we use a calculator to perform the computation. 4

r  0.0375   1    1 4  r  0.03698  3.698% Shea Savings would need to pay at least 3.698% to exceed an annual yield of 3.75%.

51. Graph y  x 3  x 2 The resulting graph is:

Chapter R: Functions, Graphs, and Models

54. Graph of x  y  5 y  2 3

The window  , , 4, 4 gives the best result to sketch the graph.

Exercise Set R.2 13 14. The correspondence is a function because every book has a unique Library of Congress number.

Exercise Set R.2 1.

The correspondence is a function because each member of the domain corresponds to only one member of the range.

The correspondence is not a function because one member of the domain, 6, corresponds to two members of the range, – 6 and – 7.

The correspondence is a function because each member of the domain corresponds to only one member of the range.

The different food item is rated with different rates. This correspondence between the food item and its cost is a function because each member of the domain corresponds to only one member of the range. The correspondence is a function because each member of the domain corresponds to only one member of the range. The correspondence is a function because each number that is two less than the number is a number.

The correspondence is a function because each number doubled is exactly one number.

The correspondence is a function because each cubed root of a number is exactly one number.

10. The correspondence is a function because each square root of a number is exactly one positive number. 11. The correspondence is not a function because there is more than one odd integer less than or equal to the given number. 12. The correspondence is not a function because there may be more than one number less than or equal to a positive number.

15. The lowest temperature of the day is considered as exactly one temperature on each day. In a single day, there is no chance to have a two or more lowest temperature. Hence, the correspondence on the date mm/dd/yyyy with the number which represents the least temperature experienced on that date is a function. 16. The highest temperature may have the chance to get in two or more days of the year. So, the correspondence on the highest temperature experienced on the date mm/dd/yyyy is not a function. 17. The same set of dosages can be given to the people of same age group or with same immunity power. So, the correspondence on the dosages given to the set of peoples is not a function. 18. Each people consume uniquely one dosage while taking medicine. So, the correspondence on the people consuming the set of dosages is a function. 19. This correspondence is a function, because a rectangles perimeter is determined by the specific length and width. 20. This correspondence is a function, because a rectangles area is determined by the specific length and width. 21. The correspondence is not a function because there may be more than one length and width to obtain the given perimeter. 22. The correspondence is not a function because there may be more than one length and width to obtain the given area. 23. a)

f  x   3x  2

x f  x

4.1 14.3

13. The correspondence is a function because every book has a unique ISBN.

4.01 14.03

4.001 14.003

4 14

14 Chapter R: Functions, Graphs, and Models b) f  x   3 x  2

26.

f  5   3  5   2  17

g  3    3   4  13 2

f  1  3  1  2  1

g 0   0  4  4 2

f  k   3  k   2  3k  2

g  1   1  4  5 2

f  x  h   3  x  h   2  3 x  3h  2

24. a)

g  x  x2  4

g  7    7   4  53 2

f  x   4x  3

g  a  h    a  h   4  a 2  2ah  h 2  4 2

f  5.1  4  5.1  3  17.4 f  5.01  4  5.01  3  17.04

g  x  h  g  x

f  5.001  4  5.001  3  17.004



f  5   4  5   3  17 x f  x

5.1 17.4

5.01 17.04

5.001 17.004

27.

f  4   4  4   3  13

f  x  f  3 

f  3  4  3  3  9 f  2   4  2   3  11

 x  5 2 1

3  5

f  1 

f  k   4  k   3  4k  3 f  x  h   4  x  h   3  4 x  4h  3

25.

2 xh  h 2  h  2x  h

5 17

b) f  x   4 x  3



x 2  2 xh  h 2  4  x 2  4

f k  

g  x   x2  3 g  1   1  3  1  3  2 g  0    0   3  0  3  3



 2 

 1  5 



 6 

1 4



1 36

 k  52

f  x  h 

  x  h   5



 x  h  5 2

28.

g 1  1  3  1  3  2 2

g  5    5   3  25  3  22

f  x 

g  a  h    a  h   3  a  2ah  h  3 2

 2  g  x  h   g  x   x  h   3   x   3  h h 2 x  2 xh  h 2  3   x 2  3  h 2 xh  h 2  h h  2x  h   h  2x  h 2

f  4  f  0  f a 

 x  32 1

  4   3

  0   3

  a   3

f  x  h 

  

7

 3



1 49



1 9

 a  32

  x  h   3



 x  h  32



Exercise Set R.2 15 f  x  h  f  x h 1 



  



 x  h  32  x  32 h 2 x  h  3  x  32    x  h  32  x  32  x  h  32  x  32





32. a. Graph k  x   x 2  3 x .

h 2

x  6 x  9  x  2hx  6 x  h 2  6h  9 h  x  h  3  x  3 2



33. Graph f  x   3 x  1 .

2hx  h 2  6h h  x  h  3  x  3 2

h  2 x  h  6  h  x  h  3  x  3 2

2 x  h  6

 x  h  3  x  3 2

h0

29. a. Graph g  x   3 x  4 . b.

34. Graph f  x   2 x  5 .

First, we choose some values for x and compute the values for f  x  , in order to form the ordered pairs that we will plot on the graph. f  1  2  1  5  7 f  0   2  0   5  5 f 1  2 1  5  3 f  2   2  2   5  1

30. a. Graph f  x   4 x  2 . b.

f  x

 x, f  x  

1

7

5

3

 1, 7   0, 5  1, 3   2, 1

2 1 Next we plot the input – output pairs from the table and, in this case, draw the line to complete the graph.

31. a. Graph h  x   x 2  x . b.

16 Chapter R: Functions, Graphs, and Models 35. Graph g  x   2 x .

36. Graph g  x   4 x .

First, we choose some values for x and compute the values for g  x  , in order to form the

f  x

2

1

2

1

 x, f  x    2, 2   1, 1  0, 2  1, 1  2, 2 

2 2 Next we plot the input – output pairs from the table and, in this case, draw the curve to complete the graph.

ordered pairs that we will plot on the graph. g  1  4  1  4 g  0   4  0   0 g 1  4 1  4 .

g  x

1

4

 x, g  x    1, 4   0, 0   0,  4 

38. Graph f  x   x 2  4 .

Next we plot the input – output pairs from the table and, in this case, draw the line to complete the graph. 39. Graph g  x    x 2  1 .

37. Graph f  x   x 2  2 .

First, we choose some values for x and compute the values for f  x  , in order to form the ordered pairs that we will plot on the graph. Choosing some values for x and evaluating the function, we have: f  2    2   2  2 2

f  1   1  2  1 2

40. Graph f  x   6  x 2 .

First, we choose some values for x and compute the values for f  x  , in order to form the ordered pairs that we will plot on the graph. f  2   6   2   2 2

f  0    0   2  2 2

f  1  6   1  5

f 1  1  2  1 2

f 0  6  0  6

f  2   2  2  2

f 1  6  1  5

f  2  6   2  2 2

Exercise Set R.2 17 x

f  x

2

1

 x, f  x    2, 2   1, 5   0, 6  1, 5   2, 2 

2 2 Next we plot the input – output pairs from the table and, in this case, draw the curve to complete the graph.

42. Graph g  x  

1 3 x . 2

41. Graph g  x   x 3 .

First, we choose some values for x and compute the values for g  x  , in order to form the ordered pairs that we will plot on the graph. g  2    2   8 3

43. The graph is a function, it is impossible to draw a vertical line that intersects the graph more than once. 44. The graph is a function, it is impossible to draw a vertical line that intersects the graph more than once.

g  1   1  1 3

g 0  0  0 3

g 1  1  1 3

45. The graph is not that of a function. A vertical line can intersect the graph more than once.

g  2   2  8 3

f  x

 x, f  x  

2

8

1

 2, 8   1, 1  0, 0  1,1  2, 8 

8 2 Next we plot the input – output pairs from the table above and, in this case, draw the curve to complete the graph.

46. The graph is a function, it is impossible to draw a vertical line that intersects the graph more than once. 47. The graph is not that of a function. A vertical line can intersect the graph more than once. 48. The graph is a function, it is impossible to draw a vertical line that intersects the graph more than once. 49. The graph is a function, it is impossible to draw a vertical line that intersects the graph more than once. 50. The graph is not that of a function. A vertical line can intersect the graph more than once. 51. The graph is a function, it is impossible to draw a vertical line that intersects the graph more than once.

18 Chapter R: Functions, Graphs, and Models 52. The graph is a function, it is impossible to draw a vertical line that intersects the graph more than once. 53. The graph is a function, it is impossible to draw a vertical line that intersects the graph more than once.

58. Graph x  y 2  2 . a) First, we choose some values for y (since x is expressed in terms of y) and compute the values for x , in order to form the ordered pairs that we will plot on the graph. For y  2; x   2   2  2 2

54. The graph is a function, it is impossible to draw a vertical line that intersects the graph more than once.

For y  1; x   1  2  1 2

For y  0; x   0   2  2 2

For y  1; x  1  2  1 2

55. The graph is not that of a function. A vertical line can intersect the graph more than once.

For y  2; x   2   2  2 2

56. The graph is not that of a function. A vertical line can intersect the graph more than once. 57. Graph x  y 2  5 a) First, we choose some values for 𝑦 (since 𝑥 is expressed in terms of 𝑦) and compute the values for 𝑥 , in order to form the ordered pairs that we will plot on the graph.

For y  2 , x   2   5  9 2

2

1

2

1

 x, y   2, 2   1, 1  2, 0   1,1  2, 2 

2 2 Next we plot the input – output pairs from the table and, in this case, draw the curve to complete the graph.

For y  1 , x   1  5  6 2

For y  0 , x   0   5  5 2

For y  1 , x  1  5  6 2

For y  2 , x   2   5  9 2

2

1

 x, y   9,  2   6,  1  5, 0   6,1  9, 2 

Plot the points and trace the curve as follows:

b) The graph is not that of a function. A vertical line can intersect the graph more than once. 59.

f  x   x2  4x f  x  h  f  x h 

 x  h  2  4  x  h    x 2  4 x  h 2



b) The graph is not that of a function. A vertical line can intersect the graph more than once.



x  2 xh  h

 4 x  4h   x 2  4 x  h

2 xh  h  4h h h  2x  h  4

h  2 x  h  4,

h0

Exercise Set R.2 19 60.

64. When x  5 , the function is defined by f  x   2 x  1; for x  0 ; therefore,

f  x   x 2  3x f  x  h  f  x

f  5   2  5   1  10  1  11 .

h 

 x  h  2  3  x  h    x 2  3x  h x  2 xh  h  3 x  3h   x  3 x  2



2 xh  h 2  3h combining  like terms h h  2 x  h  3  Factoring h  2 x  h  3, h  0

When x  5 , the function is defined by 1 f  x   x  1; for x  4 ; therefore, 2 1 5 7 f  5    5   1   1   3.5 . 2 2 2

 1, for x  2, 65. Graph f  x     x, for x  2 The function is defined such that f  2   1 and

61. To find f  1 we need to locate which piece defines the function on the domain that contains x  1 . When x  1 , the function is defined by f  x   2 x  1; for x  0 ; therefore, f   1   2   1  1  2  1  3 .

To find f 1 we need to locate which piece

for all other x-values (that is, for x  2 ), we have f  x   x . Thus, to graph this function, we graph the line given by y  x , but with an open dot at the point corresponding to x  2 . To complete the graph, we plot the point  2, 1 since f  2   1 . The graph is sketched as below:

defines the function on the domain that contains x  1 . When x  1 , the function is defined by f  x   x 2  3;

for 0  x  4 ; therefore,

f 1  1  3  1  3  2 . 2

62. To find f  0  we need to locate which piece defines the function on the domain that contains x  0. When x  0 , the function is defined by f  x   17; for x  0 ; therefore, f  0   17 .

To find f 10  we need to locate which piece defines the function on the domain that contains x  10 . When x  10 , the function is defined 1 by f  x   x  1; for x  4 ; therefore, 2 1 f 10   10   1  5  1  6 . 2

63. When x  3 , the function is defined by f  x   2 x  1; for x  0 ; therefore,

  5, for x  3, 66. Graph f  x     x  1, for x  3 The function is defined such that f  3  5 and for all other x-values (that is, for x  3 ), we have f  x   x  1 . Thus, to graph this function, we graph the line given by y  x  1 , but with an open dot at the point corresponding to x  3 . To complete the graph, we plot the point  3, 5  since f  3  5 .

f  3   2  3   1  6  1  7 .

When x  3 , the function is defined by f  x   x 2  3;

for 0  x  4 ; therefore,

f  3   3  3  9  3  6 . 2

20 Chapter R: Functions, Graphs, and Models Tabulate the values on the graph: x

 x, y 

5

4

3

2

1

 5,  4   4,  3  2,  1  0, 1 1, 2 

The graph is sketched as below:

 5, for x  0, 68. Graph f  x    3  x , for x  0 For any negative x values, the function is defined as the constant value 5. For x  0 , the function is defined as f  x   5 . Plot the coordinate point  0, 5 . Tabulate the values on the graph:

 6, for x  0, 67. Graph f  x    2  x , for x  0 For any negative x values, the function is defined as the constant value 6 . At x  0 , the function is defined as f  x   6 . Plot the

 x, y  1, 1  2, 8

Plot the points and join them to sketch the function f  x   x3 and plot the coordinate point  0, 0  .

coordinate point  0, 6  . Tabulate the values on the graph: x

 x, y  1, 1  2, 4

Plot the points and join them to sketch the function f  x   x 2 and plot the coordinate point  0, 0  with an open dot.

for x  2  7, . 69. Graph f  x    2  x  3, for x  2 First, we graph f  x   7 for x  2 . This graph consists of only one point,  2, 7  . The solid dot indicates that  2, 7  is part of the graph.

Exercise Set R.2 21

Next, we graph f  x   x 2  3 for inputs x  2 . Note that for f  x   x 2  3 f  3    3   3  6 2

f  1   1  3  2 2

  x, for x  0  for x  0 . 71. Graph g  x    4,  x  2, for x  0  First, we graph g  x    x for inputs x  0 . Creating the input – output table, we have:

f  0    0   3  3 2

f 1  1  3  2 2

f  3   3  3  6 We create the input – output table. Choosing some values for x and evaluating the function, we have: 2

f  x

 x, f  x  

3

1

2

3

2

 3, 6   1, 2   0, 3  1, 2   3, 6 

Since the input x  2 is not defined on this part of the graph, the point  2,1 is not part of the graph. The open circle indicates that  2,1 is not part of the graph.

g  x

 x, g  x  

 3, 3   2, 2  2 2  1,1 1 1 The open circle indicates that  0, 0  is not part 3

of the graph. Next, we graph g  x   4 for x  0 . This part of the graph consists of a single point. The solid dot indicates that  0, 4  is part of the graph. Next, we graph g  x   x  2 for inputs x  0 . Creating the input – output table, we have: x

g  x

 x, g  x  

1, 3   2, 4  2 4 3 5  3, 5  The open circle indicates that  0, 2  is not part 1

of the graph.

for x  3  6, 70. Graph f  x    2 .   x  5, for x  3

 2 x  3,  72. Graph g  x   5,  x  2, 

for x  1 for x  1 . for x  1

22 Chapter R: Functions, Graphs, and Models  12 x  1, for x  2  73. Graph g  x    4, for x  2 .  x  3, for x  2  First, we graph g  x   12 x  1 for inputs x  2 .

Creating the input – output table, we have: x

g  x

 x, g  x  

2

1

 2, 2   0, 1

 12

1,  12 

75. Commission. a) Graph the function 2500, for 0  x  15, 000,  S  x    0.04 x  2500, for x  15,000 For positive x values until 15000 , the function is defined as the constant value 2500 . At x  15000 , the function is defined as S  x   2500 . Plot the coordinate

15000, 2500  with a open dot point. Tabulate the values on the graph: x

 x, y 

The open circle indicates that  2, 0  is not part

15000

3100

of the graph. Next, we graph g  x    4 for x  2 . This

16000

3140

15000, 3100  16000, 3140 

part of the graph consists of a single point. The solid dot indicates that  2,  4  is part of the graph. Next, we graph g  x   x  3 for inputs x  2 .

Plot the points and join them to sketch the function S  x   0.04 x  2500 and mark the coordinate 15000, 3100  with a point.

Creating the input – output table, we have: x

g  x

 x, g  x  

 3, 0   4,1 4 1 5  5, 2  2 The open circle indicates that  2, 1 is not part 3

of the graph.

b) Monthly salary when the employee had $21000 in sales. As $21000 greater than $15000 , the function represents the salary is S  x   0.04 x  2500 . S  21000   0.04  21000   2500 S  21000   840  2500 S  21000   $3340

 x2 , for x  0  3, for g x   x  0. 74. Graph     2 x  3, for x  0 

c) Monthly salary when the employee had $14500 in sales. As $14500 lesser than $15000 , the function represents the salary is S  x   2500 . S 14500   $2500

Exercise Set R.2 23 76. Economies of scale. a) Graph the function 3.50 x, for 0  x  1, 000,  C  x   2.25 x  1250, for x  1, 000 Tabulate the values on the graph: x

 x, y 

1000

3500

2500

6875

1000, 3500   0, 0   2000, 6875 

77.

0.03   A  t   500  1   4    500 1.0075 

4 2

 500 1.061598848   530.7994239  530.80 The investment will be worth approximately $530.80 after 2 years.

At x  1000 , the function C  x   2.25 x  1250 also takes the value 3500 . Plot the points Join them to get the required graph. The graph is sketched as follows:

0.03   A  t   P 1   4   We substitute 500 in for P and 2 in for t:

78.

0.03   A  t   P 1   4  

0.03   A  t   800  1   4  

43

 800 1.0075 

 875.05 The investment will be worth approximately $875.05 after 3 years. 79.

b) Total cost for ordering 900 caps. As 900 less than 1000 , the function represents the cost of caps is C  x   3.5 x . C  900   3.5  900   $ 3.150

The total cost for ordering 900 caps is $ 3.150  $3.150 1000  $3,150 . c) Total cost for ordering 1900 caps. As 1900 greater than 1000 , the function represents the cost of caps is C  x   2.25 x  1250 . C 1900   2.25 1900   1250 C 1900   4275  1250 C 1900   $ 5,525

hw 3600 a) We substitute 170 for h and 70 for w. 170  70  s  1.818 3600 The patient’s approximate surface area is 1.818 m 2 b) We substitute 170 for h and 100 for w. 170  100  s  2.173 3600 The patient’s approximate surface area is 2.173 m 2 c) We substitute 170 for h and 50 for w. 170  50  s  1.537 3600 The patient’s approximate surface area is 1.537 m 2 s

24 Chapter R: Functions, Graphs, and Models

80.

s

85. Solve the equation for y. 2 x  y  16  4  3 y  2 x

hw 3600

a) s 

150  70 

 1.708 3600 The patient’s approximate surface area is 1.708 m 2 .

b) s 

y  16  4  3 y 4 y  20 y5 We sketch a graph of the equation.

180  70 

 1.871 3600 The patient’s approximate surface area is 1.871 m 2 .

No vertical line meets the graph more than once. Thus, the equation represents a function.

81. Hospitality occupancy rate. a) N  5400, A  200

20 N 73 A 20  5400 BOR  73  200  7.3972  7.40% b) BOR  100%, A  150 BOR 

86. First we solve the equation for y. 2 y 2  3x  4 x  5

2y2  x  5 subtract 3 x from both sides x  5 y2  divide both sides by 2 2 x5 y take the square root of both sides 2 We sketch a graph of the equation.

20 N N 73 A 73 A   100 N 20 73 150   100 N 20  54750 patients.

BOR 

82. Hospitality occupancy rate. a) N  11450, A  120

20 N 73 A 20  11450 BOR   26.14% 73  120 b) Answers may vary BOR 

83. a) Yes, each term has a specific monthly payment. b) Yes, each month payment has a specific term. c) No, the interest rate has multiple monthly payments. d) Yes, each monthly payment has a specific interest rate.

We can see that a vertical line will intersect the graph more than once; therefore, this is not a function.

87. First, we solve the equation for y.

   72x 3

3y 2

9 y 3  72 x

y 3  8x y  3 8x y  23 x  y  0 Note: since y must be non-negative to satisfy the original equation, we only graph the points for which y is non-negative.

84. a) Yes, the table represents a function. Each event is assigned exactly one scale of impact number. b) The inputs are the events; the outputs are the scale of impact numbers.

Exercise Set R.2 25 Next, we sketch a graph of the equation:

Next, we type the equation into the graphing editor.

No vertical line meets the graph more than once. Thus, the equation represents a function.

Now, we are able to look at the table. We will have to enter the appropriate values of x.

88. First we solve the equation for y.

 4 y   64 x 4 y   4 x 2

y2  x

From the table we conclude: f  3   2

y x Next, we sketch a graph of the equation.

f  2   0 f  0   2 f  4  2

91.

f  x   x 3  2 x 2  4 x  13

92.

f  x 

We can see that a vertical line will intersect the graph more than once; therefore, this is not a function.

89. Answers will vary. The vertical line test works, because you are locating the values to which each input corresponds. For a relation to be a function, each input must correspond with exactly one output. Therefore, if a vertical line intersects the graph in more than one point, that particular input corresponds to more than one output and the graph is not a function. 90.

3 2

x 4 We begin by setting up the table:

f  x  x  2  x 1  5

First, we set up the table to allow us to ask the calculator to compute specific values.

Next, we will type in the equation into the graphing editor.

26 Chapter R: Functions, Graphs, and Models

Now, we are able to look at the table:

93. Each graph is shown below. In order to graph f  x   x 3  2 x 2  4 x  13 ,

we use the window at the top of the next column.

After entering the function into the graphing editor, we get:

94. Answers will vary. Some ordered pairs are  2.978723,1.0617  ,  1.382979, 2.8438301 ,

 0, 3.1622777  , 1.7021277, 2.6651006  , and  2.6595745,1.7107494  . 95. a) f  x   5( x  2) g  x   5x  2

After entering the function into the graphing editor, we get:

In order to graph f  x  

3 x2  4

c) No, the graph is not the same function.

, we use the

standard window:

96. a) f  x   ( x  4) 2 g  x  x2  4

After entering the function into the graphing editor, we get:

c) No, the graph is not the same function. 97.

In order to graph f  x   x  2  x  1  5 , we use the standard window.

f  x   3x  6 g  x   3( x  h )

The functions represent the same function, so set them equal to each other to find the value for h. 3 x  6  3( x  h) 3 x  6  3 x  3h 6  3h 2h

Exercise Set R.2 27 98.

f  x   ( x  3) 2 g  x   x2  6x  h

The functions represent the same function, so set them equal to each other to find the value for h. ( x  3) 2  x 2  6 x  h

x2  6x  9  x 2  6x  h 9h

28 Chapter R: Functions, Graphs, and Models

Exercise Set R.3

18.

 2, 3

19.

 4, 3   0, 5

20.

 , 2   1, 4 

 1, 3 

 0, 5 

 2, 4 

 1, 2

 9, 5

b) The domain is 4, 3, 2, 1, 0,1, 2 .

 x, x  h 

c) The point on the graph with the second coordinate 2 is  2, 2  . Thus the x-value for

 9, 4 

d) The range is 2, 1, 0,1, 2, 3, 4 .

 x, x  h 

 , 0    3,  

10.

 4, 1   2, 3

11.

 5, 5 

12.

 2, 2 

21. a)

f 1  1 .

which f  x   2 is x  2 .

22. a) First, we locate 1 on the horizontal axis and then we look vertically to find the point on the graph for which 1 is the first coordinate. From that point, we look to the vertical axis to find the corresponding y-coordinate, 3. Thus, f 1  3 . b) The domain is the set of all x-values of the points on the graph. The domain is 3, 1,1, 3, 5 . c) First, we locate 2 on the vertical axis and then we look horizontally to find any points on the graph for which 2 is the second coordinate. One such point exists,  3, 2  .

13).  4, 2 

Thus the x-value for which f  x   2 is

14.

 6, 20 

x  3. d) The range is the set of all y-values of the points on the graph. The range is 2, 0, 2, 3, 4 .

15.

 3,  

16.

 , 2

17.

 10,6 

23. a) First, we locate 1 on the horizontal axis and then we look vertically to find the point on the graph for which 1 is the first coordinate. From that point, we look to the vertical axis to find the corresponding y-coordinate, 4. Thus, f 1  4 . b) The domain is the set of all x-values of the points on the graph. The domain is 5, 3,1, 2, 3, 4, 5 .

Exercise Set R.3 29 c) First, we locate 2 on the vertical axis and then we look horizontally to find any points on the graph for which 2 is the second coordinate. Three such point exists,  5, 2  ;  3, 2  ; and  4, 2  . Thus the xvalues for which f  x   2 are 5, 3, 4 . d) The range is the set of all y-values of the points on the graph. The range is 3, 2, 4, 5 24. a)

f 1  2 .

b) The domain is 6, 4, 2, 0,1, 3, 4 . c) The points on the graph with the second coordinate 2 are 1, 2  and  3, 2  . Thus the

27. a) First, we locate 1 on the horizontal axis and then we look vertically to find the point on the graph for which 1 is the first coordinate. From that point, we look to the vertical axis to find the corresponding y-coordinate, 3. Thus, f 1  3 . b) The domain is the set of all x-values of the points on the graph. These extend from -3 to 3. Thus, the domain is  x | 3  x  3 , or, in interval notation  3, 3 . c) First, we locate 2 on the vertical axis and then we look horizontally to find any points on the graph for which 2 is the second coordinate. Two such point exists,  1.4, 2  and 1.4, 2  . Thus the x-values

x-values for which f  x   2 are 1, 3 . d) The range is 5, 2, 0, 2, 5 . 25. a) First, we locate 1 on the horizontal axis and then we look vertically to find the point on the graph for which 1 is the first coordinate. From that point, we look to the vertical axis to find the corresponding y-coordinate, -1. Thus, f 1  1 . b) The domain is the set of all x-values of the points on the graph. These extend from -2 to 4. Thus, the domain is  x | 2  x  4 , or in interval notation  2, 4  . c) First, we locate 2 on the vertical axis and then we look horizontally to find any points on the graph for which 2 is the second coordinate. One such point exists,  3, 2  . Thus the x-value for which f  x   2 is x  3. d) The range is the set of all y-values of the points on the graph. These extend from -3 to 3. Thus, the range is  y | 3  y  3 , or, in interval notation  3, 3 . 26. a)

for which f  x   2 are 1.4,1.4 . d) The range is the set of all y-values of the points on the graph. These extend from -5 to 4. Thus, the range is  y | 5  y  4 , or in interval notation  5, 4  . 28. a)

f 1  2.25 .

b) The domain is  x | 4  x  3 or  4, 3 . c) The point on the graph with the second coordinate 2 appears to be  0, 2  . Thus the x-value for which f  x   2 is x  0. . d) The range is  y | 5  y  4 , or,  5, 4  . 29. a) First, we locate 1 on the horizontal axis and then we look vertically to find the point on the graph for which 1 is the first coordinate. From that point, we look to the vertical axis to find the corresponding y-coordinate, -2. Thus, f 1  2 . b) The domain is the set of all x-values of the points on the graph. These extend from -4 to 2. Thus, the domain is  x | 4  x  2 , or, in interval notation  4, 2  .

f 1  2.5 .

b) The domain is  x | 3  x  5 or  3, 5 . c) The point on the graph with the second coordinate 2 appears to be  2.25, 2  . Thus the x-value for which f  x   2 is x  2.25. d) The range is  y | 1  y  4 , or, 1, 4  .

c) First, we locate 2 on the vertical axis and then we look horizontally to find any points on the graph for which 2 is the second coordinate. One such point exists,  2, 2  . Thus the x-value for which f  x   2 are x  2.

30 Chapter R: Functions, Graphs, and Models d) The range is the set of all y-values of the points on the graph. These extend from -3 to 3. Thus, the range is  y | 3  y  3 , or in

33.

interval notation  3, 3 .

30. a)

f 1  2 .

or, in interval notation,  , 2    2,  

b) The domain is  x | 5  x  4 or  5, 4  . c) The points on the graph with the second coordinate 2 are all the points with the xvalue in the set  x | 1  x  4 . Thus the xvalues for which f  x   2 are

34.

 x | 1  x  4 , or 1, 4 . d) The range is  y | 3  y  2 , or  3, 2  . 31. a)

f 1  2 .

b) The domain is  x | 4  x  4 or  4, 4  .

6 2x Since the function value cannot be calculated when the denominator is equal to 0, we solve the following equation to find those real numbers that must be excluded from the domain of f. 2x  0 setting the denominator equal to 0 f  x 

adding x to both sides 2x Thus, 2 is not in the domain of f, while all other real numbers are. The domain of f is  x | x is a real number and x  2 ; or, in

c) The points on the graph with the second coordinate 2 are all the points with the xvalue in the set  x | 0  x  2 . Thus the xvalues for which f  x   2 are

interval notation,   , 2    2,  

 x | 0  x  2 , or  0, 2 . d) The range is 1, 2, 3, 4 . 32. a) First, we locate 1 on the horizontal axis and then we look vertically to find the point on the graph for which 1 is the first coordinate. From that point, we look to the vertical axis to find the corresponding y-coordinate, 1. Thus, f 1  1 .

3 x2 Set the denominator equal to 0 and solve. x20 x  2 The domain is  x|x is a real number and x  2 f  x 

35.

b) The domain is the set of all x-values of the points on the graph. These extend from -5 to 5. However, the open circle at the point  5, 2  indicates that 5 is not in the domain.

f  x   2x Since the function value cannot be calculated when the radicand is negative, the domain is all real numbers for which 2 x  0 . We find them by solving the inequality. 2x  0 setting the radicand  0 dividing both sides by 2 x0 The domain of f is  x | x is a real number and x  0 ; or, in interval notation,  0,  

Thus, the domain is  x | 5  x  5 , or in interval notation  5, 5  . c) First, we locate 2 on the vertical axis and then we look horizontally to find any points on the graph for which 2 is the second coordinate. We notice all the points with xvalues in the set  x | 3  x  5 Thus the xvalues for which f  x   2 are

f  x  x  2 Solve x  2  0 x2 The domain of f is  x | x is a real number and x  2 ; or, in

 x | 3  x  5 , or  3, 5  .

interval notation,  2,  

36.

d) The range is the set of all y-values of the points on the graph. The range is 2, 1, 0,1, 2

Exercise Set R.3 31 37.

f  x   x2  2x  3 We can calculate the function value for all values of x, so the domain is the set of all real numbers 

38.

f  x   x2  3 We can calculate the function value for all values of x, so the domain is the set of all real numbers 

39.

43.

adding 2 x to both sides 7  2x 7 dividing both sides by 2 x 2 7 Thus, is not in the domain of f, while all 2 other real numbers are. The domain of f is 7   x | x is a real number and x   ; or, in 2  7 7     interval notation,   ,    ,   2 2  

8 3x  6 Solve 3 x  6  0 x  2. The domain of f is  x | x is a real number and x  2 ; or, in f  x 

interval notation,   , 2    2,  

x2 40. f  x   6 x  12 Since the function value cannot be calculated when the denominator is equal to 0, we solve the following equation to find those real numbers that must be excluded from the domain of f. 6 x  12  0 setting the denominator equal to 0

6 x  12

3x  1 7  2x Since the function value cannot be calculated when the denominator is equal to 0, we solve the following equation to find those real numbers that must be excluded from the domain of f. 7  2x  0 setting the denominator equal to 0 f  x 

44.

adding 12 to both sides

dividing both sides by 6 x2 Thus, 2 is not in the domain of f, while all other real numbers are. The domain of f is  x | x is a real number and x  2 ; or, in

2x  1 9  2x Solve 9  2 x  0 9 x 2 9 is not in the domain of f, while all Thus, 2 other real numbers are. The domain of f is 9   x | x is a real number and x   ; or, in 2  9 9   interval notation,   ,    ,   2 2   f  x 

interval notation,   , 2    2,   45. 41.

f  x  x  4 We can calculate the function value for all values of x, so the domain is the set of all real numbers 

42.

g  x   2  3x Solve 2  3 x  0 2  3x 2 x 3

f  x   3x  7 We can calculate the function value for all values of x, so the domain is the set of all real numbers 

32 Chapter R: Functions, Graphs, and Models The domain of g is 2   x | x is a real number and x   ; or, in 3  2   interval notation,   ,  3 

The domain of g is  x | x is a real number and x  6, x  6 ; or, in interval notation,   , 6    6, 6    6,  

50. 46.

g  x   4  5x Since the function value cannot be calculated when the radicand is negative, the domain is all real numbers for which 4  5 x  0 . We find them by solving the inequality. 4  5x  0 setting the radicand  0 subtracting 4 from both sides 5 x  4 4 dividing both sides by 5 x 5 The domain of g is 4   x | x is a real number and x    ; or, in 5    4  interval notation,   ,    5 

47.

g  x   4 x3  5x 2  2 x We can calculate the function value for all values of x, so the domain is the set of all real numbers 

48.

g  x   2 x2  4 x  2 We can calculate the function value for all values of x , so the domain is the set of all real numbers R .

49.

g  x 

2x 2

x  25 Since the function value cannot be calculated when the denominator is equal to 0, we solve the following equation to find those real numbers that must be excluded from the domain of g. x 2  25  0 setting the denominator equal to 0

x 2  25

x   25 taking the square root or both sides x  5 Thus, 5 and 5 are not in the domain of g, while all other real numbers are. The domain of g is  x | x is a real number and x  5, x  5 ;

or, in interval notation,   , 5    5, 5    5,  

51.

52.

g  x 

x 9 We can calculate the function value for all values of x, so the domain is the set of all real numbers 

x 1 We can calculate the function value for all values of x, so the domain is the set of all real numbers 

x 1

g  x 

adding 25 to both sides

x  36 Solve x  36  0 x 2  36 2

x  6

Exercise Set R.3 33 53.

54.

g  x   2 x  3  12  5x Since the function value cannot be calculated when the radicand is negative, the domain is all real numbers for which 2 x  3  0 and 12  5 x  0 . We find them by solving the inequality. 2x  3  0 setting the radicand  0 2 x  3 subtracting 3 from both sides 3 dividing both sides by 2 x 2 setting the radicand  0 12  5 x  0 5 x  12 subtracting 12 from both sides 12 dividing both sides by -5 x 5 Thus, the domain of g is the union of these two 3 12   inequalities  x | x   and x   ; or, in 2 5   3 12  interval notation,   ,   2 5

56. The graph crosses the line y  1 at every integer value of x. Therefore, the set of x-values for which g  x   1 is

 x | 5, 4, 3, 2, 1, 0,1, 2, 3, 4, 5 . 57. First, we locate 2 on the vertical axis and then we look horizontally to find any points on the graph for which 2 is the second coordinate. We notice the point with x-value is 1. 58. First, we locate –4 on the vertical axis and then we look horizontally to find any points on the graph for which the second coordinate is greater than –4. We notice all the points with x-values in the set  x | 4  x  2 Thus the x-values for

which G  x   4 are

 x | 4  x  2 , or  4, 2  . 59. First, we locate 0 on the vertical axis and then we look horizontally to find any points on the graph for which the second coordinate is 0. We notice all the points with x-values in the set  x | x  3 and 4 Thus the x-values for which W  x   0 are 3 and 4 .

g  x   x  1  6  2x Since the function value cannot be calculated when the radicand is negative, the domain is all real numbers for which x  1  0 and 6  2 x  0 . We find them by solving the inequality. setting the radicand  0 x 1  0 subtracting 1 from both sides x  1 setting the radicand  0 6  2x  0 subtracting 6 from both sides 2 x  6 x  3 dividing both sides by -2 Thus, the domain of g is the union of these two inequalities  x | x  1 and x  3 ; or, in interval notation,  1, 3 

55. The graph of f lies on or below the x-axis when f  x   0 , so we scan the graph from left to

right looking for the values of x for which the graph lies on or below the x axis. Those values extend from -1 to 2. So the set of x-values for which f  x   0 is  x | 1  x  2 , or, in

60. First, we locate 0 on the vertical axis and then we look horizontally to find any points on the graph for which the second coordinate is greater than or equal to 0. We notice all the points with x-values in the set  x | 6  x  4 Thus the x-

values for which k  x   0 are

 x | 6  x  4 , or  6, 4  . 61. The function increases from x  8 to x  3 , as the y -value goes from 122.6 to 31.5 . The function decreases from x  3 to x  6 , as the y-value goes from 31.5 to 90 . After x  6 , the function starts increasing again. The function is increasing on the intervals  8, 3  and  6,   .

The function is decreasing on the interval  3, 6  . Therefore, f is increasing on  8, 3    6,   f is decreasing on  3, 6  .

interval notation,  1, 2  .

34 Chapter R: Functions, Graphs, and Models 62. The function is increasing, decreasing, or constant around the vertical asymptote x  5 For x  5 , the function is increasing or decreasing as x approaches 5 from the left and for x  5 , the function is increasing or decreasing as x moves right from 5 . Constant Intervals there are typically no constant intervals around a vertical asymptote. Therefore, f is decreasing on   , 5    5,   . 63. The function is increasing as the x -value goes from  to  2 . Interval  2, 7  is where function decreasing.

Interval from  7, 5  to 18, 5  as x -value increases goes from 7 to 18 as y –value remains constant at 5 . Therefore, f is increasing on  , 2  , f is decreasing on  2, 7  and, f is constant on

 7,18  . 64. The function is increasing as the x -value goes from  1 to 1, y -value increases from 0 to 4 .

66. The domain is the amount of money Marcus will spend, which is 0 to 200 or  0, 200  . The

range of the function is how much Marcus will pay in taxes, which is a minimum of T (0)  0.05(0)  0 and a maximum of T (200)  0.05(200)  10 . Therefore the range of the function will be  0,10  . 67. a) The domain in interval notation is  25,102  .

b) The range in interval notation is  0, 450  . c) Answers will vary. Since we are looking for the greatest increase in the incidence of breast cancer, we are looking for the steepest portion of the graph. It appears that the greatest increase occurs from age 50 to age 60. 68. a) The graph extends from x  0 to x  92.3 , so the domain, in interval notation, of the function N is  0, 92.3 

b) The graph extends from N  x   0 to N  x   6 million. Therefore, the range,

Interval from 1, 4  to  5, 4  as x -value increases goes from 1 to 5 as y –value remains constant at 4 . The function is decreasing as the x -value goes from  to  1 , y -value decreases from  to 0 . After x  5 , the function starts decreasing again. Therefore, f is increasing on  1,1 , f is decreasing on  , 1   5,   and, f is constant on 1, 5  . 65. The domain is the number of hours that Karen will work, which is 0 to 80 or  0, 80  . The

range of the function is how much Karen will earn, which is a minimum of P (0)  40(0)  0 and a maximum of P (80)  40(80)  3200 . Therefore the range of the function will be  0, 3200  .

in interval notation, of the function N is  0, 6, 000, 000  . c) Answers will vary. We would target the 50 year old to 60 year old age group, because that is the age group that has the most number of deaf and hard of hearing Americans. f (4)  5  3.50(3)  15.50 . The charge for the first mile is $5, and then the charge for the next 3 miles is $3.50 each, giving a fare of $15.50 for a 4-mile trip. b) f (4.25)  5  3.50(4)  19 . The 0.25 mile is considered part of a fourth additional mile after the first mile is charged at the $5 rate, so the total fare is $19.00 for a trip of at least 4 miles up to and including 5 miles. c) The fares will be $5, $8.50, $12, and so on, in increments of $3.50, up to $36.50. Thus the range is {5, 8.5, 12, 15.5, 19, 22.5, 26, 29.5, 33, 36.5}.

69. a)

70. a) S (95)  20 . It cost $20 to ship an order totaling $95. b) S (102)  20  5  15 . It cost $15 to ship an order totaling $102, since $2 is considered a part of an additional $20.

Exercise Set R.3 35

c) Orders totaling up to and including $120 will be shipped for $15, those totaling up to and including $140 will be shipped for $10, those totaling up to and including $160 will be shipped for $5. Thus, and order totaling $160.01 or more will be shipped for free. d) The shipping charges are reduced in increments of $5. Thus, the range of S is {0, 5, 10, 15, 20}.

75. Answers may vary. One example is 1 . f  x  2 x x 4





76. Answers may vary. One example is 1 f  x  . x( x  1)( x  7)

71. The domain of the function is the union of where the numerator and denominator are defined. Thus it would be where x is greater than equal to 0 and 5  x is greater than 0. x0 5 x  0 5 x The domain is the union of these sets which is  0, 5  .

77. The domain of the function would be the union of where both pieces are defined. Thus, it would be where 1  x and x  3 are both greater than equal to 0. 1 x  0 1 x

72. The domain of the function is the union of where the numerator and denominator are defined. Thus it would be where 3  x is greater than equal to 0 and x does not equal 0. x0 3 x  0 3 x The domain is the union of these sets which is  , 0    0, 3 .

78. The domain of the function is  , 5   5,   .

x3 0 x3 The domain is the union of these sets which is an empty set.

79. The domain of the function is  , 1   4,   . 80. Answer may vary. 81. Answers will vary. 82. Answers will vary.

73. The domain of the function is where the denominator is defined. Thus it would be where x x 2  9 does not equal 0.





x x 9  0 x( x  3)( x  3)  0 0, 3, 3  x The domain is the union of these sets which is  , 3    3, 0    0, 3    3,   .

74. The domain of the function is where the denominator is defined. Thus it would be where x x 2  x  12 does not equal 0.

  x  x  x  12   0 2

x ( x  4)( x  3)  0 0, 4, 3  x The domain is the union of these sets which is  , 3    3, 0    0, 4    4,   .

36 Chapter R: Functions, Graphs, and Models 4.

Exercise Set R.4 1.

Graph f  x   0.5 x

Graph y   3 x .

The function f  x   0.5 x has slope –0.5, and y-intercept  0, 0  . The function y   3 x has slope –3, and

y-intercept  0, 0  . 2.

Graph y   2 x . Using Theorem 4, The graph of y is the straight line through the origin  0, 0  and the point

When x  0, y  3  0   4  4 . When x  2, y  3  2   4  2 .

1, 2 . We plot these two points and connect

We organize these values into an input – output table. y x  x, y 

them with a straight line.

1

7

4

 1, 7   0, 4   2, 2 

2 2 Next, we plot these ordered pairs and connect them with a straight line.

The function y   2 x has slope –2, and y-intercept  0, 0  . 3.

Graph y  3 x  4 First, we make a table of values. We choose any number for x and then determine y by substitution. When x  1, y  3  1  4  7 .

Graph f  x   0.5 x . Using Theorem 4, The graph of f  x  is the straight line through the origin  0, 0  and the point 1, 0.5  . We plot these two points and

The function y  3 x  4 has slope 3, and

connect them with a straight line.

y-intercept  0, 4  . 6.

Graph y  2 x  5 .

The function f  x   0.5 x has slope 0.5, and y-intercept  0, 0  . The function y  2 x  5 has slope 2, and y-intercept  0, 5  .

Exercise Set R.4 37 7.

g  x  x  2

First, we make a table of values. We choose any number for x and then determine y by substitution. When x  0, g  0     0   2  2

Graph y  7 . The graph consists of all ordered pairs whose second coordinate is 7. This results in a horizontal line, whose y-intercept is the point  0, 7  .

When x  1, g 1   1  2  1 We organize these values into an input-output table. x g  x  x, g  x  

 0, 2  1, 1

Next, we plot these ordered pairs and connect them with a straight line.

Since the graph is horizontal, the slope is 0 and the y-intercept is  0, 7  .

10. Graph y   5 .

Since the graph is horizontal, the slope is 0 and the y-intercept is  0, 5  .

The function g  x    x  2 has slope 1 , and

y -intercept  0, 2  8.

11. First, we solve the equation for y. y  3x  6 adding 3 x to both sides y  3x  6 The slope is 3. The y-intercept is  0, 6  . 12. First, we solve the equation for y. y  4x  1 y  4x 1 The slope is 4. The y-intercept is  0,1 .

g  x   x  5.5

13. First, we solve the equation for y. 2x  y  3  0 y  2x  3 The slope is 2. The y-intercept is  0,3 .

The function g  x   x  5.5 has slope 1 , and

y -intercept  0,  5.5 

38 Chapter R: Functions, Graphs, and Models 14. First, we solve the equation for y. 2x  y  3  0

2x  y  3

19.

adding 3 to both sides

y  2 x  3 subtracting 2 x from both sides The slope is –2. The y-intercept is  0,3 .

16. First, we solve the equation for y. 2x  2 y  8  0 subtracting 8 2 x  2 y  8 2 y  2 x  8 subtracting 2 x 2 x  8 dividing both sides by 2 y 2 y   x  4 simplifying The slope is –1. The y-intercept is  0, 4  .

21.

y  3  5  x  2 

Simplifying

y  y1  m  x  x1 

y  7  7x  7 y  7x y  y1  m  x  x1 

y   3  2  x   2   y  3  2 x  4 y  2 x  7

22.

y  y1  m  x  x1 

y  2  3  x  5  y  2  3 x  15 y  3 x  13 y  y1  m  x  x1 

y   0   2  x   3 

dividing by 3

y  0  2  x  3

simplifying

y  2x  6 24.

y  y1  m  x  x1 

y   0   5  x   5   y  0  5  x  5  y  5 x  25

1 3 y   x 4 4 1 The slope is  . 4 3 The y-intercept is  0,  . 

Adding 3

y   2   3  x   5  

23.

18. First, we solve the equation for y. x  4 y  3

Substituting

y  3  2  x  2 

commutative property of equality

7 The y-intercept is  0,   . 3 

Subtracting 3

y   7   7  x  1 

17. First, we solve the equation for y. x  3y  7 subtracting 7

Substituting

y  5 x  13 20.

x7 y 3 1 7 y  x 3 3 1 The slope is . 3

y    3    5  x   2   y  3  5 x  10

15. First, we solve the equation for y. 3x  3 y  6  0 y  x2 The slope is 1. The y-intercept is  0, 2  .

3y  7  x 3y  x  7

y  y1  m  x  x1 

4

Substituting

Exercise Set R.4 39 25. a) y  y1  m  x  x1  1  x  0 2 1 y  6   x  0 2 b) y  mx  b 1 y  x   6  2 1 y  x6 2 y   6  

Substituting Simplifying

4  x  0 3 b) y  mx  b y7 

27.

Substituting

y  3  0  x  2 y 3  0 y3 28.

Simplifying Adding 3

y  y1  m  x  x1 

y2  y1 x2  x1 Substituting, we have: 21  1 m 05 20  5  4 b) y  y1  m  x  x1 

y  1  4  x   5 

Substituting

Simplifying y  1  4 x  20 Adding 1 y  4 x  21 c) The y–intercept is (0, 21).

y  8  0  x   4   y  8  0  x  4 y 8  0 y 8

y2  y1 x2  x1 Substituting, we have: 1 4 m 3  0 3  3 1 b) y  y1  m  x  x1 

32. a) m 

y2  y1 x2  x1 Substituting, we have: 1  (11) m 52 12  3 4 b) y  y1  m  x  x1 

29. a) m 

y  1  4  x   5 

Substituting

31. a) m 

y  y1  m  x  x1 

y   3  0  x   2  

y   5  2  x  1 

y  5  2x  2 Simplifying Adding 5 y  2x  3 c) The y–intercept is (0, 3).

26. a) y  y1  m  x  x1 

4 y  x7 3

y2  y1 x2  x1 Substituting, we have: 95 m 3 1 4  2 2 b) y  y1  m  x  x1 

30. a) m 

y   4   1 x   0  

Substituting

y  1  4 x  20 Simplifying Adding 1 y  4 x  19 c) The y–intercept is (0, –19).

Substituting

Simplifying y 4  x0 Adding 4 y  x4 c) The y–intercept is (0, 4).

40 Chapter R: Functions, Graphs, and Models y2  y1 x2  x1 Substituting, we have: 5  ( 3) m 3  2 2  5 2  5 b) y  y1  m  x  x1 

y2  y1 x2  x1 Substituting, we have: 12  (3) m 3  (8) 9  5 9  5 b) y  y1  m  x  x1 

33. a) m 

35. a) m 

2 Substituting  x   2 5 2 4 y3 x Simplifying 5 5 2 19 y  x Subtracting 3 5 5 19 c) The y–intercept is  0,   . 5  y   3  

y2  y1 x2  x1 Substituting, we have: 42 m 1  (11) 2  12 1  6 b) y  y1  m  x  x1 

y2  y1 x2  x1 Substituting, we have: 6  13 m 2  (5) 19  3 19  3 b) y  y1  m  x  x1 

34. a) m 

36. a) m 

1 Substituting  x  1  6 1 1 y4  x Simplifying 6 6 1 23 y  x Adding 4 6 6 23 c) The y–intercept is  0,  .  6  y   4 

9  x   8  Substituting 5 9 72 y3  x Simplifying 5 5 9 87 y   x Subtracting 3 5 5 87 c) The y–intercept is  0,   . 5   y   3   

19  x   5   Substituting 3 19 95 y  13   x  Simplifying 3 3 19 56 y   x Adding 13 3 3 56 c) The y–intercept is  0,   . 3   y  13  

y2  y1 x2  x1 Substituting, we have: 66 m 74 0  3 0

37. a) m 

Exercise Set R.4 41

b) y  y1  m  x  x1 

y   6  0  x   4 y  6  0x  0 y6

b) y  y1  m  x  x1  Substituting Simplifying Adding 6

c) The y–intercept is  0, 6  and it is a horizontal line. y2  y1 x2  x1 Substituting, we have: 1  ( 1) m 83 0  5 0 b) y  y1  m  x  x1 

y 1  0x  0 y  1

Substituting

Subtracting 1

y2  y1 x2  x1 Substituting, we have: 00 m 10  6 0  4 0 b) y  y1  m  x  x1 

39. a) m 

c) The y–intercept is  0, 0  and it is a

y2  y1 x2  x1 Substituting, we have: 73 m 6  (6) 4  0  undefined b) x  6 c) There is no y–intercept as this is a vertical line.

y2  y1 x2  x1 Substituting, we have: 94 m 44 5  0  undefined b) x  4 c) There is no y–intercept as this is a vertical line.

42. a) m 

y2  y1 x2  x1 Substituting, we have: 9  1 m 00 10  0  undefined b) x  0 c) All points are on the y–axis and this is a vertical line.

43. a) m 

y   0  0  x   6 

Substituting

y  0  0x  0 y0

Simplifying

c) The y–intercept is  0, 0  and it is a

y2  y1 x2  x1 Substituting, we have: 00 m 100  1 0  99 0

Simplifying

horizontal line.

40. a) m 

Simplifying

41. a) m 

c) The y–intercept is  0, 1 and it is a

horizontal line.

y  0  0x  0 y0

Substituting

horizontal line.

38. a) m 

y   1  0  x   3 

y   0   0  x  1 

42 Chapter R: Functions, Graphs, and Models y2  y1 x2  x1 Substituting, we have: 50  0 m 00 10  0  undefined b) x  0 c) All point are on the y–axis and this is a vertical line.

44. a) m 

y2  y1 x2  x1 Substituting, we have: 5 5 20 15   5    4 4  4 m 4 3 1 3 1 1      4 4 4 4

45. a) m 

15 4 b) y  y1  m  x  x1  

y   5 

15   1   x     Substituting 4  4 

15 15 x 4 16 15 65 y  x 4 16

y 5 

Simplifying Adding 5

c) The y–intercept is  0, 

65  . 16 

y2  y1 x2  x1 Substituting, we have: 1 1 3 5 2     5  3  15 15   15 m 1 2 1 1 1     4 4 4 4 2

47. a) m 

8 15 b) y  y1  m  x  x1  

1 8   1  y      x     Substituting 3 15    2   1 8 4 y  x Simplifying 3 15 15 8 1 y  x Adding 1/3 15 15 1 c) The y–intercept is  0,  .  15 

y2  y1 x2  x1 Substituting, we have: 5 4      15  8  23 2 3  6 6 6 m 8 15 23 2  3    20 5  4  20 20

48. a) m 

10 3 b) y  y1  m  x  x1  

y2  y1 x2  x1 Substituting, we have: 1  1 1      2  3 3  2 6 6   6 m 4 3 1 3  2  2 2 2 2  

46. a) m 

1 3 b) y  y1  m  x  x1  

 1 1 y       x   2  3 3 1 1 2 y  x 3 3 3 1 y  x 1 3

c) The y–intercept is  0, 1 .

10  4  3  y       x      Substituting 3 3  4  4 10 5 y   x Simplifying 3 3 2 10 7 y   x Adding 4/3 3 6 7 c) The y–intercept is  0,   . 6 

Substituting Simplifying Subtracting 1/3

Exercise Set R.4 43 y2  y1 x2  x1 Substituting, we have:  k  3   k  m  h  2  h

49. a) m 

54.

55.

3  2 b) y  y1  m  x  x1  3 Substituting  x   h 2 3 3h yk  x Simplifying 2 2 3 3h y  x k Adding k 2 2 3h c) The y–intercept is  0, k   . 2   y  k  

y2  y1 x2  x1 Substituting, we have:  v    v  3 m  u  3  u

50. a) m 

Substituting Simplifying Adding v

c) The y–intercept is  0, v  u  3  . 51.

52.

53.

2 ft 2   0.4. 5 ft 5 The slope of the skateboard ramp is 40%. Slope 

0.4 ft 0.4 2    0.08. 5 ft 5 25 The grade is then 8%. Slope 

6 in 2   0.667. 9 in 3 The slope (or rake) is 66.7%. Slope 

55 in 11   0.55. 100 in 20 The slope for optimum tilt is 0.55 or 55%. Slope 

57. The average price of a new car sold in 2012 is approximately $30, 510 . The average price of a new car sold in 2022 is approximately $48, 094 . Change in Price  48, 094  30,510  17,584 Change in Time  2022  2012  10 17, 584  1, 758.40 10 The average rate of change in the price of a new car sold from 2012 to 2022 is $1, 758.40 dollars per year.

3 3 1 b) y  y1  m  x  x1 

y v  x u 3 y  x u 3v

87 in  0.87. 100 in The slope for optimum tilt is 0.87 or 87%. Slope 

Average rate of change 



y   v   1 x   u  3  

56.

43.33 ft 43.33 7    0.035. 1238 ft 1238 200 Expressing the slope as a percentage, we find the head of the river is 3.5%. Slope 

58. The initial population of bacteria present in an infected sample was 145, 000 bacteria. Eight hours after the start of treatment, the population of bacteria was 105, 000 bacteria. Total change in population  105, 000  145, 000  40, 000 40, 000 Average rate of change   5, 000 8 The average rate of change in the population over the 8 hour is 5, 000 bacteria per hour. 59. a) If I, the number of inkjet cartridges required each year, is directly proportional to the number of students, s. Then there exists a constant m such that I  ms To find m, we substitute 16 for I and 2800 for s into I  ms and solve for m. 16  m  2800

16 m 2800 0.0057  m The constant of variation is m  0.0057 . The equation of variation is I  0.0057 s.

44 Chapter R: Functions, Graphs, and Models P  x  0

b) I  0.0057 s  0.0057  3100

 17.67  18 The university would need 18 inkjet cartridges if 3100 students were enrolled.

60. a) T  mw . 2.70  m  3350

2.70 m 3350 0.000806  m The constant of variation is m  0.000806 . The equation of variation is T  0.000806w . b) Substituting 3700 for w into the equation of variation, we have: T  0.000806  3700   2.98 The toll would be $2.98 for a Jeep Cherokee.

61. a) Total costs = Variables costs + Fixed Costs To produce x sets of earbuds, it costs $15 dollars per earbuds, that is the variable costs are 15x . In addition to the variable costs the fixed costs are $1, 250, 000 . The total cost is C  x   15 x  1, 250, 000 b) Revenue  Price times Quantity The price per set of earbuds is $85 , therefore, the revenue from selling x sets of earbuds is given by R  x   85 x dollars c) Profit  Revenue – Cost Using the Cost and Revenue functions found in part (a) and (b) we have: P  x  R  x  C  x P  x   85 x  15 x  1, 250, 000 

70 x  1, 250, 000  0 70 x  1, 250, 000 1, 250, 000 70 x  17,857.14 They will need to sell 17,857 sets of earbuds in order to break even. The graph is shown below, x

62. a) Total costs = Variables costs + Fixed Costs To produce x skis, it costs $80 dollars per ski, that is the variable costs are 80x . In addition to the variable costs the fixed costs are $45,000. The total cost is C  x   80 x  45, 000 The graph is shown on the next page. b) Revenue = Price times Quantity. The price of a pair of skis is $450, therefore, the revenue from selling x pairs of skis is given by R  x   450 x The graph is shown below. c) Profit = Revenue – Cost. Using the Cost and Revenue functions found in part (a) and (b) we have: P  x  R  x  C  x P  x   450 x   80 x  45, 000   370 x  45, 000 The graph is shown below

P  x   70 x  1, 250, 000 dollars

d) Substituting 250, 000 for x into the profit equation, we have P  x   70  250, 000   1, 250, 000  16, 250, 000 Total profit will be $16, 250, 000 if they sell the expected 250, 000 sets of earbuds. e) The breakeven point occurs when P  x   0 . Therefore, we set the profit function equal to 0 and solve for x .

d) Substituting 3000 for x into the profit equation, we have P  3000   370  3000   45, 000  1, 065, 000

Exercise Set R.4 45

Total profit will be $1,065,000 if they sell the expected 3000 pairs of skis. e) The breakeven point occurs when P  x   0 . Therefore, we set the profit function equal to 0 and solve for x. P  x  0 370 x  45, 000  0

b) P  x   R  x   C  x  Substituting the profit in cost functions into the equation we have: 16 x  250  R  x    4 x  250 

16 x  250  R  x   4 x  250 20 x  R  x 

370 x  45, 000 45, 000 x 370 x  121.62 x  122 They will need to sell 122 pairs of skis in order to break even.

63. a) Total costs = Variables costs + Fixed Costs The variable costs are 5x . In addition to the variable costs the fixed cost is $25,000. The total cost is C  x   5 x  25, 000

b) P  x   R  x   C  x  Substituting the profit in cost functions into the equation we have: 20 x  25, 000  R  x    5 x  25, 000 

20 x  25, 000  R  x   5 x  25, 000 25 x  R  x  Since Revenue is equal to price times quantity. Raven Entertainment charges $25 per ticket. c) The breakeven point occurs when P  x   0 . Therefore, we set the profit function equal to 0 and solve for x. Solve P  x   0

20 x  25,000  0 20 x  25,000

Since Revenue is equal to price times quantity. Jamal charges $20 per lawn. c) The break even point occurs when P  x   0 . Therefore, we set the profit function equal to 0 and solve for x. Solve P  x   0 16 x  250  0 16 x  250 x  15.625 x  16 Jamal will need to mow 16 lawns in order to break even. CS  65. a) V  t   C  t    N  Substituting 5200 for C, 1100 for S, and 8 for N into the equation we have the straightline depreciation V  t  :

 5200  1100  V  t   5200  t   8    4100   5200  t    8   5200  t  512.5   5200  512.5t b) V  t   5200  512.5t V  0   5200  512.5  0   $5200.00

x  1, 250 Raven Entertainment will need to sell 1,250 tickets in order to break even. 64. a) Total costs = Variables costs + Fixed Costs The electricity and maintenance cost are $4 dollars per lawn, that is the variable costs are 4x . In addition to the variable costs the initial fixed cost is $250. The total cost is C  x   4 x  250

V 1  5200  512.5 1  $ 4687.50 V  2   5200  512.5  2   $ 4175.00 V  3  5200  512.5  3  $3662.50 V  4   5200  512.5  4   $3150.00 V  7   5200  512.5  7   $1612.50 V  8   5200  512.5  8   $1100.00 CS  66. V  t   C  t    N  First, we figure the total cost of the improvements. 25,000 at $40 per square foot

46 Chapter R: Functions, Graphs, and Models

gives us a total cost of C   40  25, 000   1, 000, 000 . After 39 years, the salvage value S is 0. Therefore, we substitute 1,000,000 for C, 39 for N, and 0 for S into the straight-line depreciations formula. Solution is continued on the next page.  1, 000, 000  0  V  t   1, 000, 000  t   39   V  t   1, 000, 000  t  25, 641.02564  V  t   1, 000, 000  25, 641.02564t

In order to find the depreciated value of the improvements after 10 years, we substitute 10 for t. V 10   1, 000, 000  25, 641.02564 10   743,589.7436  743,589.74 The depreciated value of the improvements after 10 years is approximately $743,590.

67. V  t   640t  34,800

a) The number 640 means that the value of the Jeep Wrangler decreases by $640 per year The number 34,800 indicates the original value of the Jeep Wrangler was $34, 800. b) From the graph we see that V  t   0, we have 640t  34,800  0 640t  34,800 34,800 t 640 t  54.375 It will take 54.375 years for the Jeep Wrangler to depreciate completely. c) The domain of V is the set of all non-negative real x -values for which V is non-negative. From the graph we see that this set is  0, 54.375 , which means that 68.

after 54.375 years the value is $0. B  t   700t  3500

b) From the graph we see that B  t   0 when

t  5 . Since we have the equation, we set B  t   0 and solve for t to double check the graph. 700t  3500  0 3500  700t 5t It will take 5 years for the photocopier to depreciate completely. c) The domain of B is the set of all nonnegative real x-values for which B is nonnegative. From the graph we see that this set is  0,5 .

69. Slope is rise over run. The maximum riser on a stair is 8.25 inches, and the minimum run for a tread is 9 inches. Therefore: height of riser 8.25 in 11    0.916. Slope  length of run 9 in 12 The maximum grade of stairs in North Carolina is 91%. Don’t round up for legal reasons! 70. Slope is rise over run. The minimum vertical rise on a roof is 4 feet, and the minimum horizontal run 12 feet. Therefore: height of rise 4 ft 1    0.333. Slope  length of run 12 ft 3 The minimum pitch of the roof is about 33.3%. 71. The average rate of change can be found using the coordinates of any two points on the line. We use the given coordinates  2016,18,142 

and  2022, 22, 463  .

22, 463  18,142 2022  2016 4,321   720.166 6  720.17 The average rate of change in annual premium for a family’s health insurance was about $720.17 per year. Rate of Change 

a) The number –700 is the rate of change in the value of the photocopier per year. In other words, it tells us that the photocopier depreciates $700 per year. The number 3500 is the initial value of the photocopier. In other words, the value of the photocopier at the time of purchase was $3500.

Exercise Set R.4 47 72. The average rate of change can be found using the coordinates of any two points on the line. We use the given coordinates  2016,5277  and

 2022, 7911 . 7,911  5, 277 2022  2016 2, 634  6  439 The average rate of change in the annual premium for a single person is $439 per year. Rate of Change 

73. The average rate of change can be found using the coordinates of any two points on the line. We use the given coordinates  2017, 3520  and

 2022, 4289  . 4, 289  3,520 2022  2017 769  5  153.8 The average rate of change of the tuition and fees at public two-year colleges is approximately $153.8 per year. Rate of Change 

74. The average rate of change can be found using the coordinates of any two points on the line. We use the given coordinates  2015, 43.3  and

 2021, 63.0  . 63.0  43.3 Rate of Change  2021  2015 19.7  6  3.28 The average rate of change of organic food sales in the United States is $3.28 billion dollars per year. 75. In January 1972, the initial temperature of a single day is 54F. The final temperature rise of a single day is 49F. Change in temperature  49   54   49  54  103

The average rate of change in temperature over the 24 hour period is approximately 4.29F per hour. 76. In 1816, India averaged a temperature of 22.4C. In 2021, average temperature was 24.6C. Change in temperature  24.6  22.4  2.2 Number of years  2021 – 1816  205 2.2 Average rate of change  205  0.0107  0.011 The average rate of change in the annual temperature for India from 1816 to 2012 is 0.011C. 77. a) If R is directly proportional to T, then there is some positive constant m such that R  mT . To find m, substitute 12.51 for R, and 3 for T into the equation R  mT and solve for m. 12.51  m  3

12.51 m 3 4.17  m The variation constant m  4.17 . The equation of variation is R  4.17T . b) Using the equation of variation found in part (a), we substitute 6 for T. R  4.17  6   25.

The R-Factor for insulation that is 6 inches thick is 25. 78. The equation of variation is D  293t . The distance the impulse has to travel is 6 ft. So we solve: 6  293t 6 t 293 0.0204778157  t 0.02  t It would take an impulse 0.02 seconds to travel from the toes of the person to the brain.

Time period  24 hours 103  4.29 Average rate of change  24

48 Chapter R: Functions, Graphs, and Models 79. a) D  F   2 F  115 D  0   2  0   115  115 D  20   2  20   115  75

b) Plotting the points found in part (a) and connecting the points with a smooth curve we have:

D 10   2 10   115  135 D  32   2  32   115  179

b) For every increase of 1 F in the air temperature, the stopping distance increases by 2 ft. c) The domain should be restricted to the set of F –values for which D is nonnegative. So, we find where D  F   0 . 0  2 F  115 2 F  115 115 F 2 F  57.5 When the temperature is at least 57.5 the stopping distance is non-negative. Temperatures below 57.5 have negative stopping distances, which are not logical in the context of the problem. Since ice forms at 32 , for all temperatures above 32 there is no glare ice, so the formula does not apply. Therefore, the domain for D should be restricted to the interval  57.5, 32 . 11r  5 10 11  5   5 D 5  6 10 When traveling 5 miles per hour, the car’s reaction distance is 6 feet. 11 10   5 D 10    11.5 10 When traveling 10 miles per hour, the car’s reaction distance is 11.5 feet. 11  20   5 D  20    22.5 10 When traveling 20 miles per hour, the car’s reaction distance is 22.5 feet. 11  50   5 D  50    55.5 10 When traveling 50 miles per hour, the car’s reaction distance is 55.5 feet. 11  65   5 D  65    72 10 When traveling 65 miles per hour, the car’s reaction distance is 72 feet.

80. a) D  r  

c) For every 1–mph increase in speed, the breaking distance increases by 1.1 ft. d) Answers may vary. Negative speed has no meaning and when the speed is 0 mph, the car is already stopped. So theoretically, the domain would be (0, ) . However, realistically, the domain would have an upper bound that depended on the car’s maximum obtainable speed. 81. a) F  26   2.75  26   71.48  142.98

The female was 142.98 cm tall. b) M  26   2.89  26   70.64  145.78 The male was 145.78 cm tall. 82. a) We know that N  P  0.02 P  1.02 P. Therefore, the equation of variation is N  1.02 P. b) Substituting 200,000 for P we have N  1.02  200, 000   204, 000.

The new population is 204,000 after a growth of 2%. c) Substituting 367,200 for N we have 367, 200  1.02 P 367, 200 P 1.02 360, 000  P The previous population was 360,000. 83. a) The initial temperature in Cincinnati, Ohio is 44F. The final temperature drop over a span of 6 hours is 5F. To find the linear model for the temperature T  x   mx  b Change in temperature  5  44  39 Change in time  6 hours Change in time  6 hours

39  6.5 6 T  x   6.5x  44 m

Exercise Set R.4 49

b) To find T  3 

87. Setting V  t   0 and solving for t, we have:

V t   0

T  3  6.5  3  44 T  3  19.5  44

0  900  60t

T  3  24.5F c) Answer may vary. 84.

y2  y1 x2  x1 Substituting, we have: 13  5 m 42 8  2 4 Now find the equation of the line: y  y1  m  x  x1  m

60t  900 t  15 The domain of the depreciation model is 0  t  15 or [0, 15] where t is the number of years since 2021. 88. Answers may vary. Regions of profit will occur when total revenue is greater than total cost, or when total profit is above the x-axis. Regions of loss will occur when total revenue is less than total cost, or when total profit is below the xaxis.

y   5  4  x   2  

y  5  4x  8 y  4x  3 Use the above equation and the third point: y  4x  3 y  4(7)  3 y  25 85.

Answers may vary. The discussion should include the information on how total cost is composed of two components, fixed and variable costs. Fixed costs are the component of total cost not dependent upon output. The discussion should also include knowledge that total revenue is price times quantity sold, and total profit is total revenue minus total costs.

86. a) Graph III is appropriate, because it shows the rate before January 1 is approximately $3000 per month, and the rate after January 1 is approximately $2000 per month. b) Graph IV is appropriate, because it shows the rate before January 1 is approximately $3000 per month, and the rate after January 1 is approximately –$4000 per month. c) Graph I is appropriate, because it shows the rate before January 1 is approximately $1000 per month, and the rate after January 1 is approximately $2000 per month. d) Graph II is appropriate, because it shows the rate before January 1 is approximately $4000 per month, and the rate after January 1 is approximately –$2000 per month.

50 Chapter R: Functions, Graphs, and Models

1 2 x . First, we find the 2 vertex or the turning point. The x-coordinate of the vertex is b x 2a 0 x 0  1 2    2 Substituting 0 for x into the equation, we find the second coordinate of the vertex: 1 2 y    0  0 . 2 The vertex is  0, 0  . The y-axis (The vertical Next, we graph y  

Exercise Set R.5 1 2 1 x and y   x 2 . 2 2 1 2 Starting with y  x , we first find the vertex 2 or the turning point. The x-coordinate of the vertex is b x 2a 0  1 2  2 0 Substituting 0 for x into the equation, we find the second coordinate of the vertex: 1 2 y  0  0 . 2 The vertex is  0, 0  . The y-axis (The vertical Graph y 

line x  0 .) is the axis of symmetry. Next, we choose some x-values on each side of the vertex and compute the y-values. 1 2 1 1 When x  1, y  1  1  2 2 2 1 1 2 When x  2, y   2    4  2 2 2 1 1 1 2 When x  1, y   1  1  2 2 2 1 1 2 When x  2, y   2    4  2 2 2

y 0

1 2

1

1 2

2

line x  0 .) is the axis of symmetry. Next, we choose some x-values on each side of the vertex and compute the y-values. 1 2 1 1 When x  1, y   1   1   2 2 2 1 1 2 When x  2, y    2     4  2 2 2 1 1 1 2 When x  1, y    1   1   2 2 2 1 1 2 When x  2, y    2     4  2 2 2

y 0  12

2

1

 12

2

We plot these points and connect them with a smooth curve on the axis below.

We plot these points and connect them with a smooth curve on the axis below. The domain is  ,   or  for both.

Exercise Set R.5 51 2.

Graph y 

The vertex is  0, 1 . The y-axis (The vertical

1 2 1 x and y   x 2 . 4 4

line x  0 .) is the axis of symmetry. Next, we choose some x-values on each side of the vertex and compute the y-values.

When x  1, y  1  1  1  1  0 2

When x  2, y   2   1  4  1  3 2

When x  1, y   1  1  1  1  0 2

When x  2, y   2   1  4  1  3 y x 0 1 0 1 3 2 0 1 3 2 We plot these points and connect them with a smooth curve on the axis below.

The domain is  ,   or  for both. 3.

Graph y  x 2 and y  x 2  1 . Starting with y  x 2 , we first find the vertex or the turning point. The x-coordinate of the vertex is 0 b x  0 2a 2 1 Substituting 0 for x into the equation, we find the second coordinate of the vertex:

y   0  0 . 2

The vertex is  0,0  . The y-axis (The vertical line x  0 .) is the axis of symmetry. Next, we choose some x-values on each side of the vertex and compute the y-values. When x  1, y  1  1 2

The domain is  ,   or  for both.

When x  2, y   2   4 2

When x  1, y   1  1 2

Graph y  x 2 and y  x 2  3 .

When x  2, y   2   4 y x 0 0 2

1 2 1 2

1 4 1 4

We plot these points and connect them with a smooth curve on the axis below. Next, we graph y  x 2  1 . First, we find the vertex or the turning point. The x-coordinate of the vertex is b x 2a 0  2 1 0 Substituting 0 for x into the equation, we find the second coordinate of the vertex:

y   0   1  1 .

The domain is  ,   or  for both. 5.

Graph y  2 x 2 and y  2 x 2  3 Starting with y  2 x 2 , we first find the vertex or the turning point. The x -coordinate of the vertex is, b 0 x  0 2a 2  2  Substituting 0 for x into the equation, we find the second coordinate of the vertex: 2 y  2  0   0

52 Chapter R: Functions, Graphs, and Models The vertex is  0, 0  . The y -axis (The vertical line x  0 .) is the axis of symmetry. Next, we choose some x -values on each side of the vertex and compute the y -values. When, x  1 , y  2 1  2 2

When, x  2 , y  2  2   8 2

When, x  1 , y  2  1  2 2

When, x  2 , y  2  2   8 2

2

8

1

2 8

2

Domain:  ,   , or  , for both.

We plot these points and connect them with a smooth curve on the axis below. Next, we graph y  2 x 2  3. First, we find the vertex or the turning point. The x -coordinate of the vertex is b 0 x  0 2a 2  2 

Graph y  3 x 2 and y  3 x 2  1 Starting with y  3 x 2 , we first find the vertex or the turning point. The x -coordinate of the vertex is, b 0 x  0 2a 2  3 

Substituting 0 for x into the equation, we find the second coordinate of the vertex: 2 y  2  0   3  3

Substituting 0 for x into the equation, we find the second coordinate of the vertex:

The vertex is  0, 3  . The y -axis (The vertical

The vertex is  0, 0  . The y -axis (The vertical

line x  0 .) is the axis of symmetry. Next, we choose some x -values on each side of the vertex and compute the y -values. When, x  1 , y  2 1  3  1 2

When, x  2 , y  2  2   3  5 2

When, x  1 , y  2  1  3  1 2

When, x  2 , y  2  2   3  5 2

5

1

1 5

2

We plot these points and connect them with a smooth curve on the axis below.

y  3  0  0 2

line x  0 .) is the axis of symmetry. Next, we choose some x -values on each side of the vertex and compute the y -values. When, x  1 , y  3 1  3 2

When, x  2 , y  3  2   12 2

When, x  1 , y  3  1  3 2

When, x  2 , y  3  2   12 2

3

 12

1

3

2

 12

We plot these points and connect them with a smooth curve on the axis below.

Exercise Set R.5 53 Next, we graph y  3x 2  1 First, we find the vertex or the turning point. The x -coordinate of the vertex is b 0 x  0 2a 2  3  Substituting 0 for x into the equation, we find the second coordinate of the vertex:

y  3  0   1  1 2

The vertex is  0, 1 . The y -axis (The vertical line x  0 .) is the axis of symmetry. Next, we choose some x -values on each side of the vertex and compute the y -values.

Graph y  x and y  x  1 Starting with y  x , we choose some x -values and compute the y -values to make a table of points. When, x  2 , y  2  2 When, x  1 , y  1  1 When, x  0 , y  0  0 When, x  1 , y  1  1 When, x  2 , y  2  2

When, x  1 , y  3 1  1  4 2

2

1

When, x  1 , y  3  1  1  4

When, x  2 , y  3  2   1  13

When, x  2 , y  3  2   1  13 2

1

4

13

1

4

2

13

We plot these points and connect them with a smooth curve on the axis below. Next, we graph y  x  1 We choose some x values and compute the y -values to make a table of points. When, x  2 , y  2  1  1 When, x  1 , y  1  1  0

We plot these points and connect them with a smooth curve on the axis below.

When, x  0 , y  0  1  1 When, x  1 , y  1  1  2 When, x  2 , y  2  1  3 x

2

1

We plot these points and connect them with a smooth curve on the axis below.

Domain:  ,   , or  , for both.

54 Chapter R: Functions, Graphs, and Models We plot these points and connect them with a smooth curve on the axis below.

The domain is  ,   or  for both.

Graph y  x and y  x  2  1 .

Domain:  ,   , or  , for both. 8.

Graph y  x and y  x  3 . Starting with y  x , we choose some x-values and compute the y-values to make a table of points. When x  2, y  2    2   2 When x  1, y  1    1  1

The domain is  ,   or  for both.

10. Graph y  x and y  x  3  4 .

When x  0, y  0  0 When x  1, y  1  1 When x  2, y  2  2

x 2 1

2 1

1 2

The domain is  ,   or  for both.

11. Graph y  x3 and y  x 3  1 .

We plot these points and connect them with a smooth curve on the axis below. Next, we graph y  x  3 . We choose some xvalues and compute the y-values to make a table of points. When x  2, y  2  3  5    5   5 When x  0, y  0  3  3    3  3 When x  3, y  3  3  0  0 When x  6, y  6  3  3  3

The domain is  ,   or  for both.

When x  8, y  8  3  5  5 y x 5 2 0 3 3 0 6 3 8 5

Exercise Set R.5 55 12. Graph y  x3 and y  x3  2 .

13. Graph y  x and y  x  1 .

Starting with y  x3 , we choose some x-values and compute the y-values to make a table of points. When x  2, y   2   8 3

When x  1, y   1  1 3

Starting with y  x , we choose some x-values and compute the y-values to make a table of points. The domain of the function is the set of all nonnegative real numbers, so we choose xvalues that are in the set  0,  . When x  0, y  0  0

When x  0, y   0   0 3

When x  1, y  1  1

When x  1, y  1  1 3

When x  4, y  4  2

When x  2, y   2   8 3

When x  9, y  9  3

y 8

x 2 1

1

1 2

y 0

1 4

1 2

8 We plot these points and connect them with a smooth curve on the axis below. Next, we graph y  x3  2 . We choose some xvalues and compute the y-values to make a table of points. When x  2, y   2   2  8  2  6 3

When x   1, y   1  2  1  2  1 3

When x  0, y   0   2  0  2  2 3

9 3 We plot these points and connect them with a smooth curve on the axis below. Next, we graph y  x  1 . we choose some xvalues and compute the y-values to make a table of points. The domain of the function is the set of all positive real numbers greater than or equal to 1, so we choose x-values that are in the set 1,  When x  1, y  1  1  0  0 When x  2, y  2  1  1  1

When x  1, y  1  2  1  2  3

When x  5, y  5  1  4  2

When x  2, y   2   2  8  2  10

When x 10, y  10  1  9  3

x 2 1

y 6

x 1 2

1 2

3 10 We plot these points and connect them with a smooth curve. 1 2

The domain is  ,   or  for both.

5 10

y 0 1 2

We plot these points and connect them with a smooth curve on the axis below.

The domains are  0,  and 1,  .

56 Chapter R: Functions, Graphs, and Models 14. Graph y  x and y 

x6

Starting with y  x , we choose some x values and compute the y -values to make a table of points. The domain of the function is the set of all non-negative real numbers, so we choose x values that are in the set  0,   . When, x  0 , y  0  0 When, x  1 , y  1  1 When, x  4 , y  4  2 When, x  9 , y  9  3 x

Domains:  0,   and  6,   .

We plot these points and connect them with a smooth curve on the axis below. Next, we graph y  x  6 we choose some x values and compute the y -values to make a table of points. The domain of the function is the set of all positive real numbers greater than or equal to 1, so we choose x -values that are in the set  0,   . When, x  6 , y  6  6  0  0 When, x  5 , y  5  6  1  1

15. Graph y  x and y  x  3  4 .

The domains are  0,   and 3,   .

16. Graph y  x and y  x  2  1 .

When, x  2 , y  2  6  4  2 When, x  3 , y  3  6  9  3 x

6

5

2

The domains are  0,   and  2,   .

We plot these points and connect them with a smooth curve on the axis below.

17. a) g ( x)  x 2  6 x  5 This function is of the form y  ax 2  bx  c, a  0 , so its graph is a parabola. We have a  1 and b  6 , so the first coordinate of the vertex is b x 2a 6   3. 2 1

Exercise Set R.5 57

Substituting 3 into the equation, we find the second coordinate of the vertex: y   3  6  3  5 2

 9  18  5  4. The vertex is  3, 4  and the line of

symmetry is x  3 . b) The graph opens upward because the value of a is positive. c) To find the x–intercepts, set the equation equal to 0 and solve. 0  x2  6 x  5

0  ( x  5)( x  1) x  5,1 Therefore the intercepts are  5, 0  and 1, 0  18. a)

f ( x)  x 2  4 x  3 This function is of the form y  ax 2  bx  c, a  0 , so its graph is a parabola. We have a  1 and b  4 , so the first coordinate of the vertex is b x 2a 4   2. 2 1

Substituting  2 into the equation, we find the second coordinate of the vertex: y   2   4  2   3 2

 483  1. The vertex is  2, 1 and the line of

symmetry is x  2 . b) The graph opens upward because the value of a is positive. c) To find the x–intercepts, set the equation equal to 0 and solve. 0  x2  4 x  3

0  ( x  3)( x  1) x  3, 1 Therefore the intercepts are  3, 0  and

 1, 0  .

19. a) k (t )  t 2  8t This function is of the form y  ax 2  bx  c, a  0 , so its graph is a parabola. We have a  1 and b  8 , so the first coordinate of the vertex is b x 2a 8   4. 2 1

Substituting  2 into the equation, we find the second coordinate of the vertex: y   4   8  4  2

 16  32  16. The vertex is  4, 16  and the line of

symmetry is x  4 . b) The graph opens upward because the value of a is positive. c) To find the x–intercepts, set the equation equal to 0 and solve. 0  t 2  8t

0  t (t  8) t  0, 8 Therefore the intercepts are  8, 0  and

 0,0 . 20. a) u (t )  t 2  12t This function is of the form y  ax 2  bx  c, a  0 , so its graph is a parabola. We have a  1 and b  12 , so the first coordinate of the vertex is b x 2a 12   6. 2  1

Substituting 6 into the equation, we find the second coordinate of the vertex: y    6   12  6  2

 36  72  36. The vertex is  6,36  and the line of

symmetry is x  6 .

58 Chapter R: Functions, Graphs, and Models

b) The graph opens downward because the value of a is negative.

 1  33  ,0 Therefore the intercepts are    4 

c) To find the x–intercepts, set the equation equal to 0 and solve. 0  t 2  12t

 1  33  ,0 . and    4 

0  t (t  12) x  0, 12 Therefore the intercepts are  12, 0  and

 0,0 . 21. a) m( s )  2 s 2  s  4 This function is of the form y  ax 2  bx  c, a  0 , so its graph is a parabola. We have a  2 and b  1 , so the first coordinate of the vertex is b x 2a 1 1   . 2  2  4

1 into the equation, we find 4 the second coordinate of the vertex:

Substituting

1 1 y  2       4 4 4 1 1   4 8 4 33  . 8

b) The graph opens downward because the value of a is negative. c) To find the x–intercepts, set the equation equal to 0 and solve. 0  2s 2  s  4

Solve using the quadratic formula 1  33 4

1 into the equation, we find 3 the second coordinate of the vertex: Substituting

1 1 y  3    2    2 3 3 1 2    2 3 3 7  . 3

1 7 The vertex is  ,  and the line of 3 3 1 symmetry is x  . 3

 1 33  The vertex is  ,  and the line of 4 8  1 symmetry is x  . 4

x

22. a) v(r )  3r 2  2r  2 This function is of the form y  ax 2  bx  c, a  0 , so its graph is a parabola. We have a  3 and b  2 , so the first coordinate of the vertex is b x 2a 2 1   . 2  3 3

b) The graph opens downward because the value of a is negative. c) To find the x–intercepts, set the equation equal to 0 and solve. 0  3r 2  2r  2

Solve using the quadratic formula x

1 7 3

 1 7  , 0  and Therefore the intercepts are    3   1 7  ,0 .    3 

Exercise Set R.5 59

1 2 2 11 x  x 3 5 15 This function is of the form y  ax 2  bx  c, a  0 , so its graph is a parabola. 1 2 We have a  and b  , so the first 3 5 coordinate of the vertex is b x 2a 2 3 5   . 5 1 2  3

23. a) b( x) 

3 Substituting  into the equation, we find 5 the second coordinate of the vertex: 2

1 3 2  3  11 y       3 5 5  5  15 9 6 11    75 25 15 64  . 75

 3 64  The vertex is   ,   and the line of  5 75  3 symmetry is x   . 5 b) The graph opens upward because the value of a is positive. c) To find the x–intercepts, set the equation equal to 0 and solve. 1 2 11 0  x2  x  3 5 15 0  5 x 2  6 x  11 0  (5 x  11)( x  1) 11 x   ,1 5  11  Therefore the intercepts are   , 0  and  5 

1, 0  .

24. a) a ( x)  0.5 x 2  1.2 x  1.7 This function is of the form y  ax 2  bx  c, a  0 , so its graph is a parabola. We have a  0.5 and b  1.2 , so the first coordinate of the vertex is b x 2a 1.2 6   1.2   . 2  0.5  5

6 into the equation, we find 5 the second coordinate of the vertex:

Substituting 

6  6  17  6 y  0.5         5  5  10  5 36 36 17    50 25 10 121  . 50

 6 121  The vertex is   ,   and the line of  5 50  6 symmetry is x   . 5 b) The graph opens upward because the value of a is positive. c) To find the x–intercepts, set the equation equal to 0 and solve. 0  0.5 x 2  1.2 x  1.7

0  5 x 2  12 x  17 0  (5 x  17)( x  1) 17 x   ,1 5

 17  Therefore the intercepts are   , 0  and  5  1, 0  . 2 . x First we determine the domain. The domain is all real numbers except for 0, since substituting 0 for x would result in division by 0. Now, to find y-values, we substitute any value for x other than 0, and compute the value for y.

25. Graph y 

60 Chapter R: Functions, Graphs, and Models In interval notation  , 0    0,  

2 1  4 2 2 When x  2, y   1 2 2 When x  1, y   2 1 2 When x  1, y   2 1 2 When x  2, y   1 2 2 1 When x  4, y   4 2 When x  4, y 

4 1 4 4 When x  2, y   2 2 4 When x  2, y    2 2 4 When x  4, y    1 4

When x  4, y  

x 4 2 2 4

x 4

 12

2 1 1 2 4

1 2 2 1

y 1 2 2 1

We plot these points and connect them with a smooth curve on the axis below.

1 2

We plot these points and connect them with a smooth curve on the axis below.

Domain: ( , 0)  (0,  )

26. Graph y 

3 . x Domains:  , 0    0,   .

28. Domain: ( , 0)  (0,  )

27.

4 x First we determine the domain. The domain is all real numbers except for 0, since substituting 0 for x would result in division by 0. Now, to find y -values, we substitute any value for x other than 0, and compute the value for y . y

8 x First we determine the domain. The domain is all real numbers except for 0, since substituting 0 for x would result in division by 0. Now, to find y -values, we substitute any value for x other than 0, and compute the value for y . y

In interval notation  , 0    0,   When x  4, y  

8 2 4

Exercise Set R.5 61

8 4 2 8 When x  2, y    4 2 8 When x  4, y    2 4 y x

When x  2, y 

When x  2, y  

4 2

4

2

When x  1, y 

We plot these points and connect them with a smooth curve on the axis below.

 2 



 12

1 4

1

1 1 1 When x   , y   4 2 1 2  1  2  4   1 1 1 When x  , y  4 2 1 2 1 2 4   1 When x  1, y  2  1 1

When x  2, y 

 2



1 4

We create an input-output table. We plot these points and connect them with a smooth curve on the axis below. x 2

1

1 4

 12

Domains:  , 0    0,   .

29. Graph y 

1 4

1 2

1 1 4

1 . x 1 Domain: (  , 0)  (0,  )

Domain: ( ,1)  (1,  )

1 . x2 First we determine the domain. The domain is all real numbers except for 0, since substituting 0 for x would result in division by 0. Now, to find y-values, we substitute any value for x other than 0, and compute the value for y.

30. Graph y 

31. Graph y  3 x . First we determine the domain of the function. Since the index of the radicand is odd (3) the domain is all real numbers. We are free to choose any number for x and compute the value for y. When x  8, y  3 8  2 When x  1, y  3 1  1 When x  0, y  3 0  0 When x  1, y  3 1  1 When x  8, y  3 8  2

62 Chapter R: Functions, Graphs, and Models x

8 1

2 1

1 2

5 4 2 1

We plot these points and connect them with a smooth curve on the axis below.

y 3 2

1 2

We plot these points and draw the graph. The open circle at the point  3, 1 indicates that it is not part of the graph.

Domain: (, ) or R 32. Graph y 

1 . x

Domain: ( ,  3)  ( 3,  ) . 34. Graph g  x  

x 2  7 x  10 . x2

Domain: (, 0)  (0, ) x2  5x  6 . x3 First, we simplify the function, by factoring the numerator and removing a factor of 1 as follows: x2  5x  6 f  x  x3  x  3 x  2   x3 x3 x2 f  x   x3 1  x  2, x  3

33. Graph f  x  

The simplification assumes that x is not 3 . The number 3 is not in the domain of the original function because it would result in division by zero. Thus we can express the function as follows: y  f  x   x  2, x  3. To find function values, we substitute any value for x other than 3 and calculate the y-values.

Domain: (,  2)  (2, ) . 35. Solve x 2  2 x  1  5 . First, we put the equation into standard form. x2  2x  4  0 Next we apply the quadratic formula. x  

  2  

 2 2  4 1 4  2 1

2  4  16 2  20 2  2 5   2 2 2



2 1 5

  1 5

2 The solutions are 1  5 and 1  5 .

36. Solve x 2  2 x  2 . First, we put the equation in standard form. x2  2 x  2  0 subtract 2 from both sides The equation is in standard form with a  1, b  2, c  2

Exercise Set R.5 63

Next, we apply the quadratic formula.

The solutions are 3  10 and  3  10 .

b  b  4ac 2a Substituting the values for a, b, and c, we get: x

x

  2  

 2 2  4 1 2  2 1

2  4  8 2  12  2 2 22 3  12  2 







2 1 3

43  2 3

b  b 2  4ac 2a Substituting the values for a, b, and c, we apply the quadratic formula on the next page. x



  1 3

2 1 The solutions are 1  3 and 1  3 .

x

37. Solve x 2  4 x  3 . First, we put the equation into standard form. x2  4x  3  0 Next we apply the quadratic formula. x  

  4 

 4 2  4 1 3 2 1



  2  7



 6   4 1 1 2 1 2







4  32 8 44 2 x 8

x

b  b  4ac 2a Substituting the values for a, b, and c, we get:

6  36  4 6  40  2 2 6  2 10  40  2



2 3  10 2 1

  3  10

4  10  2 10

4  16  16 8



 6 

  4 2  4  4  1 2  4



4 1 2 42

 32  16  2  4 2    1 2 2

1 2 1 2 and . 2 2

40. Solve 4 x2  4 x  1 . First, we put the equation into standard form. 4 x2  4x  1  0 Next we apply the quadratic formula.

38. Solve x 2  6 x  1 . First, we put the equation in standard form. x2  6 x  1  0 subtract 1 from both sides The equation is in standard form with a  1, b  6, c  1 Next, we apply the quadratic formula. x

  4 

The solutions are

2 The solutions are 2  7 and  2  7 .

x



4  16  12 4  28 4  2 7   2 2 2 2 2  7

39. Solve 4 x 2  4 x  1 . First, we put the equation in standard form. 4 x 2  4 x  1  0 subtract 4 x and 1 from both sides The equation is in standard form with a  4, b  4, c  1 Next, we apply the quadratic formula.





  4 

 4 2  4  4  1 2  4

 4  16  16  4  32  4  4 2   8 8 8



4 1  2 24

  1  2

2 1  2 1  2 and . The solutions are 2 2

41. Solve 3 y 2  8 y  2  0 . The equation is in standard form with a  3, b  8, c  2 Next, we apply the quadratic formula. y

b  b 2  4ac 2a

64 Chapter R: Functions, Graphs, and Models

Substituting the values for a, b, and c, we get:  8 

y

8  4  3 2  2  3 2

9  0 x  0. x Multiplying both sides by x, we get: 9  x  x  7    0 x x 

44. Solve x  7 

8  64  24 8  40  6 6





2  4  10



x 2  7 x  9  0. This is a quadratic equation in standard form with a  1, b  7, and c  9 . Next, we apply the quadratic formula.

 40  4  10  2 10 

8  2 10  6 23

  4  10

b  b 2  4ac 2a Substituting the values for a, b, and c, we get:

The solutions are

x

4  10 4  10 . and 3 3

x

42. Solve 2 p 2  5 p  1 . First, we put the equation into standard form. 2 p2  5 p  1  0 Next we apply the quadratic formula.

p

  5  

The solutions are

5  33 5  33 and . 4 4

1 1 43. Solve 1   2 w  0 . w w Multiplying both sides by w2 we get:  1 1 w2  1    2  w2  w w 2

w  w 1 w2  w  1  0 This is a quadratic equation in standard form. Next, we apply the quadratic formula. w 

7  49  36 7  13  2 2

7  13 7  13 and . 2 2

45.

x 5  5 x1  5 x

a  a 

46.

t 7 7t

47.

t 5  5 t2

48.

y 3  3 y2

49.

y 3

50.

t 5 

a  a  m

2

2 1

1 5 1 5 and The solutions are . 2 2 1

y2 1   n a  n  a  

a  a 

5 2

b 1

a  a  m

1   n a  n  a  

3 b 5

1

b 3 



2



52.

51.

 1  4 1 1

1 1 4 1 5  2 2

 7 2  4 1 9  2 1

The solutions are

 52  4  2  1 2  2

5  25  8 5  33   4 4

  1 



 7 

1 5 b

Exercise Set R.5 65 53.

 x  3  1

 x  3 2

 y  7  1

x x

56.

x3  x 2 4

58.



y 7





a  a 

y 7

    16    4  1024

am  a n



83  83

68.

64 3

 



 3 64 m

  4

 64 3

 a a 

    8   2  4

67.

b2  b 4  b 2 ; b  0 n

 9

66. 16 2  16 2

 

n 3 1 m n  am   2  2  3; a  

  3  27

The index is 2;

a3  a 5

 

 92



1 2

65.

a  a 

x 3

57.

55.



54.

1   n a  n  a  



 

n 2 1 m n m  3  3  2; a  a   

a  a  1

 64  4

 16 59.

4 12

 x 4  x3 ; x  0 69.

60. 61.

62.

3 6

t  t 3  t2 1 m

1 t



1 x 7



 a a 

 x  7



1

 x 7

64.

 1 n   n a  a 

5

70.

 a a 



x3  4  x3  4



x2  4 x2

Solve: x20 x  2. Therefore, 2 is not in the domain. The domain of f consists of all real numbers except 2 .

 m 2

t 2

63.

f  x 

 1 n   n a  a 

2 x 2  50 x5 The domain of a rational function is restricted to those input values that do not result in division by 0. To determine the domain of 2 x 2  50 f  x  x5 We set the denominator equal to zero and solve: x5  0 f  x 

x5 Therefore, 5 is not in the domain. The domain of f consists of all real numbers except 5, in interval notation  , 5    5,  

66 Chapter R: Functions, Graphs, and Models

71.

f  x 

2x  6  0

x4  7

2x  6 x3 The domain is the set of all real numbers greater than or equal to 3, or in interval notation 3, .

x  6x  5 Solve: x2  6 x  5  0

 x  5 x  1  0 x  5 or x  1 Therefore, the numbers  5 and  1 are not in the domain. The domain of f consists of all real numbers except  5 and  1 , in interval notation  , 5   5, 1   1,  .

75. The domain of the radical function f  x   4 7  x is restricted to those input

values that result in the value of the radicand being greater than or equal to 0. In other words, the domain will be the set of real numbers that satisfy the inequality 7  x  0 .

72. The domain of a rational function is restricted to those input values that do not result in division by 0. To determine the domain of x3 f  x  2 x  5x  6 we set the denominator equal to zero and solve: x2  5x  6  0

 x  2  x  3  0

factoring

x  2  0 or x  3  0 Principle of Zero Products x  2 or x  3. Therefore, 2 and 3 are not in the domain. The domain of f consists of all real numbers except 2 and 3, in interval notation  , 2   2,3   3,  . 73. The domain of the radical function f  x   5 x  4 is restricted to those input

values that result in the value of the radicand being greater than or equal to 0. In other words, the domain will be the set of real numbers that satisfy the inequality 5 x  4  0 . To find the domain, we solve the inequality: 5x  4  0 5 x  4 4 x 5 Therefore, the domain of f consists of all real 4 numbers greater than or equal to  , or in 5  4  interval notation   ,   .  5  74.

To find the domain, we solve the inequality: 7x  0 7x Therefore, the domain of f consists of all real numbers less than or equal to 7, or in interval notation  ,7 . 76.

f  x  6 5  x Solve: 5 x  0 5 x Therefore, the domain of f consists of all real numbers less than or equal to 5, or in interval notation  ,5 .

77. We set the demand equation equal to the supply equation and solve for q. 1000  10q  250  5q 750  15q 50  q Thus, the equilibrium quantity is 50 units. To find the equilibrium price, we substitute 50 for q into either the demand equation or supply equation. We use the demand equation. p  1000  10  50  p  1000  500 p  500 The equilibrium price is $500. The equilibrium point is  50,500 .

f  x   2x  6 Solve:

Exercise Set R.5 67 78. We set demand equal to supply and solve to find equilibrium price. 8800  30q  7000  15q 1800  45q 40  q Substitute into the demand equation to find equilibrium price. p  8800  30  40   7600 .

Thus, the equilibrium quantity is 40 units and the price is $7600. The equilibrium point is  40,7600 . 79. We set the demand equation equal to the supply equation and solve for q. 4 q  q 4 2

16  q 4  q Since it is not appropriate to have a negative quantity, the equilibrium quantity is 4 thousand units or 4000. To find the equilibrium price, we substitute 4 for q into either the demand equation or supply equation. We use the demand equation. 4 p  4 p 1 The equilibrium price is 1 hundred dollars or $100. The equilibrium point is  4,1 .

80. We set the demand equation equal to the supply equation and solve for q. 5 q  q 5

25  q 2

multiply both sides by 5 x

25  q 2

take the square root of both sides

5  q Since it is not appropriate to have negative units, the equilibrium quantity is 5 thousand units or 5000. To find the equilibrium price, we substitute 5 for q into either the demand equation or supply equation. We use the demand equation. 5 p 1  5 The equilibrium price is 1 hundred dollars or $100. The equilibrium point is  5,1 .

81. We set the demand equation equal to the supply equation and solve for q.

 q  4 2  q 2  2q  6 q 2  8q  16  q 2  2q  6 8q  16  2q  6 10  10q 1 q The equilibrium quantity is 1 hundred units or 100. To find the equilibrium price, we substitute 1 for q into either the demand equation or supply equation. We use the demand equation. p  1  4 

  3   9 2

The equilibrium price is $9. The equilibrium point is 1,9  . 82. We set the demand equation equal to the supply equation and solve for q.

 q  3  2  q 2  2q  1 q 2  6q  9  q 2  2q  1 6q  9  2q  1 subtracting x 2 from both sides 8  8q 1 q The equilibrium quantity is 1 hundred units or 100. To find the equilibrium price, we substitute 1 for q into either the demand equation or supply equation. We use the demand equation. p  1  3 

  2   4 2

The equilibrium price is $4. The equilibrium point is 1, 4  . 83. We set the demand equation equal to the supply equation and solve for q. 5q  q 7 0q5

 5  q 2   q  7  Squaring both sides 2

25  10q  q 2  q  7 18  11q  q 2  0

 9  q  2  q   0 9  q  0 or 2  q  0 9  q or

2q

Factoring the quadratic

Principle of Zero Products

68 Chapter R: Functions, Graphs, and Models

Since 9 is not in the domain of the demand function, the equilibrium quantity is 2 thousand units or 2000 units. To find the equilibrium price, we substitute 2 for q into either the demand equation or supply equation. We use the demand equation. p  5   2  3 The equilibrium price is 3 thousand dollars or $3000. The equilibrium point is  2,3 . 84. We set the demand equation equal to the supply equation and solve for q. 7  q  2 q 1 0q7

 7  q 2   2 q  1 

49  14q  q 2  4q  4 45  18q  q 2  0 15  q  0 or 3  q  0

15  q or 3q Since 15 is not in the domain of the demand function, the equilibrium quantity is 3 thousand units or 3000 units. To find the equilibrium price, we substitute 3 for q into either the demand equation or supply equation. We use the demand equation. p  7   3  4 The equilibrium price is 4 thousand dollars or $4000. The equilibrium point is  3, 4  . 85.

k p Find the constant of variation by making the following substitutions. k 85, 000  1200 102, 000, 000  k Therefore, the equation of variation is: 102, 000, 000 . q p If the price drops to $850, then 102, 000, 000 x  120, 000 850 120,000 Television will be sold if the price is $850. q

86. If the number of tickets sold q is inversely proportional to the price of the ticket p, then we have: k q . p We find the constant of variation by substituting 175 for q and 20 for p. k 175  20 3500  k 3500 . The equation of variation is q  p If the price were raised to 25 we find q by substituting 25 in for p. 3500 q 25  140 The number of tickets sold would be 140 if the assumption of inverse proportionality is correct. 87. a) R  x   11.74 x0.25

R  40, 000   11.74  40, 000 

0.25

 11.74 14.14213562   166.0286722  166 The maximum range will be approximately 166 miles when the peak power is 40,000 watts.

R  50, 000   11.74  50, 000 

0.25

 11.74 14.95348781  175.5539469  176 The maximum range will be approximately 176 miles when the peak power is 50,000 watts. R  60, 000   11.74  60, 000 

0.25

 11.74 15.6508458   183.7409297  184 The maximum range will be approximately 184 miles when the peak power is 60,000 watts.

Exercise Set R.5 69 b) Plotting the points found in part (a) and connecting them with a smooth curve we see:

Pollution in 2030 t  2030  2010  20 p  20   10.007  20 

0.098

p  20   7.46  g / m 3

b) Graph of the function over the interval 0, 30 .

88. Substituting each value given for w in H  w  0.059w0.92 and using a calculator to compute the values for H we create the following table in the next column where the values for H are rounded off to one decimal. w

0 0

1000 34.0

2000 64.2

3000 93.3

4000 121.6

5000 149.3

6000 176.5

7000 203.4

H  w H  w

We plot the points in the table and graph the function.

90.

N  t   15, 240t 2 a) Number of downloads at 15 days t  15

89.

N 15   15, 240 15  2

p  t   10.007t 0.098

N 15   59, 024 downloads

a) Pollution in 2018 t  2018  2010  8 p  8   10.007  8 

0.098

Number of downloads at 35 days t  35

p  8   8.16  g / m 3

N  35   15, 240  35  2

Pollution in 2023

N  35   90,161 downloads

t  2023  2010  13 p 13  10.007 13

0.098

p 13  7.78 g / m3

Number of downloads at 45 days t  45 1

N  45   15, 240  45  2 N  45   102, 233 downloads

70 Chapter R: Functions, Graphs, and Models greater than 600,000, so there are 106  53  53 cities with a population between 300,000 and 600,000.

b) Graph of the function over the interval 0, 90 .

92.

R  t   0.0216t 6  0.369t 5  2.30t 4  6.34t 3  7.625t 2  5.31t  14

a) Revenue in 2016 t  2016  2015  1 R  t   0.0216 1  0.369 1  2.30 1 6

 6.34 1  7.625 1  5.311  14 3

 0.0216  0.369  2.30  6.34  7.625  5.31  14  16.072 billion dollars Revenue in 2019 t  2019  2015  4

91. The number of cities N with a population greater than S is inversely proportional to S . k N S k 317  100, 000 31, 700, 000  k The equation of variation is 31, 700, 000 N S To estimate the number of cities with a population between 350,000 and 500,000 we find N when S is 350,000 and when S is 500,000. 31, 700, 000 N  90.57  91 350, 000 31, 700, 000 N  63.4  63 500, 000 There are 91 cities with a population greater than 350,000 and 63 cities with a population greater than 500,000, so there are 91  63  28 cities with a population between 350,000 and 500,000. To estimate the number of cities with a population between 300,000 and 600,000 we find N when S is 300,000 and when S is 600,000. 31, 700, 000 N  105.6  106 300, 000 31, 700, 000 N  52.8  53 600, 000 There are 106 cities with a population greater than 300,000 and 53 cities with a population

R  t   0.0216  4   0.369  4   2.30  4  6

 6.34  4   7.625  4   5.31 4   14 3

 88.4736  377.856  588.8  405.76  122  21.24  14  19.582 billion dollars Revenue in 2021 t  2021  2015  6 R  t   0.0216  6   0.369  6   2.30  6   6.34  6  6

 7.625  6   5.31 6   14 2

 1007.7696  2869.344  2980.8  1369.44  274.5  31.86  14  21.574 billion dollars

b) Revenue at 2021 is 21.574 billion dollars Advertising spending 236 million or 0.236 billion dollars Revenue Advertising spending 21.574  0.236  91.4 c) Answer may vary ROAS 

Exercise Set R.5 71 93. a) Advertising spending 4,000 ROAS is 6.38 Revenue ROAS  Advertising spending Revenue 6.38  4, 000 Revenue  6.38  4, 000 Revenue  25,520 ROAS is 4.85 Revenue ROAS  Advertising spending Revenue 4.85  4, 000 Revenue  4.85  4, 000 Revenue  19, 400 ROAS is 1.81 Revenue ROAS  Advertising spending Revenue 1.81  4, 000 Revenue  1.81 4, 000 Revenue  7, 240 b) Answer may vary f  t   t 2  12

94.

f  3  32  12  9  12 f  3  21 thousand cell or 21, 000 Percent error  100

Estimate  Exact value Exact value

 100

21, 000  20,500 20,500

 100

500 20,500

 100  0.0244   2.44%

b) f  6   6 2  12  36  12 f  3  48 thousand cell or 48, 000 Percent error  100

Estimate  Exact value Exact value

 100

48, 000  49, 200 49, 200

 100

1200 49, 200

 100  0.0244   2.44% c) Answer may vary

95.

Answers will vary. A function has at most one y-intercept. If a graph has more than one yintercept, then the vertical line x  0 intersect the graph in more than one point. Thus, the graph fails the vertical line test, and the graph does not represent a function.

96.

Answers will vary. A rational function is the quotient of two polynomial functions. Every polynomial function can be expressed as a rational function with a denominator of 1, so every polynomial function is a rational function. However, most rational functions are not polynomial functions.

97. We graph f  x   x3  x on our calculator.

Using the ZERO feature we find the 3 zeros are –1, 0, 1. Algebraically we find the zeros by solving the equation: x3  x  0





x x2  1  0 x  x  1 x  1  0 x  0 or x  1  0 or x  1  0 x  0 or x  1 or The zeros are –1, 0, 1.

x  1

72 Chapter R: Functions, Graphs, and Models 98.

f  x   2 x3  x2  14 x  10 Enter the function into your calculator.

Using the window:

We see the graph:

Using the window:

Now using the ZERO feature on the calculator, we approximate the zeros. The zeros are 1.489 and 5.673.

We see the graph:

101. f  x   7  x 2  1 Using the window:

Now using the ZERO feature on the calculator, we approximate the zeros. The zeros are –1.831, –0.856, 3.188. 99.

1  x4  x7 4 2 Using the window: f  x 

We get the graph:

The zeros are  2.449 and 2.449. 102. f  x   x  1  x  2  5

Enter the function into your calculator. The zeros are 1.5 and 9.5. 100. f  x   x4  4 x3  36 x2  160 x  300

Enter the function into your calculator.

Exercise Set R.5 73

Using the standard window:

We see the graph:

Now using the ZERO feature on the calculator, we approximate the zeros. The zeros are –2 and 3.

We see that the graph intersects the x-axis between 1  x  2 . The zeros of this function are all real numbers in  1, 2

103. f  x   x  1  x  2

Enter the function into your calculator.

105. f  x   x8  8x7  28x6  56 x5  70 x4  56 x3 28 x 2  8 x  1 It is difficult to choose a window that shows the entire graph and all of the function’s zeros. We use the window:

Using the standard window:

This results in the following graph:

We see the graph:

We see that the graph does not cross the x-axis, there are no real zeros.

You can see that it difficult to see all of the zeros. It is recommended to zoom in around each zero and then use the ZERO feature to approximate each zero. The zeros are: –10.153, –1.871, –0.821, –0.303, 0.098, 0.535, 1.219, and 3.297.

104. f  x   x  1  x  2  3

Enter the function into your calculator.

74 Chapter R: Functions, Graphs, and Models 106. Solve x 2  x  1 . First, we put the equation into standard form. x2  x  1  0 Next we apply the quadratic formula.

x 

  1 

 12  4 1 1 2 1

1 1 4 1 5  2 2

The only positive solution is

1 5 . 2

107. Solve x  x  1 . First, we put the equation into standard form.

 x 2   x  1  x2  x  1

x2  x  1  0 Next we apply the quadratic formula. x 

  1 

 12  4 1 1 2 1

1 1 4 1 5  2 2

The only positive solution is

1 5 . 2

1 1 . x First, we put the equation into standard form. 1 x  x  x     1    x  1 

108. Solve x 

x2  1  x x2  x 1  0 Next we apply the quadratic formula.

x 

  1 

 12  4 1 1 2 1

1 1 4 1 5  2 2

The only positive solution is

1 5 . 2

109. All three problems translate into the same equation x 2  x  1  0 .

Exercise Set R.6 75 2.

Exercise Set R.6

a) When x  3, f ( x)  43  0.015625 When x  2, f ( x)  42  0.0625 When x  1, f ( x)  41  0.25

a) When x  3, g ( x)  5

3

 0.008

When x  0, f ( x)  40  1

When x  2, g ( x)  52  0.04

When x  1, f ( x)  41  4

When x  1, g ( x)  51  0.2

When x  2, f ( x)  42  16

When x  0, g ( x)  5  1

When x  3, f ( x)  43  64

When x  1, g ( x)  51  5 x

When x  2, g ( x)  52  25

3

When x  3, g ( x)  53  125

2 1

f ( x)  5 x 0.008 0.04 0.2

1 2

5 25 125

3

2 1 0

1 2 3

b) The percent increase is the percent between any two consecutive values. Therefore, the percent increase for this function is: 125  25 100   4  400% . 25 25

f ( x)  4 x 0.0156 0.0625 0.25

1 4 16 64

b) The percent increase is the percent between any two consecutive values. Therefore, the percent increase for this function is: 64  16 48   3  300% . 16 16 c) The y–intercept will be the point (0,1). d)

c) The y–intercept will be the point (0,1). d)

Chapter R: Functions, Graphs, and Models

a) k  x   8 1.15 

When x  3, k  3  8 1.15 

When x  2,

 5.2601

k  2   8 1.15 

When x  1,

3

2

 6.0491 k  1  8 1.15 

1

 6.9565

When x  0,

k  0   8 1.15 

8

When x  1,

a) h  x   2 1.24  When x  3,

k 1  8 1.15 

h  3  2 1.24 

 9.2

When x  2,

k  2   8 1.15 

When x  2,

k  3  8 1.15 

When x  1,

k  x   8 1.5

3

5.2601 6.0491 6.9565 8 9.2 10.58 12.167

2 1

0 1 2

 1.0490

b) The percent increase is the percent between any two consecutive values. Therefore, the percent increase for this function is: 9.2  8  8  0.15  15% c) The y-intercept will be the point  0,8 

2

 1.3007 h  1  2 1.24 

 12.167

3

h  2   2 1.24 

 10.58

When x  3,

1

 1.6129

When x  0,

h  0   2 1.24 

2

When x  1,

h 1  2 1.24 

 2.48

When x  2,

h  2   2 1.24   3.0752

When x  3,

h  3   2 1.24 

 3.8132

h  x   2 1.24 

3

1.0490 1.3007 1.6129

2 1

2.48 3.0752 3.8132

Exercise Set R.6 77 d)

b) The percent increase is the percent between any two consecutive values. Therefore, the percent increase for this function is: 2.48  2  2  0.24  24% c) The y-intercept will be the point  0, 2  6.

3 a) When x  3, m( x)  3  2.15  0.3019

When x  2, m( x)  3  2.152  0.649 When x  1, m( x)  3  2.151  1.3953 When x  0, m( x)  3  2.150  3 When x  1, m( x)  3  2.151  6.45 When x  2, m( x)  3  2.152  13.8675 When x  3, m( x)  3  2.153  29.8151 x

3

2 1 5.

3 a) When x  3, n( x)  2  3.04  0.0712

When x  2, n( x)  2  3.04

2

1 2

 0.2164

When x  1, n( x)  2  3.041  0.6579

f ( x)  3  2.15 x 0.3019 0.649 1.3953 3 6.45 13.8675 29.8151

When x  0, n( x)  2  3.04  2 When x  1, n( x)  2  3.041  6.08 When x  2, n( x)  2  3.042  18.4832 When x  3, n( x)  2  3.043  56.1889 x

3

2 1 0

1 2 3

f ( x)  2  3.04 x 0.0712 0.2164 0.6579

b) The percent decrease is the percent between any two consecutive values. Therefore, the percent decrease for this function is: 29.8151  (13.8675)  1.15  115% . 13.8675 c) The y–intercept will be the point (0,–3). d)

2

6.08 18.4832 56.1889

b) The percent decrease is the percent between any two consecutive values. Therefore, the percent decrease for this function is: 56.1889  (18.4832)  2.04  204% . 18.4832 c) The y–intercept will be the point (0,–2).

Chapter R: Functions, Graphs, and Models

a) When x  3, p ( x)  13,996.2094 When x  2, p ( x)  13, 296.3989 When x  1, p ( x)  12, 631.579 When x  0, p ( x)  12, 000 When x  1, p ( x)  11, 400 When x  2, p( x )  10,830 When x  3, p ( x)  10, 288.5 x

3

2 1 0

1 2 3

f ( x)  75, 000, 000(0.875) x 111,953,352.8 97,959,183.7 85, 714, 285.7 75, 000, 000 65, 625, 000 57, 421,875 50, 244,140.6

3

2 1 0

1 2 3

f ( x)  12, 000(0.95) x 13,996.2094 13, 296.3989 12, 631.579 12, 000 11, 400 10,830 10, 288.5

b) The percent decrease is the percent between any two consecutive values. Therefore, the percent decrease for this function is: 50, 244,140.63  57, 421,875  0.125 . 57, 421,875  12.5% c) The y–intercept will be the point (0,75,000,000).

b) The percent decrease is the percent between any two consecutive values. Therefore, the percent decrease for this function is: 10, 288.5  10,830  0.05  5% . 10,830

c) The y–intercept will be the point (0,12000). d)

a) When x  3, r ( x)  79.0555r0 When x  2, r ( x)  18.4199r0

When x  1, r ( x)  4.2918r0 When x  0, r ( x)  r0 When x  1, r ( x)  0.233r0 When x  2, r ( x)  0.0543r0 8.

a) When x  3, q ( x )  111,953,352.8 When x  2, q ( x)  979,591,83.67 When x  1, q ( x)  857,142,85.71 When x  0, q( x)  75, 000, 000 When x  1, q ( x)  65, 625, 000 When x  2, q ( x)  57, 421,875 When x  3, q ( x)  50, 244,140.63

When x  3, r ( x)  0.0126r0 x

f ( x)  r0 (0.233) x

3

79.0555r0

2 1

18.4199r0

1 2

0.233r0 0.0543r0

0.0126r0

4.2918r0

Exercise Set R.6 79 b) The percent decrease is the percent between any two consecutive values. Therefore, the percent decrease for this function is: 0.0126r0  0.0543r0  0.767  76.7% . 0.0543r0

11.

54  625  log 5 625  4

12.

43  64  log 4 64  3

13.

7 2 

c) The y–intercept will be the point (0, r0 ) .

1  1   log 7    2 49  49 

14. 102  0.01  log10 0.01  2

10. a) When x  3, s ( x)  2.3515s0 When x  2, s ( x)  1.7683s0

15.

36  6  (36) 2  6 1  log 36 6  2

16.

25  5  (25) 2  5 1  log 25 5  2

When x  1, s ( x)  1.3298s0 When x  0, s( x)  s0

17.

216  6  (216) 3  6 1  log 216 6  3

18.

32  2  (32) 5  2 1  log 32 2  5

19.

p n  Q  log p Q  n

20.

r v  z  log r z  v

21.

log 4 1024  5  45  1024

22.

log 2 128  7  27  128

23.

log100, 000  5  log10 100, 000  5

When x  1, s ( x)  0.752 s0 When x  2, s ( x)  0.5655s0 When x  3, s ( x)  0.4253s0 x

f ( x)  s0 (0.752) x

3

2.3515s0

2 1

1.7683s0

1 2

0.752s0 0.5655s0

0.4253s0

1.3298s0

b) The percent decrease is the percent between any two consecutive values. Therefore, the percent decrease for this function is: 0.4253s0  0.5655s0  0.248  24.8% . 0.5655s0

 105  100, 000

24. c) The y–intercept will be the point (0, s0 ) .

log10, 000, 000  7  log10 10, 000, 000  7  107  10, 000, 000

1 1  16 2  4 2

25.

log16 4 

26.

log100 10 

1 1  100 2  10 2

Chapter R: Functions, Graphs, and Models

27.

log15 4 15 

28.

log 7 3 7 

36. First rewrite the function in exponential form: log 25 25  x  25 x  25 Now solve the resulting equation: 25x  25

1 1  15 4  4 15 4

1 1  73  3 7 3

29.

log 9 1  0  90  1

30.

log 1  0   0  1

25x  251 x 1

31. First rewrite the function in exponential form: log 2 x  10  210  x Now solve the resulting equation: x  210  1024 . 32. First rewrite the function in exponential form: log 2 x  8  28  x Now solve the resulting equation: x  28  256 . 33. First rewrite the function in exponential form: log x 125  3  x3  125 Now solve the resulting equation: x3  125 .

x3  53

34. First rewrite the function in exponential form: log x 64  2  x 2  64 Now solve the resulting equation: x 2  64 .

x 8

x  2 38. First rewrite the function in exponential form: 1  1  log 5    x  5x  25 25   Now solve the resulting equation: 1 5x  25 5x  52

x  2 39. First rewrite the function in exponential form:

x5

37. First rewrite the function in exponential form: 1  1  log 6    x  6 x  36 36   Now solve the resulting equation: 1 6x  36 6 x  62

 

log 7 7 2  x  7 x  7 2 Now solve the resulting equation:

7x  7 2 x 2

40. First rewrite the function in exponential form: log 6 6  x  6 x  6

x8

 

35. First rewrite the function in exponential form: log13 13  x  13x  13 Now solve the resulting equation: 13x  13

13x  131 x 1

Now solve the resulting equation: 6 x  6 x  41. By the change of base formula, we have: log 8 log 5 8   1.29203 log 5 42. By the change of base formula, we have: log 7 log 2 7   2.80735 log 2

Exercise Set R.6 81 43. By the change of base formula, we have: log16 log3 16   2.52372 log 3 44. By the change of base formula, we have: log14 log 7 14   1.35621 log 7 45. By the change of base formula, we have: log 25 log 4 25   2.32193 log 4 46. By the change of base formula, we have: log 30 log8 30   1.63563 log 8 47. By the change of base formula, we have: log 9 log 2 9   3.16993 log 2

53. First rewrite the function in exponential form: 5 x  50  log 5 50  x Now solve the resulting equation using the change of base formula: log 50 log 5 50   2.43068 . log 5 54. First rewrite the function in exponential form: 9 x  63  log 9 63  x Now solve the resulting equation using the change of base formula: log 63 log9 63   1.88562 . log 9 55. First rewrite the function in exponential form:

1.055 x  1.78  log1.055 1.78  x Now solve the resulting equation using the change of base formula: log1.78 log1.055 1.78   10.76961 . log1.055

48. By the change of base formula, we have: log 45 log 9 45   1.73249 log 9

56. First rewrite the function in exponential form:

49. By the change of base formula, we have: log 3 log 6 3   0.61315 log 6

Now solve the resulting equation using the change of base formula: log 3.5 log1.0325 3.5   39.16960 . log1.0325

50. By the change of base formula, we have: log 2 log9 2   0.31546 log 9 51. First rewrite the function in exponential form: 11x  33  log11 33  x Now solve the resulting equation using the change of base formula: log 33 log11 33   1.45816 . log11 52. First rewrite the function in exponential form: 7 x  28  log 7 28  x Now solve the resulting equation using the change of base formula: log 28 log 7 28   1.71241 . log 7

1.0325 x  3.5  log1.0325 3.5  x

57. First rewrite the function in exponential form:

 0.87  x  0.59  log0.87 0.59  x Now solve the resulting equation using the change of base formula: log 0.59 log 0.87 0.59   3.78878 . log 0.87 58. First rewrite the function in exponential form:

 0.45 x  0.22  log0.45 0.22  x Now solve the resulting equation using the change of base formula: log 0.22 log 0.45 0.22   1.89620 . log 0.45

Chapter R: Functions, Graphs, and Models

59. First rewrite the function in exponential form: 19 3  6 x  19  6 x  3  19   log 6    x  3 Now solve the resulting equation using the change of base formula:  19  log    19   3   1.03018 . log 6    log 6  3 60. First rewrite the function in exponential form: 42 5  7 x  42  7 x  5  log 7  8.4   x Now solve the resulting equation using the change of base formula: log  8.4  log 7  8.4    1.09369 . log 7 61.

65.

 32   log a    2  log a 32  log a 2  2 log a 3  log a 2  2(0.766)  0.483  1.049

66.

 0.283

67.

 0.483  0.766  1  2.249

68.

 

log a (4a )  log a 22 a

 log a 22  log a a

 1.532

 2(0.483)  1

log a 8  log a 23

 1.966

 3log a 2  3(0.483)

69.

 1.449

log a 12  log a  3  4 



 log a 3  22

 log a 3  2 log a 2  0.766  2(0.483)



 log a 2  32

 0.483

70.

1 log a    log a 31 3   log a 3  (0.766)

 1.732 log a 18  log a  2  9 

1 log a    log a 21 2   log a 2  (0.483)



 log a 3  log a 22

64.

log a (6a)  log a  2  3  a   log a 2  log a 3  log a a

log a 9  log a 3

 2(0.766)

63.

3 log a 1.5  log a   2  log a 3  log a 2  0.766  0.483

 2 log a 3

62.

9 log a 4.5  log a   2

 0.766



71. The sketch of m( x)  log5 x is:

 log a 2  log a 32  log a 2  2 log a 3  0.483  2(0.766)  2.015 The domain of the function is  x | x  0 .

Exercise Set R.6 83 72. The sketch of f ( x)  log3 x is:

77. a) Based on the given information, the initial population is 900, or A0  900 and the base of the function is 104.5% because the population is increasing or the value 1.045. Therefore, the function would be defined as:

A  t   900 1.045 . t

The domain of the function is  x | x  0 .

b) A  5   900 1.045 

 1121.56 which means that in 2025 the student population will be about 1122.

73. The sketch of v ( x)  log 3  x  5  is:

The domain of the function is  x | x  5 . 74. The sketch of t ( x)  log 4  x  3 is:

The domain of the function is  x | x  3 . 75. The sketch of k ( x )  log10  2 x  1 is:

1200  900 1.045 

1200 t  1.045  900 t 4 log    log 1.045  3 4 log    t log 1.045  3 4 log   3 t log1.045 6.535  t The population will first reach 1200 students after 6.535 years or in the 7th year.

78. a) Based on the given information, the initial population is 12,500, or P0  12,500 and the base of the function is 102.75% because the population is increasing or the value 1.0275. Therefore, the function would be defined as:

P  t   12,500 1.0275 . t

b) P  8   12,500 1.0275 

1  The domain of the function is  x | x   . 2 

76. The sketch of r ( x)  log 2  2  3x  is:

 15, 529.76 which means that in 2028 the population of Parker Valley will be about 15,530.

2  The domain of the function is  x | x   . 3 

Chapter R: Functions, Graphs, and Models c)

20, 000  12,500 1.0275 

b) q 11  2500  0.976 

 1913.76 which means that in 2028 the population of Tonopah Acres will be about 1914.

20, 000 t  1.0275  12,500 t 8 log    log 1.0275 5 8 log    t log 1.0275  5 8 log   5  t log1.0275 17.324  t The population will first reach 20,000 residents after about 17.3 years.

79. a) Based on the given information, the initial sample size is 50 mg, or M 0  50 and the base of the function is 98.75% because the sample is decaying or decreasing or the value 0.9875. Therefore, the function would be defined as:

M  t   50  0.9875 . t

1250  2500  0.976  1250 t   0.976  2500 log  0.5   log  0.976 

log  0.5   t log  0.976 

log  0.5 

t log 0.976 28.53  t The population of Tonopah Acres will be 1250 after about 28.5 years.

81. a) Based on the given information, the initial enrollment is 550 and the enrollment after 3 years is 1750, or E0  550 and E3  1750 . We will use this information to find the base of the function:

b) M 15   50  0.9875 

 41.4 which means that after 15 days the mass of the sample will be about 41.4 mg.

20  50  0.9875 

E  t   550  r 

1750  550  r 

3.181818182  r 3 ln 3.181818182  3ln r ln 3.181818182  ln r 3

20 t   0.9875  50 log  0.4   log  0.9875 

ln 3.181818182 3

r 1.471  r Thus, the function will be: e

E  t   550 1.471

log  0.4   t log  0.9875 

log  0.4 

t log 0.9875 72.8  t The sample will have 20 mg remaining after about 72.8 days.

b) E 10   550 1.471

 26, 090.92 which means that in 10 years there will be about 26,091 students enrolled in the online program.

80. a) Based on the given information, the initial population is 2500, or q0  2500 and the base of the function is 97.6% because the population is decreasing or the value 0.976. Therefore, the function would be defined as:

q  t   2500  0.976  . t

Exercise Set R.6 85 c)

5000  550 1.471

c) N  t   280

5000 t  1.471 550 t  100  log    log 1.471 11    100  log    t log 1.471  11   100  log    11   t log1.471 5.72  t The enrollment in the online program will reach 5000 students after 5.72 years or in the 6th year.

82. a) Based on the given information, the initial subscribers is 22.5 million and the growth rate is 26.8% We will use this information to find the base of the function If t  1  N  t   22.5  26.8%  22.5  28.53 N  t   22.5  r  N 1  22.5  r 

28.53  22.5r 28.53 r 22.5 r  1.268

Thus the function will be t N  t   22.5 1.268  b) N  t   22.5 1.268 

N  t   22.5 1.268 

280  22.5 1.268 

12.44  1.268 

log 12.44   log 1.268 

log 12.44   t log 1.268  t

log 12.44  log 1.268 

 10.61

d) Answer may vary 83. Based on the given information, the initial deposit is $1,500 and we need to find the time for the balance to be doubled or $3,000. P0  1,500, Pt  3, 000 We also know that the base of the function is 4.05% or 1.0405 because the balance is increasing. We will use this information to find the time needed: t P  t   P0 1.0405  3, 000  1, 500 1.0405 

3, 000 t  1.0405  1, 500 ln  2   t ln 1.0405  ln  2 

ln 1.0405 

t

t  17.459

N  8   22.5 1.268 

N  8   150.36

Thus, it will take approximately 17.46 years for Alisha’s money to double.

This means after 8 years, there are about 150.36 million subscribers.

Chapter R: Functions, Graphs, and Models

84. Based on the given information, the initial deposit is $5,000 and we need to find the time for the balance to be tripled or $15,000, or P0  5, 000 and Pt  15, 000 . We also know that the base of the function is 103% or 1.03 because the balance is increasing. We will use this information to find the time needed:

86. Based on the given information, the initial value is $50 and we need to find the time for the balance to drop to $45, or P0  50 and Pt  45 . We also know that the base of the function is 99% or 0.99 because the balance is decreasing. We will use this information to find the time needed:

P  t   5, 000 1.03

P  t   50  0.99 

15, 000  5, 000 1.03

45  50  0.99 

t t

45  0.99t 50 ln  0.9   t ln 0.99

15, 000  1.03t 5, 000 ln  3  t ln1.03 ln  3

ln  0.9 

t ln 0.99 10.5  t Thus, it will take approximately 10.5 week for the stock to drop to $45.

t

ln1.03 37.2  t Thus, it will take approximately 37.2 years for Jenny’s money to triple. 85. Based on the given information, the initial value is $12,000 and we need to find the time for the value to drop to $10,000, or P0  12, 000 and Pt  10, 000 . We also know that the base of the function is 97% or 0.97 because the value is decreasing. We will use this information to find the time needed: P  t   12, 000  0.97 

10, 000  12, 000  0.97 

87. Let x  log a M and y  log a N , then a x  M and a y  N . Now use the fact that ax

M . N M Since a x  y  , it follows that N M  log a    log a a x  y N M  log a    x  y N M  log a    log a M  log a N N a x y 





10, 000  0.97t 12, 000 5 ln    t ln 0.97 6 5 ln   6 t ln 0.97 6t Thus, it will take approximately 6 months for the property value to drop to $10,000.

88.



   M   M  log a M b  log a  M M   b times    log a M  log a M  log a M    log a M 

 

b times

 b log a M

Exercise Set R.6 87 89.

94. a) f  x   mx  b

log a M  y y

a M log b a y  log b M y log b a  log b M

90.

y

log b M log b a

log a M 

log b M log b a

 1  N log a    log a   M M    N

b) g  x   a  x  h   k 2

M   log a 1  log a   N M   0  log a   N N M  log a     log a   M  N 91.

8090 ?9080 log 8090 ? log 9080 90 log 80?80 log 90 171.28  156.34 Therefore, 8090  9080 .

f  x   ab x , b  1

92. 125200 ?200125

log125200 ? log 200125 200 log125?125log 200 419.38  287.63 Therefore, 125200  200125 . 93. Use a graphing calculator to see the graph of the function and the domain will be  ,  1  1,  .

Chapter R: Functions, Graphs, and Models d) A  t   P 1  r 

e) h  x   logb x, b  1

f) For part (c), the base b must be positive and not equal to 1 to ensure meaningful behaviour of the exponential function. For part (d), while the principal P must be positive, the rate r can be any real number, though r  1 leads to trivial or impractical results in most scenarios.

Exercise Set R.7 89

Exercise Set R.7

10. The data falls but not at a constant rate, so an exponential with 0  a  1 could be used to model the data.

11. a) We find the linear function f  x   mx  b that contains the data points

The data is decreasing at a constant rate, so a linear function f  x   mx  b could be used to model the data.

The data falls, possibly to a low point, so a quadratic function f  x   ax 2  bx  c, a  0 or an exponential with 0  a  1 could be used to model the data.

The data rises first and then falls over the domain, so a quadratic function f  x   ax 2  bx  c, a  0 could be used to model the data.

The data appears to first falls, then rises, falls again, and finally rises, so a polynomial function that is neither quadratic nor linear would be the most appropriate model for the data. The data rises over the domain, so an exponential function could be used to model the data due to the rapid and accelerating increase. However, if the data shows more complexity, such as fluctuations or multiple changes in direction, a polynomial function that is neither quadratic nor linear could also be considered.

The data is increase in the average NBA salary over the years, so a linear function f  x   mx  b could be used to model the data.

The data shows a general upward trend with a slight dip in the 6th year. The overall trend is linear. So a linear function f  x   mx  b could be used to model the data.

The data first rises, then falls, and rises again over the domain. This implies that a polynomial function that is neither quadratic nor linear or exponential would be best to model the data. The data falls first and then rises over the domain, so a quadratic function f  x   ax 2  bx  c , a  0 could be used to model the data.

 0, 5.2  and  6, 8.5  . We find the slope containing these two points first. 8.5  5.2 3.3 m   0.55 60 6 Next, we use the slope and one of the points and substitute into the point-slope equation to find the equation of the line. y  y1  m  x  x1  y  5.2  0.55  x  0  y  5.2  0.55 x y  0.55 x  5.2 The linear function that models the data is: f  x   0.55 x  5.2

Alternatively, we substitute in to the equation y  mx  b to obtain a system of equations. 5.2  m  0  b 1 8.5  m  6  b  2 Subtracting each side of Equation (1) from each side of Equation (2) we get: 3.3  6m 3.3 m 6 0.55  m Now substitute m  0.55 in for either Equation (1) or (2) and solve for b. We use Equation (1). 5.2   0.55  0   b 5.2  0  b 5.2  b Therefore, the linear function that fits the data is f  x   0.55 x  5.2

b) In 2025, x  9.

f  9   0.55  9   5.2  10.15

The average salary in 2025 will be approximately $10.15 million. In 2030, x  14. f 14   0.55 14   5.2  12.9

The average salary in 2030 will be approximately $12.9 million.

Chapter R: Functions, Graphs, and Models c) Let f ( x)  15.0 15.0  0.55 x  5.2 9.8  0.55 x 17.82  x The average salary will reach $15.0 million in the year 2016  18  2034.

12. a) We will use the points  0, 346.8  and

 6, 382.9  . First, we find the slope: 382.9  346.8 36.1   6.01667 60 6 Next, since we chose the y-intercept, we can substitute into the slope-intercept equation. y  mx  b m

y  6.01667 x  346.8 b) In 2028, x  2028  2016  12 Substituting 12 for x we have: y  6.01667(12)  346.8  419.00004 The trade deficit with Japan in 2028 will be approximately $419 billion. 13. a) We will use the points (1, 55, 775), (3, 60,336), (5, 68, 703), and (7, 70, 784) . First, we find the slope: n  xy     x   y  m 2 n  x 2      x  

  x   1  3  5  7  16   y   55, 775  60,336  68, 703  70, 784  255,598   xy   (1 55, 775)  (3  60,336)  (5  68, 703)  (7  70, 784)  1, 075, 786

  x   1  3  5  7  84 2

Substitute the values into the slope formula,  4 1, 075, 786   16  255,598 m  4 84   162  4,303,144  4, 089,568 213,576   2669.7 336  256 80 Next, since we chose the y-intercept,  y  m x b n 255,598   2669.7 16   53, 220.7 b 4 m

we can substitute into the slope-intercept equation. y  mx  b

y  2669.7 x  53, 220.7 b) y  ax 2  bx  c

55, 775  a 12  b 1  c 60,336  a  32  b  3  c 68, 703  a  52  b  5  c 70, 784  a  7 2  b  7  c The system of equation is a  b  c  55, 775

9a  3b  c  60,336 25a  5b  c  68, 703 49a  7b  c  70,784 Solving this system we get: a  155, b  3909.7, and c  51,515.7 . Therefore, the function is: y  155 x 2  3909.7 x  51,515.7 c) In 2027, x  2027  2014  13 Substituting 13 for x for the linear model and quadratic model: y  2669.7(13)  53, 220.7  87,926.8 y  155(13) 2  3909.7(13)  51,515.7  76,146.8 In 2020, the median household income, rounded to the nearest dollar in 2027 for both models, will be approximately $87926.8 and $76146.8.

d) Answers may vary. 14. a) Consider the general quadratic function y  ax 2  bx  c Using the points  0, 0  ,  2, 200  , and  3,167  , we substitute

each point into the general quadratic function to obtain the system of equations 0  a  02  b  0  c 200  a  2 2  b  2  c 167  a  32  b  3  c Which gives us: 0c

200  4a  2b  c 167  9a  3b  c From the first equation we see that c  0.

Exercise Set R.7 91

We substitute this value for c into the other two equations and our system is reduced to: 200  4a  2b 167  9a  3b Solving this system of equations, we get: 133 566 a ,b  ,c  0 3 3 This gives us the quadratic function: 133 2 566 f  x   x  x0 3 3 133 2 566  x  x 3 3 Writing the equation with proper fractions we have: f  x   44 13 x 2  188 23 x . f  x   44.333x 2  188.667 x .

b) f  4   44 13  4   188 23  4   45.3 . 2

Approximately 45.3 mg of albuterol will be in the bloodstream after 4 hours. c) Answers will vary. It does not make sense to use this function for t  6 because albuterol will leave the blood stream entirely after about 4 hours. 6 is not in the domain of this function. 15. a) Answers will vary depending on which points are used to find the function. We will use the points  27, 6.7  and

 70, 70.8 . We substitute in to the equation y  mx  b to obtain a system of equations. 6.7  m  27  b 70.8  m  70  b

1  2

b) Substituting in 50 for x we have: y  1.4907  50   33.549  41.001 . Approximately 41% of 50-yr-old woman have high blood pressure c) The estimate is higher than the actual value. 16. a) Answers will vary depending on which points are used to find the function. We will use the points  27,9.1 and

 70, 63.9  . We substitute in to the equation y  mx  b to obtain a system of equations. 9.1  m  27  b

1  2

63.9  m  70  b Subtracting each side of Equation (1) from each side of Equation (2) we get: 54.8  43m 1.2744  m Now substitute m  1.2744 in for either Equation (1) or (2) and solve for b. We use Equation (1). 9.1  1.2744  27   b 9.1  34.409  b 25.309  b Therefore, the linear function that fits the data is y  1.2744 x  25.309 .

b) Substituting in 40 for x we have: y  1.274  40   25.309  25.67 . Approximately 26% of 40-yr-old men have high blood pressure c) The estimate is slightly higher than the actual value.

Subtracting each side of Equation (1) from each side of Equation (2) we get: 64.1  43m 1.4907  m Now substitute m  1.491 in for either Equation (1) or (2) and solve for b. We use Equation (1). 6.7  1.4907  27   b 6.7  40.249  b 33.549  b Therefore, the linear function that fits the data is y  1.4907 x  33.549 .

Chapter R: Functions, Graphs, and Models

17. a) First, we enter the data into the statistic editor on the calculator, letting L1 be the values for x and L2 be the values for y.

18. a) First, we enter the data into the statistic editor on the calculator, letting L1 be the values for x and L2 be the values for y.

Using the exponential regression feature we get:

The exponential function that fits the data is

The exponential function that fits the data is x y  88.6504 1.29948 

y  109.66 1.02952 

b) Substituting 45 in for x we get: 45 y  109.66 1.02952 

 406.08 The equivalent buying power of $100 (in 1980 dollars) in 2025 will be approximately $406.08.

b) Substituting 9 in for x we get: 9 y  88.6504 1.29948 

 936.71 The price of one share of Microsoft stack in 2027 will be approximately $936.71. c) Answers will vary.

c) Answers will vary.

Exercise Set R.7 93 19. a) First, we enter the data into the statistic editor on the calculator, letting L1 be the values for x and L2 be the values for y.

Using the exponential regression feature we get:

The exponential function that fits the data is x y  1.467  0.983 b) Substituting 60 in for x we get: 60 y  1.467  0.983 Using the exponential regression feature we get:

 0.524 The population of Detroit in 2030 will be approximately 0.524 million people. 21. a) First, we enter the data into the statistic editor on the calculator, letting L1 be the values for x and L2 be the values for y.

The exponential function that fits the data is x y  11.5043 1.0195  b) Substituting 60 in for x we get: 60 y  11.5043 1.0195 

 36.652 The population of Texas in 2030 will be approximately 36.652 million people.

Using the exponential regression feature we get:

20. a) First, we enter the data into the statistic editor on the calculator, letting L1 be the values for x and L2 be the values for y.

The exponential function that fits the data is y  2137.54 1.8723 

Chapter R: Functions, Graphs, and Models Substituting 10 in for x we get: 10 y  2137.54 1.8723  1,131,536.65 The global weekly average of daily new cases at week 10 will be approximately 1,131,536 cases.

22. a) First, we enter the data into the statistic editor on the calculator, letting L1 be the values for x and L2 be the values for y.

b) First, we enter the data into the statistic editor on the calculator, letting L1 be the values for x and L2 be the values for y.

Using the exponential regression feature we get:

Using the linear regression feature we get:

The exponential function that fits the data is y  2.15372  0.86372 

The linear function that fits the data is y  14, 470.9 x  9897.17 Substituting 10 in for x we get: y  14, 470.9(10)  9897.17

 134,811.83 The global weekly average of daily new cases at week 10 will be approximately 134,812 cases.

March 2023 corresponds to month 15 (since March is the third month, and 2023 is one year after 2022). Substituting 15 in for x we get: 15 y  2.15372  0.86372 

 0.239 The global weekly average number of new confirmed cases daily for March 2023 will be approximately 0.239 million cases.

c) Answers may vary.

Exercise Set R.7 95 b) First, we enter the data into the statistic editor on the calculator, letting L1 be the values for x and L2 be the values for y.

Using the quadratic regression feature we get:

The quadratic function that fits the data is y  0.017238 x 2  0.382783 x  2.58311 March 2023 corresponds to month 15 (since March is the third month, and 2023 is one year after 2022). Substituting 15 in for x we get: y  0.017238(15) 2  0.382783(15)  2.58311

 0.720 The global weekly average number of new confirmed cases daily for March 2023 will be approximately 0.720 million cases.

24.

Answers will vary. The domain must be restricted to the nonnegative numbers since negative speed has no meaning. A car traveling at 0 miles per hour would have a breaking distance of 0 feet, so 0 could be included in the domain. Realistically, there will also be an upper limit on the domain. The upper limit would be based upon the maximum speed of the automobile.

25. a) The slope between the first and second point is as follows: 3 1 2 m  2 1 0 1 The slope between the first and third points is as follows: 4 1 3 m   1.5 20 2 The slope between the second and third points is as follows: 43 1  1 m 2 1 1 The average of the three slopes is: 2  1.5  1 4.5   1.5 3 3 b) Using the linear regression with these three points will yield the equation y  1.5 x  1.16666 . c) Both slopes are the same.

26. Answers will vary. During the late spring and early summer, the days grow longer and thus the number of hours of daylight increases. After the summer solstice, the hours of daylight begin to decrease. Since the data values increase then decrease, a quadratic function with a  0 would be appropriate to model the number of hours of daylight for the dates April 22 to August 22.

c) Answers may vary.

23. Answers will vary. With the small amount of data, it is difficult to determine what trend the data is taking on. Since linear functions are easier to find than other polynomial functions, we might want to trade off accuracy for simplicity.

27. a) First, we enter the data into the statistic editor on the calculator, letting L1 be the values for x and L2 be the values for y.

Chapter R: Functions, Graphs, and Models Using the quadratic regression feature we get:

Next, we use the quartic regression feature on the calculator to fit the data to the model.

The quadratic function that fits the data is y  0.00648 x 2  0.884 x  33.217 b) Substituting 50 in for x we get: y  0.00648(50) 2  0.884(50)  33.217  5.217  5.2 The number of moving violations per million miles driven for drivers aged 50 will be approximately 5.2. c) The exponential function that fits the data is x y  24.068  0.975  . Substituting 50 in for x we get: y  24.068(0.975)50  6.7869  6.8 The number of moving violations per million miles driven for drivers aged 50 will be approximately 6.8. d)

Answers will vary.

28. a) First, we enter the data into the statistic editor on the calculator, letting L1 be the values for x and L2 be the values for y.

The quartic function that fits the data is: y  0.492045 x 4  10.5684 x 3  62.6549 x 2  109.461x  341.277 b) To predict the trade deficit in 2028, we substitute 12 in for x and compute y. y  0.492045(12) 4  10.5684(12)3  62.6549(12) 2  109.461(12)  341.277  691.65

The trade deficit in 2028 will be approximately 691.65 billion dollars. c) Answers will vary.

29. a) 49a  7b  c  626 25a  5b  c  1121 24a  2b  495 (eq 4) b) 49a  7b  c  626 81a  9b  c  967 32a  2b  341 (eq 5) c) 24a  2b  495 32a  2b  341 8a  836 a  104.5 d) 32(104.5)  2b  341 3344  2b  341 2b  3003 b  1501.5 e) 25(104.5)  5(1501.5)  c  1121 2612.5  7507.5  c  1121 4895  c  1121 c  6016

Chapter R Test 97 6.

The average rate of change will be the slope of the line. Use the points (0, 30) and (3, 9) and the slope formula: y  y1 9  30 21 m 2    7 x2  x1 30 3 Thus the average rate of change is –7 hundred dollars per year or –$700/year.

The average rate of change will be the slope of the line. Use the points (0, 0) and (6, 3) and the slope formula: y  y1 3  0 3 1 m 2    x2  x1 6  0 6 2 Thus the average rate of change is 0.5lbs/bag.

Using the direct proportion equation; R  kN

Chapter R Test 1.

Use the compound interest formula with i = 0.65, n = 1, and A = 798.75

i  A  P 1   n 

11 

0.065   798.75  P  1   1   798.75  P 1  0.065 

798.75  P 1.065  750  P Cecelia should deposit $750 in order to have $798.75 at the end of 1 year.

1500  k (30) 1500 k 30 50  k Thus the equation will be R  50 N .

f ( x)   x 2  5 a) f (3)  ( 3) 2  5  9  5  4 9.

b) f ( x  h)  ( x  h) 2  5  ( x 2  2 xh  h 2 )  5   x 2  2 xh  h 2  5 4 2 x  , it is already in 5 3 4 the slope–intercept form, thus m  and the 5 2   y–intercept is  0,   . 3 

Given the equation y 

We are given the slope and a point, so use the point–slope form to determine the equation of the line: y  y1  m  x  x1  1 y  7   x  3 4 1 3 y7  x 4 4 1 31 y  x 4 4

a) The cost of producing the cards will be: C ( x )  0.08 x  8000 . b) The revenue from selling the cards will be R ( x)  0.50 x . c) The profit from selling x cards will be: P ( x )  R ( x)  C ( x)  (0.50 x)  (0.08 x  8000)  0.42 x  8000 d) At breakeven, the profit is 0, therefore 0  0.42 x  8000 8000  0.42 x 19, 047.62  x Thus, approximately 19,048 cards must be produced and sold to break even.

10. The equilibrium point is where supply and demand are the same.

 q  8  2  q 2  q  13 q 2  16q  64  q 2  q  13 51  17 q 3q

We are given two points, so first find the slope using the slope formula: y  y1 4  2 6 1 m 2    x2  x1 3  (9) 12 2

Chapter R: Functions, Graphs, and Models

Now substitute the value of q into either function to find the equilibrium price. p  3  8

 (5) 2  25 quantity = 3000 and price = 25 Thus the equilibrium point is (3, 25).

11. Yes, the graph is a function. Based on the vertical line test, any vertical line drawn will only cross the graph in exactly one point.

17. Using the logarithm to exponential theorem: 1 log 9  x 3 1 9x  3 2x 3  31 2 x  1 1 x 2 18.

12. No, the graph is not a function. Based on the vertical line test, a vertical line drawn at x  0 will cross the graph in more than one point. 13. a) Based on the graph, the function value when x  1 is –4. b) The domain of the function is all real numbers or  ,   in interval notation. c) The function has a value of 4 when x  3 and when x  3 . d) The range of the function is all real numbers greater than or equal to –5 or  5,   in

interval notation. 14.

19. The graph in interval notation is  c, d  . 20. The function is defined where the denominator is not equal to 0 and when the radical expression is greater than 0 or 3x  6  0 3 x  6 x  2 Therefore, the domain of the function, in interval notation is  2,   . 21. The function is not defined when the denominator is equal to 0 or x 2  5 x  14  0

 x  7  x  2  0 1

15.

16.

 35



t

t 2



1 t



x  7, 2 Therefore, the domain of the function, in interval notation is  , 7    7, 2    2,   .

 12

22.

5 3

Chapter R Test 99

The zero of the function is

23.

5  Therefore, the domain is  ,  and the zero 3  2   is  798 , 0  . 3  

1.5  1.041

ln1.5  t ln1.041

d) y  1.94(30) 2  102.74(30)  1253.49  1.94(900)  102.74(30)  1253.49  2589.69 A 30-year-old physically active female will need approximately 2590 calories. e) Answer will vary.

7

2396 x 3 2 798  x 3

b) Yes, based on the scatter plot from part a. c) Using the quadreg function on the calculator, the function that fits the data is y  1.94 x 2  102.74 x  1253.49 .

7   5  3x 

0.25



24. a) The annual percent increase is 0.041 or 4.1%. b) 75, 000  50, 000(1.041) t

25. a) The scatter plot will be:

0.25

7 4  5  3x 2401  5  3 x

 1 The y–intercept, based on the graph, is  0,  .  2

ln1.5 t ln1.041 10.09  t It will take approximately 10.09 years for the investment to reach $75,000.

0   5  3x 

28. Answers vary, but one possible solution is: y   x  3   x  1 x  4  . 29. The average rate of change is similar to finding the slope. Therefore, we have a slope of 3  and points of (1, 9) and (5, y). 7 3 y  9  7 5 1 7( y  9)  3(4) 7 y  63  12 7 y  51 51 y 7 30.

26. Use the power rule and then simplify the result.  12 4 1 1 1 2 64 3  64 3  2   3 2 16 64 3 64

Based on the graph above; Zeros:  8,  10

27. The domain of the function is where 5  3x  0

Range:  1,  

 

Domain:  ,  

5  3x 5 x 3

100 Chapter R: Functions, Graphs, and Models 31. a) Using the data from exercise 25 and the QUADREG function on the calculator, the equation of the functions will be y  1.51x 2  79.98 x  1436.93 . b) y  1.51(30) 2  79.98(30)  1436.93  1.51(900)  79.98(30)  1436.93  2477.33 A 30–year old physically active woman will need approximately 2480 calories. c) Answers will vary.

Chapter 1 Differentiation 13. The notation lim is read “the limit, as x x 2

Exercise Set 1.1 1.

approaches 2 from the right.”

The limit of the sequence is 2.8 and using the notation will give the expression x  2.8.

x 3

approaches 3 from the left”.

The limit of the sequence is 0.3 and using the notation will give the expression x  0.3.

15. The notation lim is read “the limit, as x

The limit of the sequence is 3 and using the notation will give the expression x  3.

16. The notation lim1 is read “the limit, as x

The limit of the sequence is 4.9 and using the notation will give the expression x  4.9 . The limit of the sequence is

2 and using the 3

notation will give the expression x 

14. The notation lim is read “the limit, as x

The limit of the sequence is

2 . 3

4 and using the 3

notation will give the expression x 



4 . 3

The limit of the sequence is 5.4 and using the notation will give the expression x  5.4.

The limit of the sequence is 0.3 and using the notation will give the expression x  0.3.

The limit of the sequence is 1 and using the notation will give the expression x  1.

10. The limit of the sequence is 0 and using the notation will give the expression x  0. 11. As x approaches 0, the value of x  5 approaches 5. 12. We solve the equation: 2 x  6 x  3 Therefore, As x approaches 3, the value of x approaches 6.

x 5

approaches 5”. x 2

approaches

1 ”. 2

17. The notation lim f  x  is read “the limit, as x x4

approaches 4, of f  x  . ” 18. The notation lim g  x  is read “the limit, as x x 1

approaches 1, of g  x  . ” 19. The notation lim F  x  is read “the limit, as x x 5

approaches 5 from the left, of F  x  . ” 20. The notation lim G  x  is read “the limit, as x x4

approaches 4 from the right, of G  x  . ” 21. a) As inputs x approach 1 from the left, outputs f  x  approach 3. Thus the limit from the left is 3. That is, lim  f  x   3. x 1

b) As inputs x approach 1 from the right, outputs f  x  approach 3. That is, lim f  x   3.

x 1

c) From parts (a) and (b) we find lim f  x   3. x 1

102 Chapter 1: Differentiation 22. a) As inputs x approach 3 from the left, outputs f  x  approach 1. That is, lim f  x   1.

26. We have lim F  x   4 and lim F  x   4. x2

Therefore, lim F  x   4.

x 3

x2

b) As inputs x approach 3 from the right, outputs f  x  approach 2. Thus the limit from the right is 2. That is, lim f  x   2.

27. As inputs x approach 2 from the left, outputs F  x  approach 4. Thus the limit from the left is 4. That is, lim  F  x   4.

x 3

x 2

c) From parts (a) and (b) we know that lim f  x   1 and lim f  x   2. Since the x 3

As inputs x approach 2 from the right, outputs F  x  approach 2. Thus the limit from the right

x 3

limit from the left, 1, is not the same as the limit from the right, 2, lim f  x  does not

is 2. That is, lim  F  x   2. x 2

x 3

Since the limit from the left, 4, is not the same as the limit from the right, 2, we have lim F  x  does not exist.

exist. 23. a) As inputs x approach 4 from the right, outputs g  x  approach 1 . Thus the limit from the right is 1 . That is, lim g  x   1.

x 2

28. We have lim  F  x   0 and lim  F  x   0. x 5

Therefore, lim F  x   0.

x4

lim g  x   1 .

x 5

29. As inputs x approach 4 from the left, outputs F  x  approach 2. Thus the limit from the left is

x  4

2. That is, lim F  x   2.

c) Since the limit from the left, 1 , is the same as the limit from the right, 1 , we have lim g  x   1 .

x4

As inputs x approach 4 from the right, outputs F  x  approach 2. Thus the limit from the right

x4

is 2. That is, lim F  x   2.

24. a) As inputs x approach 2 from the left, outputs g  x  approach 4. Thus the limit

x4

Since the limit from the left, 2, is the same as the limit from the right, 2, we have lim F  x   2.

from the left is 4. That is, lim  g  x   4. x 2

b) c)

lim g  x   2.

x4

30. We have lim F  x   0 and lim F  x   0.

x 2

x 6

Therefore, lim F  x   0.

lim g  x  does not exist.

x 6

x 2

25. As inputs x approach 3 from the left, outputs F  x  approach 5. Thus the limit from the left is

31. As inputs x approach 2 from the right, outputs F  x  approach 2. Thus the limit from the right is 2. That is, lim  F  x   2.

5. That is, lim  F  x   5.

x 2

x 3

As inputs x approach 3 from the right, outputs F  x  approach 5. Thus the limit from the right

32. As inputs x approach 2 from the left, outputs F  x  approach 4. Thus the limit from the left is 4. That is, lim  F  x   4.

is 5. That is, lim  F ( x)  5.

x 2

x 3

Since the limit from the left, 5, is the same as the limit from the right, 5, we have lim F  x   5 . x 3

x 5

33.

lim F ( x)

x 

As inputs x approaches negative infinity, F(x) decreases without bound. lim F ( x)   x 

Exercise Set 1.1 103 34.

lim F  x 

42. We have lim G  x   0 and lim G  x   0 .

x 

x 3

As inputs x approaches positive infinity, F  x  decreases without bound. lim F  x   

Therefore, lim G  x   0 . x 3

43.

x 

x 3

lim G ( x)

x 

As inputs x approaches negative infinity, G  x 

35. As inputs x approach 2 from the left, outputs G (x ) approach 1. Thus the limit from the left is

decreases without bound. lim G  x   

1. That is, lim  G  x   1 .

x 

x 2

As inputs x approach 2 from the right, outputs G (x ) approach 1. Thus the limit from the right

44.

As inputs x approaches positive infinity, G  x 

is 1. That is, lim  G  x   1 . x 2

Since the limit from the left, 1, is the same as the limit from the right, 1, we have lim G  x   1 . x 2

36. We have lim G  x   3 and lim G  x   3 . x 0

Therefore, lim G  x   3 .

increases without bound. lim G  x    x 

45. As inputs x approach 3 from the left, outputs H  x  approach 0. Thus the limit from the left is 0. That is, lim  H  x   0 .

x 0

x 3

As inputs x approach 3 from the right, outputs H  x  approach 0. Thus the limit from the right

x 0

37. As inputs x approach 1 from the left, outputs G  x  approach 4. Thus the limit from the left is

is 0. That is, lim  H  x   0 . x 3

4. That is, lim G  x   4 .

Since the limit from the left, 0, is the same as the limit from the right, 0, we have lim H  x   0 .

x 1

38. 39.

lim G  x 

x 

lim G  x   1

x 3

x 1

46.

lim G  x   0

lim H  x   1 .

x 2

x 3

40. As inputs x approach 1 from the left, outputs G  x  approach 4. Thus the limit from the left is

47. As inputs x approach 2 from the right, outputs H  x  approach 1. Thus the limit from the right is 1. That is, lim  H  x   1 .

4. That is, lim G  x   4 . x 1

x 2

As inputs x approach 1 from the right, outputs G  x  approach 1 . Thus the limit from the right is 1 . That is, lim G  x   1 .

48. We have lim  H  x   1 and lim  H  x   1 . x 2

Therefore, lim H  x   1 .

x 1

Since the limit from the left, 4, is not the same as the limit from the right, 1 , we have lim G  x  does not exist. x 1

x 2

49. As inputs x approach 1 from the left, outputs H  x  approach 4. Thus the limit from the left is 4. That is, lim H  x   4 .

41. As inputs x approach 3 from the right, outputs G  x  approach 0. Thus the limit from the right is 0. That is, lim G  x   0 . x 3

x 2

x 1

50.

lim H  x   2 .

x 1

104 Chapter 1: Differentiation 51.

lim H  x   1 .

x 3

52. As inputs x approach 1 from the left, outputs H  x  approach 4. Thus the limit from the left is 4. That is, lim H  x   4 . The solution is continued on the next page. As inputs x approach 1 from the right, outputs H  x  approach 2 . Thus the limit from the right is 2 . That is, lim H  x   2 . x 1

Since the limit from the left, 4, is not the same as the limit from the right, 2 , we have lim H  x  does not exist. x 1

53. As inputs x approach 3 from the right, outputs H  x  approach 1. Thus the limit from the right is 1. That is, lim H  x   1 .

x 3

54. We have lim H  x   1 and lim H  x   1 . Therefore, lim H  x   1 .

x 3

55.

59. We have lim f  x   2 and lim f  x   2 . Therefore, lim f  x   2 .

x 0

60. As inputs x approach 3 from the left, outputs f  x  increase without bound. We say that the limit from the left is infinity. That is, lim  f  x    . x 3

As inputs x approach 3 from the right, outputs f  x  decrease without bound. We say that limit from the right is negative infinity. That is, lim  f  x   . x 3

Since the function values as x  3 from the left increase without bound, and the function values as x  3 from the right decrease without bound, the limit does not exist. We have, lim f  x  does not exist. x 3

61. We have lim f  x    and lim f  x   .

lim H  x 

x 1

x 

As inputs x approaches negative infinity, H  x  decreases without bound. lim H  x   

x 1

Therefore, lim f  x  does not exist. x 1

62. As inputs x approach 3 from the left, outputs f  x  approach 0. Thus the limit from the left is

x 

56.

x 1

x 0

x 1

x 3

Since the limit from the left, 1, is the same as the limit from the right, 1, we have lim f  x   1 .

0. That is, lim f  x   0 .

lim H ( x)

x 

As inputs x approaches positive infinity, H  x  inecreases without bound. lim H ( x)  

Thus the limit from the right is 0. That is, lim f  x   0 .

x 

57. We have lim f  x   1 and lim f  x   1 . x2

Therefore, lim f  x   1 .

x 3

As inputs x approach 3 from the right, outputs f  x  approach 0.

x2

x 3

Since the limit from the left, 0, is the same as the limit from the right, 0, we have lim f  x   0 . x 3

58. As inputs x approach 1 from the left, outputs f  x  approach 1. Thus the limit from the left is 1. That is, lim  f  x   1 .

63. We have lim  f  x   0 and lim  f  x   0 . x 2

Therefore, lim f  x   0 . x 2

x 1

As inputs x approach 1 from the right, outputs f  x  approach 1. Thus the limit from the right is 1. That is, lim  f  x   1 . x 1

x 2

Exercise Set 1.1 105 64. As inputs x approach 4 from the left, outputs f  x  approach 3. Thus the limit from the left is

Since the limit from the left is the same as the limit from the right, we have lim f  x   0 .

3. That is, lim  f  x   3 .

x 0

Find lim f  x  . x 2

x 4

As inputs x approach 2 from the left, outputs f  x  approach 2. We have, lim  f  x   2 .

As inputs x approach 4 from the right, outputs f  x  approach 3. Thus the limit from the right

x 2

As inputs x approach 2 from the right, outputs f  x  approach 2. We have, lim  f  x   2

is 3. That is, lim  f  x   3 .

x 2

x 4

Since the limit from the left is the same as the limit from the right, we have lim f  x   2 .

Since the limit from the left, 3, is the same as the limit from the right, 3, we have lim f  x   3 .

x 2

x 4

68. 65. As inputs x get more and more negative, output f  x  get closer and closer to 2. lim f  x   2 .

f  x   x2

x 

66. As inputs x get larger and larger, outputs f  x  get closer and closer to 1. We have lim f  x   1 . x 

67. Defining f  x   x as a piecewise defined

Find lim f  x  .

function we have:   x, x  0 f  x   .  x, x  0

x 1

We have lim  f  x   1 and lim  f  x   1 . x 1

x 1

Therefore, lim f  x   1 . x 1

We graph the function by creating an inputoutput table.

Find lim f  x  .

x f  x

Therefore, lim f  x   0 .

−2 2

−1 1

0 0

1 1

x0

We have lim f  x   0 and lim f  x   0 . x 0

2 2

Next, we plot the points from the table and draw the graph.

x 0

69.

g ( x)  x 2  2 Creating an input-output table we have, x

g ( x)

2 6

1 3

0 2

1 3

2 6

Next we plot the points and draw the graph. Find lim f  x  . x0

As inputs x approach 0 from the left, outputs f  x  approach 0. We have, lim f  x   0 . x 0

As inputs x approach 0 from the right, outputs f  x  approach 0. We have, lim f  x   0 x 0

106 Chapter 1: Differentiation As inputs x approach െ3 from the right, outputs g  x  approach 0. We have, lim  G  x   0. x 3

Since the limit from the left is the same as the limit from the right, we have, lim G  x   0. x 3

Find lim g  x  . x 0

As inputs x approach 0 from the left, outputs g  x  approach −3. We have, lim  g  x   3. x 0

As inputs x approach 0 from the right, outputs g  x  approach −3. We have, lim g  x   3. x 0

Find lim g  x  . x 0

As inputs x approach 0 from the left, outputs g  x  approach 2. We have, lim g  x   2. x 0

As inputs x approach 0 from the right, outputs g ( x) approach 2. We have, lim g  x   2. x 0

Since the limit from the left is the same as the limit from the right, we have, lim g  x   2. Find lim g  x  .

x 0

Since the limit from the left is the same as the limit from the right, we have, lim g  x   3. x 0

3 71. F  x   x2 Since x  2 makes the denominator zero, we exclude the value 2 from the domain. Creating an input-output table we have, x

F (x )

x 1

As inputs x approach −1 from the left, outputs g  x  approach 3. We have, lim  g  x   3.

1 –3

1.5 –6

2.5 6

3 3

3.5 2

4 1.5

Next we plot the points and draw the graph.

x 1

As inputs x approach െ1 from the right, outputs g  x  approach 3. We have, lim  g  x   3. x 1

Since the limit from the left is the same as the limit from the right, we have, lim g  x   3. x 1

70.

g  x   | x | 3 Next we plot the points and draw the graph.

Find lim F  x  . x 3

As inputs x approach 3 from the left, outputs F  x  approach 3. We have, lim F  x   3. x  3

As inputs x approach 3 from the right, outputs F  x  approach 3. We have, lim F  x   3. x  3

Find lim g  x  . x 3

As inputs x approach −3 from the left, outputs g ( x) approach 0. We have, lim  G  x   0. x 3

Since the limit from the left is the same as the limit from the right, we have, lim F  x   3. x 3

Find lim F  x  . x4

Exercise Set 1.1 107 As inputs x approach 4 from the left, outputs F  x  approach 1.5. We have,

As inputs x approach −2 from the right, outputs G  x  increase without bound. We have,

lim F  x   1.5.

lim G  x   .

x  4

x   2

As inputs x approach 4 from the right, outputs F  x  approach 1.5. We have,

Since the function values as x   2 from the left decrease without bound, and the function values a x   2 from the right increase without bound, the limit does not exist. We have, lim G  x  does not exist.

lim F  x   1.5.

x  4

Since the limit from the left is the same as the limit from the right, we have, lim F  x   1.5.

x  2

x4

72.

5 x2 Since x  2 makes the denominator zero, we exclude the value 2 from the domain. Creating an input-output table we have, G  x 

–2

–1

G ( x)

∞

2.5

1.6

1.25

Next we plot the points and draw the graph.

73.

1  2x x Since x  0 makes the denominator zero, we exclude the value 0 from the domain. Creating an input-output table we have f  x 

x f  x

−1 3

−0.5

−0.1

4

12

0.1 8

0.5 0

1 1

Next we plot the points and draw the graph.

Find lim f  x  . x 

As inputs x get larger and larger, outputs f  x  get closer and closer to 2 . We have

Find lim G  x  .

lim f  x   2 .

x  1

x 

As inputs x approach െ1 from the left, outputs G  x  approach 5. We have, lim G  x   5.

Find lim f  x  .

As inputs x approach −1 from the right, outputs G  x  approach 5. We have, lim G  x   5.

As inputs x approach 0 from the left, outputs f  x  decrease without bound. We have,

Since the limit from the left is the same as the limit from the right, we have lim G  x   5.

x  0

x  1

x  1

x  1

Find lim G  x  . x  2

As inputs x approach −2 from the left, outputs G  x  decrease without bound. We have, lim G  x   .

x   2

x0

lim f  x    .

As inputs x approach 0 from the right, outputs f  x  increase without bound. We have, lim f  x   

x  0

Since the function values as x  0 from the left decrease without bound, and the function values as x  0 from the right increase without bound, the limit does not exist. We have, lim f  x  does not exist. x 0

108 Chapter 1: Differentiation 74.

f  x 

Find lim g  x  .

1  3x x

x 3

As inputs x approach 3 from the left, outputs g  x  decrease without bound. We have, lim g  x    .

x 3

As inputs x approach 3 from the right, outputs g  x  increase without bound. We have, lim g  x    .

x 3

Since the function values as x  3 from the left decrease without bound, and the function values as x  3 from the right increase without bound, the limit does not exist. We have, lim g  x  does not exist.

Find lim f  x  . x 

As inputs x get larger and larger, outputs f  x  get closer and closer to 3 . We have lim f  x   3 .

x 3

x 

Find lim f  x  .

76.

x0

As inputs x approach 0 from the left, outputs f  x  decrease without bound. We have, lim f  x    .

x  0

As inputs x approach 0 from the right, outputs f  x  increase without bound. We have, lim f  x   

x  0

4x  9 x2 Since x  2 makes the denominator zero, we exclude the value 2 from the domain. Creating an input-output table we have g  x 

x g ( x)

−3 3

−2.5

−2.1 −1.9 6 14

−1.5 −1 6 5

Next we plot the points and draw the graph.

x 0

75.

g  x 

2x  5 x 3

Find lim g  x  . x

As inputs x get larger and larger, outputs g  x  get closer and closer to 2 . We have

As inputs x get larger and larger, outputs g  x  get closer and closer to 4 . We have lim g  x   4 .

x 

lim g  x   2 .

x 

0 9 2

Exercise Set 1.1 109 Find lim g  x  .

Find lim F  x  .

As inputs x approach 2 from the left, outputs g  x  decrease without bound. We have,

As inputs x approach 1 from the right, outputs F  x  approach 1. That is,

x 2

x 1

lim g  x    .

lim F  x   1 .

x 2

x 1

As inputs x approach 2 from the right, outputs g  x  increase without bound. We have,

Find lim F  x  x 1

lim  g  x    .

Since the limit from the left, 3, is not the same as the limit from the right, 1, we have lim F  x  does not exist.

x 2

Since the function values as x  2 from the left decrease without bound, and the function values as x  2 from the right increase without bound, the limit does not exist. We have, lim g  x  does not exist. x 2

77.

x 1

78.

 x  3, G  x    x  1,

for x  2 for x  2.

2 x  1, for x  1 F  x   for x  1.  x, We create an input-output table for each piece of the function. For x  1 x

F  x

1 1

0.9

2.8

We have lim G  x   1 and lim G  x   3 . x2

Therefore, lim G  x  does not exist. x2

We plot the points and draw the graph. Notice we draw an open circle at the point 1, 3  to indicate that the point is not part of the graph. For x  1 x

F  x

We plot the points and draw the graph. Notice we draw a solid circle at the point 1,1 to indicate that the point is part of the graph.

x2

79.

 x  4, for x  3 g  x    x  3, for x  3. We create an input-output table for each piece of the function. For x  3 x 0

g  x

1 3

2 2

2.9 1.1

We plot the points from the table draw the graph. Notice we draw an open circle at the point  3,1 to indicate that the point is not part of the graph. For x  3 x 3.1

g  x

Find lim F  x  . x 1

As inputs x approach 1 from the left, outputs F  x  approach 3 . That is, lim F  x   3 .

0.1

4 1

We plot the points and draw the graph. Notice we draw an open circle at the point  3, 0  to indicate that the point is not part of the graph.

x 1

110 Chapter 1: Differentiation

We have lim  F  x   1 and lim  F  x   1. x 1

Therefore, lim F  x   1 .

x 1

82. Find lim g  x  . x 3

As inputs x approach 3 from the left, outputs g  x  approach 1. That is, lim g  x   1 .

x 3

Find lim g  x  . x 3

As inputs x approach 3 from the right, outputs g  x  approach 0. That is, lim g  x   0 .

x  3

x 3

Since the limit from the left, 1, is not the same as the limit from the right, 0, we have lim g  x  does not exist. x 3

3x  4, f  x    x  2,

For x  1 x G  x

3 9

2 4

1.1 1.21

We plot the points and draw the graph. Notice we draw an open circle at the point  1,1 to indicate that the point is not part of the graph. For x  1 x

Find lim g  x 

80.

for x  1  x 2 , G  x    x  2, for x  1. We create an input-output table for each piece of the function.

G  x

0.9

1.1

1 3

We plot the points and draw the graph. Notice we draw an open circle at the point  1,1 to indicate that the point is not part of the graph.

for x  1 for x  1.

Find lim G  x  . x 1

We have lim f  x   1 and lim f  x   1 . x 1

Therefore, lim f  x   1 .

x 1

81.

2 x  3, F  x   3  x ,

As inputs x approach 1 from the left, outputs G  x  approach 1 . We have, lim G  x   1 .

x 1

for x  1 for x  1.

As inputs x approach 1 from the right, outputs G  x  approach 1 . We have, lim G  x   1

x 1

Since the limit from the left is the same as the limit from the right, we have lim G  x   1 . x 1

Exercise Set 1.1 111

Find lim H  x 

 x  1, for x  0  83. H  x    2, for 0  x  1 3  x, for x  1.  We create an input-output table for each piece of the function. For x  0 x 0.5 0.1 1 H  x 0 0.5 0.9

x 1

As inputs x approach 1 from the left, outputs H  x  approach 2 . That is, lim H  x   2 .

x 1

As inputs x approach 1 from the right, outputs H  x  approach 2. That is, lim H  x   2 .

x 1

Since the limit from the left, 2, is the same as the limit from the right, 2, we have lim H  x   2 .

We plot the points and draw the graph. Notice we draw an open circle at the point  0,1 to indicate that the point is not part of the graph. For 0  x  1 , the function has value of 2. We draw a solid circle at the point  0, 2  to indicate the point is part of the graph and we draw an open circle at 1, 2  to indicate that the point is

x 0

84.

 2  x,  G  x    x2 ,  9, 

for x  1 for  1  x  3 for x  3.

not part of the graph. For x  1 x

H  x

1 2

2 1

1

We plot the points and draw the graph. Notice we draw a solid circle at the point 1, 2  to

We have lim  G  x   1 and lim  G  x   1 .

indicate that the point is part of the graph.

x 1

Therefore, lim G  x   1 .

x 1

We have lim G  x   9 and lim G  x   9 . x 3

Therefore, lim G  x   9 . x 3

85.

Find lim H  x 

lim C  x   $4.40.

x0

x  0.2

As inputs x approach 0 from the left, outputs H  x  approach 1 . That is,

lim C  x  does not exist.

x  0.2

lim H  x   1 .

x  0

As inputs x approach 0 from the right, outputs H  x  approach 2. That is,

lim C  x   $3.70.

x  0.2

86.

lim C  x   $4.40.

x  0.25

lim C  x   $4.40.

lim H  x   2 .

x  0.25

Since the limit from the left, 1, is not the same as the limit from the right, 2, we have lim H  x  does not exist.

x  0.25

lim C  x   $4.40.

x 0

x0

87.

lim C  x   $5.10.

x  0.6

lim C  x   $5.80.

x  0.6

lim C  x  does not exist.

x  0.6

x 3

112 Chapter 1: Differentiation 88.

lim p  x   $1.50.

x  2

97.

lim p  x   $1.74.

lim p  x  does not exist.

x2

lim p  x   $1.26.

x 1

lim p  x   $1.50.

x 1

lim p  x  does not exist. x 1

90.

lim p  x   $1.74.

x  2.6

lim  p  x   $1.74.

x  2.6

lim

x  2.6

91. 92. 93.

p  x   $1.74.

lim p  x   $1.98.

x  3.4

lim p  x  does not exist.

x 3

lim r  x   10%

x  9000

lim

r  x   10%

lim

r  x   10%

lim

r  x   12%

lim

r  x   22%

x  9000 x  9000

94.

x  44,725 x  44,725

lim

x  44,725

95.

lim r  x   12% lim

r  x   does not exist.

lim

r  x   10%

lim

r  x   12%

lim

r  x  does not exist.

x 15,700 x 15,700 x 15,700

lim

r  x   22%

lim

r  x   22%

lim

r  x  does not exist.

lim

r  x   10%

x  65,000

98.

x  59,850 x 12,000

99. As inputs x approach 2 from the right, outputs f  x  approach 4. We have, lim f  x   4 . In order for lim f  x  to exist

x  2

x2

we need lim f  x   4 . We will use the letter c x2

for the unknown in the equation; therefore, 1 lim  x   c  4 . x 2 2 Substitute 2 in for x to get the equation: 1  2   c  4 and solving for c we get 2 1 c  4 c  3. Therefore, in order for the limit to exist as x approaches 2, the function must be:  1 x  3 for x  2 f  x   2 for x  2.   x  6 100. As inputs x approach 2 from the left, outputs f  x  approach 0. We have, lim f  x   0 . In order for lim f  x  to exist

x  2

x2

we need lim f  x   0 . We will use the letter c x2

x 15,000

x  95,375

96.

r  x  does not exist.

r  x   22%

x  65,000

x  2

89.

lim

x  65,000

for the unknown in the equation and this gives us 3 lim  x   c  0 x  2 2 Substitute 2 in for x to get the equation: 3  2   c  0 and solving for c we get 2 3 c  0 c  3. Therefore, in order for the limit to exist as x approaches 2, the function must be:   12 x  1 for x  2 f  x   3 for x  2.  2 x   3

Exercise Set 1.1 113 101. As inputs x approach 2 from the left, outputs f  x  approach 5 . We have, lim f  x   5 . In order for lim f  x  to exist x2

x2

we need lim f  x   5 . We will use the letter x2

c for the unknown in the equation and this gives





a) b) c) d)

lim f  x   4

x 2

lim f  x   4

x 2

lim f  x   4

x 2

lim f  x   4

x  2

lim f  x   4

us lim  x 2  c  5

Substitute 2 in for x to get the equation:

  2   c  5 and solving for c we get

g) lim f  x   4  f  2  ; yes

x 2 2

x  2

lim f  x   4  3  f  2  ; no

x 2 x2

  4   c  5 c  1. Therefore, in order for the limit to exist as x approaches 2, the function must be:  x 2  9 for x  2 f  x   2 for x  2.   x   1

 x 2  2, for x  0 104. Graph f  x    2 for x  0. 2  x , Using the calculator we enter the function into the graphing editor as follows:

102. a) The limit of the sequence is 0.

b) The limit of the sequence is 2. c) Answers will vary.  3, 103. Graph f  x    2  x ,

When you select the table feature you get: for x  2 for x  2.

Using the calculator we enter the function into the graphing editor as follows: The calculator graphs the function

Notice, when you select the table feature you get: Using the trace feature, we find the limits. Find lim f  x  . x 0

As inputs x approach 0 from the left, outputs f  x  approach 2. We have, The calculator graphs the function:

lim f  x   2.

x  0

As inputs x approach 0 from the right, outputs f  x  approach 2. We have, lim f  x   2

x  0

Using the trace feature, we find the limits.

114 Chapter 1: Differentiation

Since the limit from the left, 2 , is not the same as the limit from the right, 2, we have lim f  x  does not exist.

The calculator graphs the function:

x0

Find lim f  x  . x 2

As inputs x approach 2 from the left, outputs f  x  approach 2. We have, lim  f  x   2.

x 2

As inputs x approach 2 from the right, outputs f  x  approach 2. We have, lim f  x   2

x 2

Since the limit from the left is the same as the limit from the right, we have lim f  x   2. x 2

105. Graph g  x  

x 1

As inputs x approach 1 from the left, outputs f  x  increase without bound. We have, lim f  x   .

x 1

As inputs x approach 1 from the right, outputs f  x  decrease without bound. We have, lim f  x   

20 x 2 3

Using the trace feature on the calculator we find the limits. Find lim f  x  .

x 1

x  2 x  5x

Since the function values as x  1 from the left increase without bound, and the function values as x  1 from the right decrease without bound, the limit does not exist. lim f  x  does not exist. x 1

Using the trace feature on the graph, we have: lim g  x   0 x 

lim g  x   0.

x  

106. Graph f  x  

x5 2

x  4x  5 Using the calculator we enter the function into the graphing editor.

Using the following window:

Find lim f  x  . x 5

As inputs x approach 5 from the left, outputs 1 f  x  decrease to the value . We have, 6 1 lim f  x   . x  5 6 As inputs x approach 5 from the right, outputs 1 f  x  increase to the value . We have, 6 1 lim f  x   . x  5 6 Since the function values as x  5 from the left and the function values as x  5 from the right 1 are the same value, the limit is the value . We 6 1 have, lim f  x   . x 5 6

Exercise Set 1.2 115

x →3

By limit principle L2, the statement is true.

By limit principle L1, lim 7 = 7 x →3

By limit principle L2, 2

2 lim  g ( x = g ( x= ) lim ) [5= ]2 25. x →1 x →1 Therefore, the statement is true.

By limit principle L6, the statement is true.

The statement is true.

A function can be discontinuous at a point even if the function value exists at that point. Therefore, the statement is false.

x →5

) ( 5) − 6 ( 5) + 9= 4 .

x→2

(

)

= 2 ( 2) − 3( 2) + 4 ( 2) −1 = 2 (16 ) − 3 ( 8 ) + 8 − 1

16.

(

lim 3 x + 4 x − 3 x + 6

x →−1

17. It follows from the Theorem on Limits of Rational Functions that we can find the limit by substitution: 2

 x 2 − 9  ( 3) − 9 lim  = x →3 x − 2  3− 2   9−9 = 3− 2 = 0.

It follows from the Theorem on Limits of Rational Functions that we can find the limit by substitution: lim ( 3 x + 2 ) = 3 (1) + 2

18.

lim ( 4 x − 5= 5 3. ) 4 ( 2 ) −= x→2

11. It follows from the Theorem on Limits of Rational Functions that we can find the limit by substitution:

(

)

lim x 2 − 4 =− ( 1) − 4 = 1− 4 = −3.

(

)

lim x 2 + 3 =( −2 ) + 3 =7.

x →−2

= 10.

x→4

By the definition of continuity, the statement is true.

)

= 3 ( −1) + 4 ( −1) − 3 ( −1) + 6

This statement is false. If lim F ( x ) exists but is

x →−1

lim 2 x 4 − 3 x3 + 4 x − 1

= 5.

12.

(

lim x 2 − 6 x + 9=

15. It follows from the Theorem on Limits of Rational Functions that we can find the limit by substitution:

x →1

10.

= 32 − 24 + 8 − 1 = 15.

not equal to F (4) , then F is not continuous. 8.

) ( 3) − 4 ( 3) + 7 =9 − 12 + 7 = 4.

14.

Therefore, the statement is a false. 3.

(

lim x 2 − 4 x + 7 =

Exercise Set 1.2

19.

lim

x →3

x 2 − 25 x2 − 5

( 3)2 − 25 9 − 25 −16 = = = −4 . ( 3)2 − 5 9 − 5 4

x 2 − 25 x →5 x − 5 First we will simplify the function by factoring the numerator and canceling common factors. ( x − 5)( x + 5) x 2 − 25 lim = lim x →5 x − 5 x →5 x −5 = lim ( x + 5 ) lim

x →5

13. It follows from the Theorem on Limits of Rational Functions that we can find the limit by substitution:

= 5+5 = 10.

116 Chapter 1: Differentiation 20. We verify the expression yields an indeterminate form by substitution: lim

x →3

22.

x − 9 ( 3) − 9 = x −3 ( 3) − 3 2

0 = . 0 This is an indeterminate form.

lim

x2 − 2 x − 8

x2 − 4 First we will simplify the function by factoring the numerator and canceling common factors. x 2 − 2 x − 8 ( x − 4 )( x + 2 ) = lim x →−2 ( x − 2 )( x + 2 ) x2 − 4 x →−2

 x−4 = lim   x →−2  x − 2  −2 − 4 = −2 − 2 −6 = −4 3 = . 2

In order to find the limit we will simplify the function by factoring the numerator and canceling common factors. Then we will apply the Theorem on Limits of Rational Functions to the simplified function. ( x − 3)( x + 3) x2 − 9 lim = lim x →3 x − 3 x →3 x −3 = lim ( x + 3)

simplifying, assuming x ≠−3

= 3+3 = 6.

substitution

x →3

21. We verify the expression yields an indeterminate form by substitution: lim

x →1

x2 + 5x − 6 x2 − 1

(1)2 + 5 (1) − 6 (1)2 − 1

0 = . 0 This is an indeterminate form. In order to find the limit we will simplify the function by factoring the numerator and denominator and canceling common factors. Then we will apply the Theorem on Limits of Rational Functions to the simplified function. ( x − 1)( x + 6 ) x2 + 5x − 6 = lim lim 2 1 → x →1 x ( x − 1)( x + 1) x −1 = lim

x →1

( x + 6) ( x + 1)

1+ 6 = 1+1 7 = . 2

simplifying, assuming x ≠1

substitution

23.

lim

2 x 2 − x − 21

9 − x2 First we will simplify the function by factoring the numerator and denominator and canceling common factors. ( 2 x − 7 )( x + 3) 2 x 2 − x − 21 lim = lim 2 x →−3 x →−3 ( 3 − x )( 3 + x ) 9− x x →−3

= lim

( 2x − 7)

3− x 2 ( −3) − 7

x →−3

3 − ( −3)

= −

13 . 6

24. We verify the expression yields an indeterminate form by substitution:

lim

x → −2

3 x 2 + x − 10 x2 − 4

3 ( − 2 ) + ( −2 ) − 10

( − 2 )2 − 4

3(4) − 2 − 10 4−4 0 = . 0 This is an indeterminate form. In order to find the limit we will simplify the function by factoring the numerator and denominator and canceling common factors. Then we will apply the Theorem on Limits of Rational Functions to the simplified function.

Exercise Set 1.2 117

lim

3 x + x − 10

x → −2

x −4

( 3x − 5 )( x + 2 ) ( x − 2 )( x + 2 ) ( 3x − 5 ) simplifying, = lim assuming x ≠−2 x → −2 ( x − 2 ) 3 ( − 2) − 5 −6−5 = lim

x →9

9− x x −3

−2−2 11 = . 4

−4

(

)

(( 2) + 2 ( 2) + 4) 2

= 12. 26. We verify the expression yields an indeterminate form by substitution: 3

x3 − 1 (1) − 1 lim = x →1 x − 1 (1) − 1

)

( x − 1) x + x + 1 x3 − 1 = lim lim x →1 x − 1 x →1 ( x − 1)

(

)

= lim x 2 + x + 1 x →1

(1)2 + (1) + 1

simplifying, assuming x ≠1

substitution

= 3. 27.

lim

( x − 3)( x + 3)   

x −3

( x + 3) = − ( 9 + 3) 28. We verify the expression yields an indeterminate form by substitution: x −5 25 − 5 lim = x → 25 x − 25 ( 25) − 25

0 = . 0 This is an indeterminate form. In order to find the limit we will simplify the function by factoring the denominator and canceling common factors. Then we will apply the Theorem on Limits of Rational Functions to the simplified function. x −5 x −5 lim = lim x → 25 x − 25 x → 25 x −5 x +5

(

0 = . 0 This is an indeterminate form. In order to find the limit we will simplify the function by factoring the numerator and canceling common factors. Then we will apply the Theorem on Limits of Rational Functions to the simplified function. 2

x −3

x →9

= −6.

( x − 2) x2 + 2 x + 4 x3 − 8 lim = lim x→2 x − 2 x→2 x−2 2  = lim x + 2 x + 4   x→2 

(

− ( x − 9)

= lim  − x →9 

substitution

25. First we will simplify the function by factoring the numerator and canceling common factors.

(

= lim

− = lim  x →9  

x → −2

lim

9− x

x→2 x − 3 First we will simplify the function by factoring the numerator and canceling common factors.

= lim

x → 25

( x + 5)

)(

25 + 5

)

simplifying, assuming x ≠ 25

substitution

1 . 10

29. We verify the expression yields an indeterminate form by substitution:

( −1)2 + 5 ( −1) + 4 lim = x →−1 x 2 + 2 x + 1 ( −1)2 + 2 ( −1) + 1 x2 + 5x + 4

118 Chapter 1: Differentiation lim

x2 + 5x + 4

x →−1 x

+ 2x +1

( x + 1)( x + 4 ) x →−1 ( x + 1)( x + 1)

= lim

32.

 x + 4  simplifying, = lim   assuming x ≠−1 x →−1  x + 1  −1 + 4 = substitution −1 + 1 3 = . 0 Substitution yields division by zero. Therefore, x2 + 5x + 4 does not exist. lim 2 x →−1 x + 2 x + 1

30.

lim

 7 + 5x  = lim   x →∞  3    7 = . 3

x + 3 x − 10

Therefore,  7x + 5  7 lim  = . x →∞  3 x  3

x→2 x 2 − 4 x + 4

First we will simplify the function by factoring the numerator and canceling common factors. ( x − 2 )( x + 5) x 2 + 3 x − 10 lim 2 = lim x→2 x − 4 x + 4 x → 2 ( x − 2 )( x − 2 )

33.

 x + 5 = lim  x→2  x − 2   7 = . 0

 x+2  lim  

x →−∞  6 x − 10 

First we will divide all terms by the highest power in the denominator, or x , and use the 1 fact that lim   = 0 . x →−∞  x  x + 2x x+2 = lim 6xx 10 lim x →−∞ 6 x − 10 x →−∞ − x x

Therefore, x 2 + 3 x − 10 does not exist. lim 2 x→2 x − 4 x + 4 31.

 7x + 5  lim   x →∞  3 x  First we will divide all terms by the highest power in the denominator, or x , and use the 1 fact that lim   = 0 . x →∞  x  7x +5 7x + 5 = lim x 3 x x lim x →∞ 3 x x →∞ x

 1 + 2x  = lim  10  x →−∞  6 − x    1 = . 6

 5x − 2  lim  

x →∞  4 x + 1 

First we will divide all terms by the highest power in the denominator, or x , and use the 1 fact that lim   = 0 . x →∞  x  5x −2 5x − 2 = lim 4xx 1x lim x →∞ 4 x + 1 x →∞ +x x  5 − 2x  = lim   x →∞  4 + 1  x  5 = . 4

Therefore,  5x − 2  5 lim  = . x →∞  4 x + 1  4

Therefore,  x+2  1 lim  = . x →−∞  6 x − 10  6 34.

 12 x  lim  

x →−∞  4 x − 7 

First we will divide all terms by the highest power in the denominator, or x , and use the 1 fact that lim   = 0 . x →−∞  x  12 x 12 x = lim 4 x x 7 x →−∞ 4 x − 7 x →−∞ −x x

lim

 12  = lim   x →−∞  4 − 7  x  =

12 = 3. 4

Exercise Set 1.2 119 Therefore,  12 x  lim   = 3. x →−∞  4 x − 7 

lim

3x 2 + 5

x →∞

= lim

x →∞

x2 + 22x + 32 x2 x x 3 x2 + 52 x2 x

38.

lim

x →∞ 10 x 2 + x + 1

= lim

 5 + 5x + 22  x  = lim  x →∞ 10 + 1 + 1  x x2   5+0+0 1 = = . 10 + 0 + 0 2 Therefore,  5x2 + 5x + 2  1 lim  = . x →∞ 10 x 2 + x + 1    2

37.

lim

+ 12 x2 x2

x →∞

2 x2 + 6 x − 1 x3 + x + 7

= lim

2 x2 + 63x − 13 x3 x x

x →∞ x3 + x + 7 x3 x3 x3

 2x + 62 − 13  x x  = lim  x →∞  1 + 1 + 7  x2 x3   0+0+0 = = 0. 1+ 0 + 0 Therefore,  2 x2 + 6 x − 1  lim  3  = 0 . x →∞  x + x+7 

5 x2 + 52x + 22 x2 x x

x →∞ 10 x 2 + x + 1 x2 x2 x2

 2 x2 + 6 x − 1  lim  3  x →∞  x + x+7  First we will divide all terms by the highest power in the denominator, or x3 , and use the  1  fact that lim  n  = 0 . x →∞  x  x →∞

 5x2 + 5x + 2  lim   x →∞ 10 x 2 + x + 1    First we will divide all terms by the highest power in the denominator, or x 2 , and use the  1  fact that lim  n  = 0 . x →∞  x  5x2 + 5x + 2

= lim x

 8x + 12  x  = lim  x →∞  1    0+0 = = 0. 1 Therefore,  8x + 1  lim  2  = 0 . x →∞  x 

1 + 2x + 32  x  = lim  x →∞  3 + 5  2 x   1+ 0 + 0 1 = = . 3+ 0 3 Therefore,  x2 + 2 x + 3  1 lim   = . 2 x →∞  3x + 5  3

36.

x →∞

 x2 + 2 x + 3  35. lim   2 x →∞  3x + 5  First we will divide all terms by the highest power in the denominator, or x 2 , and use the  1  fact that lim  n  = 0 . x →∞  x  x2 + 2 x + 3

8x + 1

39.

 x3 + x 2 + x + 1  lim   2 x →−∞  x + x +1  First we will divide all terms by the highest power in the denominator, or x 2 , and use the  1  fact that lim  n  = 0 . x →∞  x  lim

x →−∞

x3 + x 2 + x + 1 x2 + x + 1

 8x + 1  lim  2  x  First we will divide all terms by the highest power in the denominator, or x 2 , and use the  1  fact that lim  n  = 0 . x →∞  x  x →∞ 

= lim

x →−∞

2 x3 + x2 + x2 + 12 x2 x x x

x2 + x2 + 12 x2 x x

 x + 1 + 1x + 12  x  = lim  x →−∞  1 + 1 + 1  x x2   −∞ + 1 + 0 + 0 = = −∞. 1+ 0 + 0

120 Chapter 1: Differentiation Therefore,  x3 + x 2 + x + 1  lim   = DNE . 2 x →−∞  x + x +1  40.

x→2

By limit principle L2, 2 −9 lim x=

lim

x →5

lim x 2 − lim 16

( lim x ) − 16

x →5

= 4−9 −5.

Therefore, lim x 2 − 9 does not exist. x→2

lim x 2 = − 16

44.

(

lim x 2 − 16

x →3

lim x 2 − lim16

By L3

( lim x ) −16

By L2 and L1

x →3

x →3 2

)

By L2

By L3

= ( 5 ) − 16

9 − 16

−7.

Therefore, lim x 2 − 16 does not exist. 45.

= 9 = 3.

lim

( x − 16 )

lim x 2 − lim 16

By L3

2 − 16 x=

lim

x →− 4−

25 − 16

x →− 4−

= ( 4 ) − 16

x→4

By limit principle L2,

(

lim x 2 − 9

x→4

)

x→4

( lim x ) − 9 2

x→4

46. By L2 and L1

16 − 16 = 0 = 0.

lim x 2 − lim 9 By L3

By L2

   lim − x  − 16  x →− 4 

lim x 2 − 9

x→4

By L2

= ( 3) − 16

By L2 and L1

x →5

2 lim x= −9

)

42.

x →2

x →3

By L2 and L1

Therefore,   x5 lim  3  = DNE . x →∞ x + 4 x + 10   

(

( lim x ) − 9

x →2

= ( 2) − 9

(∞ ) 2 = ∞. 1+ 0 + 0

lim x 2 − 16

By L3

  x2  = lim  x →∞ 1 + 4 + 10   x 2 x3 

lim x 2 = − 16

)

lim x 2 − lim 9 x →2

x →∞ x3 + 4 x + 10 x3 x3 x3

x→2

x5 x3

= lim

(

lim x 2 − 9

x→2

  x5 lim  3  x →∞ x + 4 x + 10    First we will divide all terms by the highest power in the denominator, or x3 , and use the  1  fact that lim  n  = 0 . x →∞  x  x →−∞ x 3 + 4 x + 10

41.

lim x 2 − 9

43.

lim

x →3+

x2 − 9

By limit principle L2,

= ( 4) − 9 = 16 − 9 = 7.

By L2 and L1

Exercise Set 1.2 121 x2 − 9 lim =

x →3+

(

lim x 2 − 9

x →3+

)

53. a)

lim x 2 − lim 9 By L3

x →3

=  lim x  − 9  x →3+ 

By L2 and L1

= ( 3) − 9 = 9−9

x →1−

Therefore, lim g ( x ) does not exist. x →1

g (1) = 7

Since the limit of g ( x ) as x approaches 1

does not exist, the function is not continuous at x = 1. lim g ( x ) = 6 and lim g ( x ) = 6. d) x →−2+

x →−2+

Therefore, lim g ( x ) = 6.

= 0 = 0.

x →−2

47. The function is not continuous over the interval, because g ( x ) is not continuous at x = 1. As x

g ( −2 ) = 6

The function g ( x ) is continuous at x = −2, because 6, 1) g ( −2 ) exists, g ( −2 ) =

approaches 1 from the left, g ( x ) approaches 7.

However, as x approaches 1 from the right g ( x ) approaches 3. Therefore, g ( x ) is not continuous at 1.

3) 54. a)

48. The function is not continuous over the interval, because it is not continuous at x = 1. The limit does not exist as x approaches 1, furthermore, f (1) does not exist. Therefore the function is not continuous at at x = 1.

lim g ( x ) exists, lim g ( x ) = 6, and x →−2

x →−2

lim g ( x )= 6= g ( −2 ) .

x →−2

lim f ( x ) = ∞ and lim f ( x ) = −∞.

x →−1+

x →1−

Therefore, lim f ( x ) does not exist. x →1

f (1) is undefined or does not exist.

Since is f (1) undefined and the limit of

f ( x ) as x approaches 1 does not exist, the

49. The function is not continuous over the interval, because h ( x ) is not continuous at x = −2. The

function is not continuous at x = 1. d) lim f ( x ) = 2 and lim f ( x ) = 2. x →−2+

words h ( −2 ) does not exist. Therefore, k ( x )

is not continuous at −2.

x →−2

f ( −2 ) = 2

The function g ( x ) is continuous at x = −2, because 2, 1) f ( −2 ) exists, f ( −2 ) =

50. The function k ( x ) is continuous over the interval.

51. The function t ( x ) is continuous over the

interval. 52. The function is not continuous over the interval, because r ( x ) is not continuous at x = 1. As x approaches 1 from the left, r ( x ) approaches 2. However, as x approaches 1 from the right r ( x ) approaches −2. Therefore, r ( x ) is not

x →−2+

Therefore, lim f ( x ) = 2.

function is not defined at x = −2, in other

continuous at 1.

lim g ( x ) = 3 and lim g ( x ) = 7.

x →1+

55. a)

lim f ( x ) exists, lim f ( x ) = 2, and

x →−2

lim f ( x )= 2=

x →−2

f ( −2 ) .

lim h ( x ) = 3 and lim h ( x ) = 3.

x →−1+

x →1−

Therefore, lim h ( x ) = 3 x →1

h (1) = 1 exists

Since the limit of h ( x ) as x approaches 1

exist, the function is at x = 1. lim h ( x ) = −3 and lim h ( x ) = 1. x →−2+

x →−2+

Therefore, lim f ( x ) does not exist. x →−2

122 Chapter 1: Differentiation e)

h ( −2 ) = 1

The function h ( x ) is not continuous at

58. a)

x = −2, because the limit does not exist. 56. a)

lim k ( x ) = 0 and lim k ( x ) = 0. x →1−

x →−1+

x →1−

Therefore, lim r ( x ) does not exist. x →1

r (1) = 2

Since the limit of r ( x ) as x approaches 1

does not exist, the function is not continuous at x = 1. d) lim r ( x ) = −2 and lim r ( x ) = −2.

Therefore, lim k ( x ) = 0 x →1

k (1) = 0 exists

Since the limit of f ( x ) as x approaches 1

exist, the function is at x = 1. d) lim k ( x ) = 1.5 and lim k ( x ) = 1.5. x →−3+

x →−3+

lim r ( x ) does not exist and lim r ( x ) = 2.

x →1+

Therefore, lim k ( x ) = 1.5.

x → 4−

x → 4+

Therefore, lim r ( x ) = −2. x →−2

r ( 4 ) = −2

The function r ( x ) is continuous at x = 4, because 1) r ( 4 ) exists, r ( 4 ) = −2,

x →3

f ( 3 ) = 1.5

The function g ( x ) is continuous at x = 3,

lim r ( x ) exists, lim r ( x ) = −2, and

because 1) f ( 3 ) exists, f ( 3 ) = 1.5,

lim r ( x ) =−2 =r ( −2 ) .

57. a)

lim f ( x ) exists, lim f ( x ) = 1.5, and

lim f ( x= = ) 1.5

x →3

f ( 3) .

x →1−

Therefore, lim t ( x ) = −3

t (1) = −3 exists

Since the limit of t ( x ) as x approaches 1 x →−2+

Therefore, lim t ( x ) = 1.

t ( −2 ) = 1

The function t ( x ) is continuous at x = −2, because 1, 1) t ( −2 ) exists, f ( −2 ) = 3)

lim t ( x ) exists, lim t ( x ) = 1, and x →−2

lim t ( x )= 1= t ( −2 ) .

x →−2

b) As inputs x approach 3 from the left, outputs G ( x ) approach 1 . Thus, c) Since the limit from the left, 1, is not the same as the limit from the right, 3, the limit does not exist. lim G ( x ) does not exist. d) G ( 3 ) = 1

x →3

e) The function G ( x ) is not continuous at

x →−2

x → 3+

x → 3−

exist, the function is at 𝑥𝑥 = −3. d) lim t ( x ) = 1 and lim t ( x ) = 1.

59. a) As inputs x approach 3 from the right, outputs G ( x ) approach 3 . Thus,

lim G ( x ) = 1 .

x →1

x →−2+

x→4

lim G ( x ) = 3 .

lim t ( x ) = −3 and lim t ( x ) = −3.

x →−1+

x→4

x = 3 because the limit does not exist as x approaches 3. f) The function G ( x ) is continuous at x = 0 ,

because 1) G ( 0 ) exists, 2) lim G ( x ) exists, and x →0

3) lim G ( x ) = G ( 0 ) . x →0

g) The function G ( x ) is continuous at x = 2.9, because

1) G ( 2.9 ) exists, 2) lim G ( x ) exists, and x → 2.9

3) lim G ( x ) = G ( 2.9 ) . x → 2.9

Exercise Set 1.2 123 60. a) b)

lim C ( x ) = 1

x →2+

lim C ( x ) = −1

x →2−

c) lim C ( x ) does not exist. x→2

d) C ( 2 ) = 1 e) The function C ( x ) is not continuous at x = 2 because the limit does not exist as x approaches 2. f) The function C ( x ) is continuous at x = 1.95, because

lim C ( x ) =−1 =C (1.95 ) .

x →1.95

61. The function g ( x ) is continuous at x = 4 , because: 1) g ( 4 ) exists, g ( 4 ) = 4 x→4

62. First we find the function value when x = 5 . f ( 5= 2 13 . Hence, f ( 5 ) exists. ) 3 (5) − = Next, we find the limit as x approaches 5. It follows from the Theorem on Limits of Rational Functions that we can find the limit by 2 13 substitution: lim f ( x= ) 3 (5) − = x →5

f ( 5 ) and the

function is continuous at x = 5 . 1 is not continuous at x 1 x = 0 because G ( 0 ) = is undefined. 0

63. The function G ( x ) =

64. The function F ( x ) =

x is not continuous at

x = −1 because F ( −1) = −1 is undefined on

the real numbers. 65. The function f ( x ) is continuous at x = 4 , because: 1) f ( 4 ) exists, 3) lim f ( x )= 3= x→4

Since the limit from the left, 5, is the same as the limit from the right, 5. The limit exists. We have: lim g ( x ) = 5 . Therefore, we have lim g ( x )= 5= g ( 3 ) .

67. To verify this we take the limit x approaches 3 from the left and the limit as x approaches 3 from the right. As x approaches 3 from the left we have 1 lim g= ( x ) ( 3) + 4 x → 3− 3 lim− g ( x ) = 5 x →3

As 𝑥𝑥 approaches 3 from the left we have lim+ g= ( x ) 2 ( 3) − 1 x →3

lim g ( x )= 6 − 1

x → 3+

lim g ( x ) = 5

x → 3+

Since the limit from the left 5 is same as the limit from the right 5, the limit exist. Therefore, lim g ( x ) exists. x →3

Yes, the limit exists and equals the function value at x = 3.

68. The function is not continuous at x = 4 because the limit does not exist as x approaches 4. lim G ( x= 1 3 ) 12 ( 4 ) += x →4−

lim G ( x ) =− ( 4 ) + 5 =1

2) lim f ( x ) exists, and x→4

lim g ( x= 1 5. ) 2 ( 3 ) −=

x → 3+

x →3

x→4

Therefore, lim f ( x= ) 13=

−

Thus the function is continuous at x = 3 .

3) lim g ( x )= 4= g ( 4 ) .

x →5

1 4 5. ( 3) + = 3 x →3 As the inputs x approach 3 from the left, the outputs g ( x ) approach 5, that is, lim g ( x= )

x →3

2) lim g ( x ) exists, lim g ( x ) = 4 , and x→4

66. First we find the function value when x = 3 . 1 g ( 3 ) = ( 3 ) + 4 =1 + 4 = 5 . Hence, g ( 3 ) 3 exists. Next, we find the limit as x approaches 3. As the inputs x approach 3 from the right, the outputs g ( x ) approach 5, that is,

f ( 4) .

x →4+

lim G ( x ) ≠ lim G ( x )

x →4−

x →4+

124 Chapter 1: Differentiation However, when x = 2 , the output G ( 2 ) is

Therefore, lim G ( x ) does not exist. x→4

Furthermore, the function is not defined at x = 4 , so G ( 4 ) does not exist. 69. The function is not continuous at x = 4 because the function is not defined at x = 4 . Therefore, g ( 4 ) does not exist. 70.

 x + 2, x ≤ 3 F ( x) =   2 x − 5, x > 3 To check: The function is continuous at x = 3 . Left limit of the function f ( x ) approaches

x = 3: The function f ( x ) defined at the left of x = 3

is f ( x )= x + 2 . lim− f ( x ) = lim− ( x + 2 ) = 3 + 2 = 5

x →3

Right limit of the function f ( x ) approaches x = 3: The function f ( x ) defined at the right of x = 3 is f ( x= ) 2x − 5 .

lim f ( x= )

x → 3+

lim ( 2 x − 5= 5 1 ) 2 ( 3) −=

x → 3+

As left limit and right limit are not equal, the limit of the function does not exist at x = 3 . Hence, the function f ( x ) is discontinuous at x =5.

71. The function is not continuous at x = 2 . To verify this, we take the limit as x approaches 2. Using the Theorem on Limits of Rational Functions, we simplify the function near 2 by factoring the numerator and canceling common factors. x2 − 4 lim G ( x ) = lim x→2 x→2 x − 2 ( x − 2 )( x + 2 ) = lim x→2 x−2 = lim ( x + 2 ) x→2

= 2+2 = 4

defined to be 5. That is, G ( 2 ) = 5 . Therefore,

lim G ( x ) = 4 ≠ 5 = G ( 2 ) . Thus the function is x→2

not continuous at x = 2 . 72. The function is not continuous at x = 1 because lim F ( x ) = 2 ≠ 4 = F (1) . x →1

73. The function G ( x ) is continuous at x = 4 , because: 1) G ( 4 ) exists, 2) lim G ( x ) exists, and x→4

3) lim G ( x )= 5= G ( 4 ) . x→4

74. First we find the function value when x = 5 . f ( 5 )= ( 5 ) + 1= 6, f ( 5 ) exists. Next we find the limit as x approaches 5. To find the limit as x approaches 5 from the left, we first simplify the rational function by factoring the numerator and canceling common factors. x2 − 4x − 5 lim f ( x ) = lim x−5 x →5 − x →5 − ( x − 5 )( x + 1) = lim x−5 x →5 − = lim ( x + 1) x →5 −

= 5 +1 =6 To find the limit as x approaches 5 from the right, we can use substitution. f ( x ) lim x + 1 lim = x →5 +

x →5 +

= 5 +1 =6 Therefore, the limit exists. lim f ( x ) = 6 . x →5

Thus we have, lim f ( x )= 6= x →5

f (5) .

Therefore, the function in continuous at x = 5 .

Therefore, lim G ( x ) = 4 . x→2

Exercise Set 1.2 125 75. The function is not continuous at x = 5 because g ( 5 ) does not exist. g (5) =

79. Yes, the function is continuous over the interval ( −4, 4 ) . Since the function is defined for every value in the interval, the Theorem on Limits of Rational Functions tells us lim g ( x ) = g ( a ) for

1 2

( 5 ) − 7 ( 5 ) + 10

x→a

all values a in the interval. Thus g ( x ) is

1 25 − 35 + 10 1 = 0 =

continuous over the interval.

76. The function f ( x ) is continuous at x = 3 ,

80. Yes, the function is continuous over the interval ( −5, 5 ) . Since the function is defined for every value in the interval, the Theorem on Limits of Rational Functions tells us lim F ( x ) = F ( a )

because: 1) f ( 3 ) exists,

x→a

for all values a in the interval. Thus F ( x ) is

2) lim f ( x ) exists, and

continuous over the interval.

x →3

3) lim f ( x ) =−1 = f ( 3 ) .

81. Yes, because lim g ( x ) = g ( a ) for all a such

x →3

x→a

77. The function is not continuous at x = 2 , because G ( 2 ) does not exist.

that 1 < a < ∞ , and lim g ( x ) = g (1) .

82. Yes, because lim h ( x ) = h ( a ) for all a such

G ( 2) =

x →1+

x→a

( 2) − 6 ( 2) + 8

that −3 < a < ∞ , and lim h ( x= ) h ( −3 ) . x →−3+

1 = 0

83. Yes, because lim F ( x ) = F ( a ) for all a such x→a

78. First we find the function value when x = 4 . 1 F ( 4) = 2 ( 4 ) − 7 ( 4 ) + 10

1 16 − 28 + 10 1 = −2 1 = − 2 Hence, F ( 4 ) exists.

that −5 < a < 5 , and lim F ( x= ) F ( −5 ) and lim F ( x ) = F ( 5 ) .

x →5 −

84. Yes, because lim G ( x ) = G ( a ) for all a such

x→a

that −3 < a < 3 , and lim G ( x= ) G ( −3 ) and lim G ( x ) = G ( 3 ) .

x →−3+

x → 3−

85.

x → 20 +

lim p ( x ) does not exist.

x → 20

86. 1

1 lim F ( x ) = 2 = − =F ( 4 ) . x→4 2 ( 4 ) − 7 ( 4 ) + 10

Therefore, the function is continuous at x = 4 .

lim= p ( x ) 1.50(20) = 30

x → 20 −

lim= = 25 p ( x ) 1.25(20)

Next we find the limit. Applying the Theorem on Limits of Rational Functions we have: 1 1 lim F ( x ) = = − . 2 x→4 2 ( 4 ) − 7 ( 4 ) + 10 We now have,

x →−5+

p ( x ) 0.08(100) = = 8 lim

x →100 −

lim = p ( x ) 0.06(100) = 6

x →100 +

lim p ( x ) does not exist.

x →100

126 Chapter 1: Differentiation 87.

lim T = = 120 ( t ) 2(60)

97. In order for the function to be continuous at x = 20 the limit as x approaches 20 from the left must equal the limit as x approaches 20 from the right. The limit from the left of the function is: = lim= (1.5 x ) 1.50 ( 20 ) 30 . Therefore, the

x → 60 −

lim T ( t ) =300 − 3(60) =120

x → 60 +

lim T ( t ) = 120

x → 60

88.

x → 20 −

limit of p(x) as x approaches 20 from the right must be equal to 30. We set the right hand limit equal to 30: lim (1.25 x + k ) = 30 .

lim = P ( t ) 0.04 ( 5 ) + 5

t → 5−

= 0.04 × 625 + 5 = 25 + 5 = 30

x → 20 +

We allow x to approach 20. By Limit Property L3, we have: 1.25 ( 20 ) + k = 30

lim P ( t ) = −2 ( 5 ) + 40 +

t →5

30 = −10 + 40 = lim P ( t ) = 30

t →5

P ( t ) is intended to represent the population in thousands, thus requiring multiplication by 1000. lim− P ( t ) = 30, 000 bacteria. t →5

lim+ P ( t ) = 30, 000 bacteria.

Solving this equation for k yields: 25 + k = 30

k = 5. Therefore, k must equal 5 in order for the price function to be continuous at x = 20. 98.= lim 0.08 = (100 ) 8. In order for T ( t ) to be t →100 −

continuous, the limit from the right must equal 8. Therefore, lim 0.06 x + k = 8

t →5

lim P ( t ) = 30, 000 bacteria.

t →5

89. Yes, because the function is continuous over the 1 1   domain  −∞,    , ∞  . 2 2   90. Yes, because the function is continuous over the domain ( −∞, 2 )  ( 2, ∞ ) . 91. Yes, because the function is continuous over the domain ( −∞, ∞ ) . 92. Yes, because the function is continuous over the domain ( −∞, −1)  (1, ∞ ) . 93. No, because the function is not continuous at the value x = 0 . 94. No, because the function is not continuous at the value x = 1 . 95. Yes, because the function is continuous over the domain ( −∞, ∞ ) . 96. Yes, because the function is continuous over the domain ( −∞, ∞ ) .

t →100 +

0.06 (100 ) + k = 8

By Limit Property L3

6+k = 8 k=2 The constant k must equal 2 in order for the function to be continuous at t = 100. 99. a) As the inputs x approach 0 from the left, the outputs approach −1 . We see this by looking at a table: x −0.1 −0.01 −0.001 −1 −1 −1 x x

lim

= −1 . x As the inputs x approach 0 from the right, the outputs approach 1 . We see this by looking at a table: x 0.001 0.1 0.01 1 1 1 x x →0

−

x lim

x →0

x x

= 1.

Exercise Set 1.2 127 Since the limit from the left, −1 , is not the same as the limit from the right, 1 , the limit x does not exist. lim does not exist. x →0 x b) The limit as x approaches −2 from the left x3 + 8 is: lim 2 = −3. x →−2 − x − 4 The limit as x approaches −2 from the right x3 + 8 is: lim 2 = −3. x →−2 + x − 4 Since the limit from the left is the same as the limit from the right, the limit exists and x3 + 8 is: lim 2 = −3 . x →−2 x − 4 Another approach would be to simplify the function, by factoring the numerator and denominator and canceling common factors. ( x + 2) x2 − 2x + 4 x3 + 8 lim = lim x →−2 x 2 − 4 x →−2 ( x + 2 )( x − 2 )

(

)

102. 0.5 , or

1 2

103. 6 104. 0.5, or

1 2

105. −0.2887, or −

106. 0.378, or

107. 0.75, or

7 3 4

108. 0

x − 2x + 4 x →−2 x−2

= lim

101. c = 9.

assuming x ≠ −2

( −2 ) 2 − 2 ( −2 ) + 4 = ( −2 ) − 2

109. 0.25, or

1 4

12 −4 = −3. 100. For the function to be continuous, y = −3 when x = 0 and y = 16 when x = 4 . Using the first condition yields the equation: a (0) + b =−3 b = −3 Using the second condition and the value for b yields the equation: a (4) − 3 = 16 =

4a = 19 19 a= 4

1 2 3

128 Chapter 1: Differentiation 11. We must first find the y values that correspond to the given x values, therefore: f (4)  3(4)  2 and f (6)  3(6)  2

Exercise Set 1.3

78  72 6  3 53 2 or the temperature rose 3 degrees/hr. The average rate of change is

The average rate of change is

6  3 or Dani 2

hiked 3 mi/hr. 3.

28  7 or The average rate of change is 4 Marcus delivered 7 packages/hr.

The average rate of change is 3000  2500 500   100 or the population of 2023  2018 5 Felton grew by 100 people/yr.

The average rate of change is

125  25 or 5

Tanya scored 25 points/game. 6.

The average rate of change is 180  150 30   7.5 or Chris grew 7.5 cm/yr. 17  13 4

The average rate of change is 20, 000, 000  5, 000, 000 or Rendini industries 4 had revenue of $5,000,000/month.

The average rate of change is 4  6 2   0.333 or unemployment 6 6 changed by 0.333 percentage point/month.

The average rate of change is

53.25  3.55 or 15 Juan spent $3.55/gallon on gasoline.

126  4.2 or 30 Shannon spent $4.2/day on electricity for April.

10. The average rate of change is

 20  14 So the average rate of change will be 20  14 6  3 64 2 12. We must first find the y values that correspond to the given x values, therefore: g (8)  5(8)  4 and g (11)  5(11)  4

 36  51 So the average rate of change will be 51  36 15  5 11  8 3 13. We must first find the y values that correspond to the given x values, therefore: F (1)  2(1) 2 and F (2)  2(2) 2 8 2 So the average rate of change will be 82 6  2 2  (1) 3 14. We must first find the y values that correspond to the given x values, therefore: G ( 2)  3(2) 2 and G (0)  3(0) 2 0  12 So the average rate of change will be 0  (12) 12  6 0  (2) 2 15. We must first find the y values that correspond to the given x values, therefore: 1 1 h(4)  and h(8)  8 4 So the average rate of change will be 1  14  18 1 8   84 4 32 16. We must first find the y values that correspond to the given x values, therefore: 3 3 k (1)   3 and k (8)  8 1 So the average rate of change will be 3  3  21 21 3 8  8   8 1 7 56 8

Exercise Set 1.3 129 17. We must first find the y values that correspond to the given x values, therefore: f (0)  (0) 2  2(0) and f (6)  (6) 2  2(6) 0  48 So the average rate of change will be 48  0 48  8 60 6 18. We must first find the y values that correspond to the given x values, therefore: g (4)  (4) 2  4(4) and  32 g (0)  (0) 2  4(0) 0 So the average rate of change will be 0  (32) 32  8 0  (4) 4 19. We must first find the y values that correspond to the given x values, therefore: 1 1 r (4)  (4) 2  (4)  9 and 2 4 1 1 r (8)  (8) 2  (8)  34 2 4 So the average rate of change will be 34  9 25   6.25 84 4 20. We must first find the y values that correspond to the given x values, therefore: 1 1 s (4)  (4) 2  (4) and 10 2 18  5 1 1 s (7)  (7) 2  (7) 10 2 84  10 So the average rate of change will be 84 48  185 48 10  10   1.6 74 3 30

21. a)

f  x   5x 2

so, f  x  h  f  x h

5  x  h   5x 2 2



 5x  10 xh  5h   5 x 2

  

h 2

10 xh  5h h h 10 x  5h 

h h   10 x  5h  h  10 x  5h  5  2 x  h 

b) The difference quotient column in the table can be completed using the simplified difference quotient. 10 x  5h Simplified difference quotient  10  5   5  2   60 substitiuting 5 for x and 2 for h;

 10  5   5 1  55 substitiuting 5 for x and 1 for h;

 10  5   5  0.1  50.5 substitiuting 5 for x and 0.1 for h;

 10  5   5  0.01  50.05 substitiuting 5 for x and 0.01 for h.

The completed table is:

f  x  h  f  x

2 1 0.1 0.01

h 60 55 50.5 50.05

x 5 5 5 5

22. a)

f  x   4x2

so,

f  x  h  4  x  h

substituing x  h for x



 4 x 2  2 xh  h 2 2

 4 x  8 xh  4h



130 Chapter 1: Differentiation Then f  x  h  f  x Difference quotient h

 4 x  8xh  4h   4 x  2



8 xh  4h h h  8 x  4h 

Factoring the numerator

10 xh  5h 2 h h  10 x  5h 

h h   10 x  5h  h  10 x  5h  5  2 x  h 

Substituting



h   8 x  4h  Removing a factor = 1. h  8 x  4 h, Simplified difference quotient 

b) The difference quotient column in the table can be completed using the simplified difference quotient.  10 x  5h Simplified difference quotient  10  5   5  2   60

or 4  2 x  h 

substitiuting 5 for x and 2 for h;

 10  5   5 1  55

b) The difference quotient column in the table can be completed using the simplified difference quotient. 8 x  4h Simplified difference quotient

substitiuting 5 for x and 1 for h;

 10  5   5  0.1  50.5 substitiuting 5 for x and 0.1 for h;

 8  5   4  2   48

 10  5   5  0.01  50.05

substitiuting 5 for x and 2 for h;

substitiuting 5 for x and 0.01 for h.

 8  5   4 1  44

The completed table is:

substitiuting 5 for x and 1 for h;

 8  5   4  0.1  40.4 substitiuting 5 for x and 0.1 for h;

 8  5   4  0.01  40.04 substitiuting 5 for x and 0.01 for h.

The completed table is:

23. a)

f  x  h  f  x

5 5 5 5

2 1 0.1 0.01

h 48 44 40.4 40.04

24. a)

f  x  h  f  x h



5 5 5 5

2 1 0.1 0.01

h 60 55 50.5 50.05

f  x   4 x 2



substituing x  h for x

  4 x  2 xh  h 2

  4 x  8 xh  4h





 5x  10 xh  5h   5x 2



Then f  x  h  f  x Difference quotient h



f  x  h  f  x

f  x  h   4  x  h 

5  x  h   5 x 2 2

so,

f  x   5 x 2

so,

Exercise Set 1.3 131

 4 x  8xh  4h    4 x   2

Substituting

  

8 xh  4h 2 h h  8 x  4 h 

 

h h    2 x  h  1 h Simplified difference quotient  2x  h  1

Factoring the numerator

h h   8 x  4h  h

Removing a factor = 1. Simplified difference quotient

 8 x  4h, or  4  2 x  h 

b) The difference quotient column in the table can be completed using the simplified difference quotient. 2 x  h  1 Simplified difference quotient  2  5    2   1  11

b) The difference quotient column in the table can be completed using the simplified difference quotient.  8 x  4h Simplified difference quotient

substitiuting 5 for x and 2 for h;

 2  5   1  1  10 substitiuting 5 for x and 1 for h;

 8  5   4  2    48

 2  5    0.1  1  9.1

substitiuting 5 for x and 2 for h;

substitiuting 5 for x and 0.1 for h;

 8  5   4 1   44

 2  5    0.01  1  9.01

substitiuting 5 for x and 1 for h;

substitiuting 5 for x and 0.01 for h.

 8  5   4  0.1   40.4

The completed table is:

substitiuting 5 for x and 0.1 for h;

 8  5   4  0.01   40.04 substitiuting 5 for x and 0.01 for h.

The completed table is:

25. a)

5 5 5 5

f  x  h  f  x h  48  44  40.4  40.04

0.1 0.01

2 xh  h 2  h h h  2 x  h  1

26. a)

5 5 5 5

2 1

10 9.1 9.01

f  x   x2  x

We substitute x  h for x f  x  h   x  h   x  h 2





 x 2  2 xh  h 2  x  h

f  x  x  x

Then f  x  h  f  x Difference quotient h

 x  h   x  h   x  x  2

 x 2  2 xh  h 2  x  h Then f  x  h  f  x Difference quotient h



h 11

0.1 0.01

 x  2 xh  h  x  h    x  x  2



 x  2 xh  h  x  h    x  x  2

f  x  h  f  x

132 Chapter 1: Differentiation  

2 xh  h 2  h h h  2 x  h  1 h

Factoring the numerator

h   2 x  h  1 Removing a factor = 1. h Simplified difference quotient  2x  h  1 

b) The difference quotient column in the table can be completed using the simplified difference quotient. 2 x  h  1 Simplified difference quotient  2  5    2   1  13

 9x   9  x  h    x  x  h     x  x  h       h 9 h x  x  h  h 9 h 1   x  x  h h 

h  9    h  x  x  h  



9 x  x  h

substitiuting 5 for x and 2 for h;

 2  5   1  1  12 substitiuting 5 for x and 1 for h;

 2  5    0.1  1  11.1 substitiuting 5 for x and 0.1 for h;

 2  5    0.01  1  11.01 substitiuting 5 for x and 0.01 for h.

b) The difference quotient column in the table can be completed using the simplified difference quotient. 9 Simplified difference quotient  x  x  h 

The completed table is:

27. a)

9 9  5 5  2 35 substitiuting 5 for x and 2 for h;

f  x  h  f  x

5 5 5 5

2 1

0.1 0.01

11.1 11.01

f  x 

Simplified difference quotient



h 13

9 x

substitiuting 5 for x and 1 for h;

9 9 6    5  5  0.1 25.5 17 substitiuting 5 for x and 0.1 for h;



Then f  x  h  f  x Difference quotient h  9  9    xh  x   h x   9  x  h   9         x  h x   x  x  h    h

9 9 3   5  5  1 30 10

9 9 60   5  5  0.01 25.05 167 substitiuting 5 for x and 0.01 for h.

The completed table is:

0.1

0.01

f  x  h  f  x h 9  35 3  10 6  17 60  167

Exercise Set 1.3 133 28. a)

2 x We substitute x  h for x 2 f  x  h  xh Then f  x  h  f  x Difference quotient h  2  2    xh  x  h x   2  x  h   2         x  h x   x  x  h    multiplying by 1 h  2x   2  x  h    x  x  h     x  x  h       h 2h x  x  h  adding fractions h 2h x  x  h h  h h 1 1 2h 1   multiplying by the reciprocal x  x  h h

f  x 

h  2      h  x  x  h   

2 x  x  h

Removing a factor = 1.

Simplified difference quotient

b) The difference quotient column in the table can be completed using the simplified difference quotient. 2  Simplified difference quotient x  x  h 

2 2  5 5  2 35



2 2 4   5  5  0.1 25.5 51 substitiuting 5 for x and 0.1 for h;



2 2 40   5  5  0.01 25.05 501 substitiuting 5 for x and 0.01 for h.

The completed table is:

29. a)

f  x  h  f  x

0.1

0.01

h 2  35 1  15 4  51 40  501

f  x   2x  3 f  x  h  f  x h

 

Difference quotient

 2  x  h   3   2 x  3 h

 2 x  2h  3    2 x  3 

2h  h 2

Simplified difference quotient

b) The difference quotient is 2 for all values of x and h. Therefore, the completed table is:

5 5 5 5

2 1

substitiuting 5 for x and 2 for h;

2 2 1    5  5  1 30 15 substitiuting 5 for x and 1 for h;

0.1 0.01

f  x  h  f  x h 2 2 2 2

134 Chapter 1: Differentiation 30. a)

f  x   2 x  5

36 x 2  36 xh  12h 2

We substitute x  h for x f  x  h   2  x  h   5

 36  5   36  5   2   12  2   1308 2

substitiuting 5 for x and 2 for h;

2 x  2 h  5

 36  5   36  5 1  12 1  1092 2

Then f  x  h  f  x Difference quotient h  2 x  2h  5    2 x  5   h 2h  h Simplified difference quotient  2

31. a)

5 5 5 5

2 1

 36  5   36  5  0.1  12  0.1  918.12 2

0.1 0.01

f  x   12 x

substitiuting 5 for x and 0.1 for h;

 36  5   36  5  0.01  12  0.01  901.8012

The completed table is:





 

 12 x 3

 

12 x  3x 2 h  3 xh 2  h 3  12 x 3

h 36 x h  36 xh  12h 3  h



h 36 x 2  36 xh  12h 2

h 1308 1092 918.12 901.8012

0.1 0.01



substituing x  h for x 2



 1  x 3  3 x 2 h  3 xh 2  h 3

Then f  x  h  f  x Difference quotient h



1  x  3x h  3xh  h   1  x   3

h 3 x 2 h  3 xh 2  h 3  h







2 1

 1  x  3x h  3 xh 2  h 3

h  36 x 2  36 xh  12h 2 h  36 x 2  36 xh  12h 2 

5 5 5 5

f  x  h  f  x

so,





f  x  h  1   x  h



f  x   1  x3

32. a)

Then f  x  h  f  x Difference quotient h 

substitiuting 5 for x and 0.01 for h.

12  x  h 

f  x  h  f  x h 2 2 2 2

substitiuting 5 for x and 1 for h;

b) The difference quotient is  2 for all values of x and h. Therefore, the completed table is shown at the top of the next column.



b) The difference quotient column in the table can be completed using the simplified difference quotient.





h  3 x 2  3 xh  h 2 h  3 x 2  3 xh  h 2 

Simplified difference quotient



h 3 x 2  3 xh  h 2

Factoring the numerator

 Removing a factor = 1.

Simplified difference quotient

b) The difference quotient column in the table can be completed using the simplified difference quotient.

Exercise Set 1.3 135  2  5    0.1  4  6.1

 3 x 2  3 xh  h 2  3  5   3  5   2    2   109 2

substitiuting 5 for x and 0.1 for h;

 2  5    0.01  4  6.01

substitiuting 5 for x and 2 for h;

substitiuting 5 for x and 0.01 for h.

 3  5   3  5  1  1  91 2

The completed table is:

substitiuting 5 for x and 1 for h;

 3  5   3  5   0.1   0.1  76.51 2

substitiuting 5 for x and 0.1 for h;

5 5 5 5

 3  5   3  5   0.01   0.01  75.1501 2

substitiuting 5 for x and 0.01 for h.

The completed table is:

33. a)

5 5 5 5

2 1

f  x  h  f  x h 109 91 76.51 75.1501

0.1 0.01

f  x   x2  4x

Then f  x  h  f  x Difference quotient h

 x  h   4  x  h    x  4x   2

 x  2 xh  h  4 x  4h    x  4 x  2

  

h 2

2 xh  h  4h h h  2x  h  4

h h    2x  h  4 h  2 x  h  4 Simplified difference quotient

b) The difference quotient column in the table can be completed using the simplified difference quotient. 2x  h  4 Simplified difference quotient  2  5   2   4  8

f  x  h  f  x

34. a)

8 7 6.1 6.01

2 1

0.1 0.01

f  x   x 2  3x

We substitute x  h for x f  x  h   x  h  3 x  h 2





 x 2  2 xh  h 2  3x  3h 2

 x  2 xh  h  3x  3h Then f  x  h  f  x Difference quotient h

 x  2 xh  h  3x  3h    x  3x   2

h 2 xh  h 2  3h  h h  2 x  h  3 Factoring the numerator  h h    2 x  h  3  Removing a factor = 1. h Simplified difference quotient  2x  h  3

b) The difference quotient column in the table can be completed using the simplified difference quotient. 2 x  h  3 Simplified difference quotient  2  5   2   3  9 substitiuting 5 for x and 2 for h;

 2  5   1  3  8 substitiuting 5 for x and 1 for h;

 2  5    0.1  3  7.1

substitiuting 5 for x and 2 for h;

 2  5   1  4  7

substitiuting 5 for x and 0.1 for h;

 2  5    0.01  3  7.01

substitiuting 5 for x and 1 for h;

substitiuting 5 for x and 0.01 for h.

136 Chapter 1: Differentiation

Using the values from the previous page, the completed table is:

x 5 5 5 5 35. a)

f  x  h  f  x



  x  h   3  x  h   5    x  3x  5  

 

 x  2 xh  h  3x  3h  5    x  3x  5   2

2 xh  h  3h h h  2 x  h  3

h h    2 x  h  3 h  2 x  h  3 Simplified difference quotient

b) The difference quotient column in the table can be completed using the simplified difference quotient. 2 x  h  3 Simplified difference quotient  2  5   2   3  9

 x  2 xh  h  4 x  4h  3   x  4 x  3



h 2 xh  h  4h h h  2x  h  4 h

h   2 x  h  4  Removing a factor = 1 h  2 x  h  4 Simplified difference quotient 

b) The difference quotient column in the table can be completed using the simplified difference quotient. 2x  h  4 Simplified difference quotient  2  5    2   4  16 substitiuting 5 for x and 2 for h;

 2  5   1  4  15 substitiuting 5 for x and 1 for h;

 2  5    0.1  4  14.1 substitiuting 5 for x and 0.1 for h;

 2  5    0.1  3  7.1 substitiuting 5 for x and 0.1 for h;

 2  5    0.01  3  7.01 substitiuting 5 for x and 0.01 for h.

The completed table is:

2 1

0.1 0.01

substitiuting 5 for x and 0.01 for h.

The completed table is:

substitiuting 5 for x and 1 for h;

5 5 5 5

Factoring the numerator

 2  5    0.01  4  14.01

 2  5   1  3  8

substitiuting 5 for x and 2 for h;

 x  2 xh  h  4 x  4h  3 Then f  x  h  f  x Difference quotient h 2



Then f  x  h  f  x Difference quotient h



 x 2  2 xh  h 2  4 x  4h  3

f  x   x 2  3x  5



f  x  h   x  h  4  x  h  3

8 7.1 7.01

0.1 0.01

f  x   x2  4x  3

We substitute x  h for x

h 9

2 1

36. a)

f  x  h  f  x

h 5 16 2 5 15 1 5 0.1 14.1 5 0.01 14.01 (NOTE: 37 – 44 answers vary based on estimation of points used)

h 9 8 7.1 7.01

Exercise Set 1.3 137 37. From 2015 to 2018: 2015: The percentage is approximately 2. 2018: The percentage is approximately 1.8. Change in percentage Average rate of change  Change in time 1.8  2 = 2018  2015 0.2  3  0.07 percentage point/year

From 2017 to 2020: 2017: The percentage is approximately 2.2. 2020: The percentage is approximately −3.4. Change in percentage Average rate of change  Change in time 3.4  2.2 = 2020  2017 5.6  3  1.87 percentage point/year

From 2017 to 2020: 2017: The percentage is approximately 1. 2020: The percentage is approximately −6.7. Change in percentage Average rate of change  Change in time 6.7  1 = 2020  2017 7.7  3  2.57 percentage point/year

From 2015 to 2021: 2015: The percentage is approximately 5. 2021: The percentage is approximately 2.9. Change in percentage Average rate of change  Change in time 2.9  5 = 2021  2015 2.1  6  0.35 percentage point/year

From 2015 to 2021: 2015: The percentage is approximately 2. 2021: The percentage is approximately 4.1. Change in percentage Average rate of change  Change in time 4.1  2 = 2021  2015 2.1  6  0.35 percentage point/year

39. From 2015 to 2018: 2015: The percentage is approximately 3. 2018: The percentage is approximately 2.6. Change in percentage Average rate of change  Change in time 2.6  3 = 2018  2015 0.4  3  0.13 percentage point/year

38. From 2015 to 2018: 2015: The percentage is approximately 5. 2018: The percentage is approximately 4.3. Change in percentage Average rate of change  Change in time 4.3  5 = 2018  2015 0.7  3  0.23 percentage point/year

From 2017 to 2020: 2017: The percentage is approximately 3.4. 2020: The percentage is approximately −4.4. Change in percentage Average rate of change  Change in time 4.4  3.4 = 2020  2017 7.8  3  2.6 percentage point/year

138 Chapter 1: Differentiation From 2015 to 2021: 2015: The percentage is approximately 3. 2021: The percentage is approximately 4.8. Change in percentage Average rate of change  Change in time 4.8  3 = 2021  2015 1.8  6  0.3 percentage point/year

41. From 2015 to 2018: 2015: The percentage is approximately 1.4. 2018: The percentage is approximately 0.6. Change in percentage Average rate of change  Change in time 0.6  1.4 = 2018  2015 0.8  3  0.27 percentage point/year

40. From 2015 to 2018: 2015: The percentage is approximately 2.6. 2018: The percentage is approximately 2.2. Change in percentage Average rate of change  Change in time 2.2  2.6 = 2018  2015 0.4  3  0.13 percentage point/year

From 2017 to 2020: 2017: The percentage is approximately 2.5. 2020: The percentage is approximately −4.5. Change in percentage Average rate of change  Change in time 4.5  2.5 = 2020  2017 7  3  2.33 percentage point/year

From 2017 to 2020: 2017: The percentage is approximately 2.5. 2020: The percentage is approximately −3.1. Change in percentage Average rate of change  Change in time 3.1  2.5 = 2020  2017 5.6  3  1.87 percentage point/year

From 2015 to 2021: 2015: The percentage is approximately 1.4. 2021: The percentage is approximately 0.1. Change in percentage Average rate of change  Change in time 0.1  1.4 = 2021  2015 1.3  6  0.22 percentage point/year

From 2015 to 2021: 2015: The percentage is approximately 2.6. 2021: The percentage is approximately −1.87. Change in percentage Average rate of change  Change in time 1.6  2.6 = 2021  2015 1  6  0.17 percentage point/year

42. From 2015 to 2018: 2015: The percentage is approximately 0.6. 2018: The percentage is approximately −3.2. Change in percentage Average rate of change  Change in time 3.2  0.6 = 2018  2015 3.8  3  1.27 percentage point/year

Exercise Set 1.3 139 From 2017 to 2020: 2017: The percentage is approximately 0.9. 2020: The percentage is approximately 1.1. Change in percentage Average rate of change  Change in time 1.1  0.9 = 2020  2017 0.2  3  0.07 percentage point/year From 2015 to 2021: 2015: The percentage is approximately 0.6. 2021: The percentage is approximately 1.9. Change in percentage Average rate of change  Change in time 1.9  0.6 = 2021  2015 1.3  6  0.22 percentage point/year 43. From 2015 to 2018: 2015: The percentage is approximately 1. 2018: The percentage is approximately 1. Change in percentage Average rate of change  Change in time 11 = 2018  2015 0  3  0 percentage point/year From 2017 to 2020: 2017: The percentage is approximately −9. 2020: The percentage is approximately −8.1. Change in percentage Average rate of change  Change in time 8.1  (9) = 2020  2017 0.9  3  0.3 percentage point/year

Change in percentage Change in time 1.2  1 = 2021  2015 2.2  6  0.37 percentage point/year

Average rate of change 

44. From 2015 to 2018: 2015: The percentage is approximately 1.8. 2018: The percentage is approximately 1.9. Change in percentage Average rate of change  Change in time 1.9  1.8 = 2018  2015 0.1  3  0.03 percentage point/year From 2017 to 2020: 2017: The percentage is approximately −0.4. 2020: The percentage is approximately −5.3. Change in percentage Average rate of change  Change in time 5.3  (0.4) = 2020  2017 4.9  3  1.63 percentage point/year From 2015 to 2021: 2015: The percentage is approximately 1.8. 2021: The percentage is approximately 1.8. Change in percentage Average rate of change  Change in time 1.8  1.8 = 2021  2015 0  6  0 percentage point/year

From 2015 to 2021: 2015: The percentage is approximately 1. 2021: The percentage is approximately −1.2.

140 Chapter 1: Differentiation 45. Calculate the rate of change from 2015 to 2017: Value in 2017  Value in 2015 2017  2015 335 miles  270 miles Rate of change  2017  2015 65 miles   32.5 miles per year 2 years Rate of change 

Calculate the rate of change from 2017 to 2018: Value in 2018  Value in 2017 2018  2017 335 miles  335 miles Rate of change  2017  2015 0 mile   0 mile per year 1 year Rate of change 

Calculate the rate of change from 2018 to 2021: Value in 2021  Value in 2018 2021  2018 405 miles  335 miles Rate of change  2021  2018 70 miles   23.33 miles per year 3 years Rate of change 

Average Rate of Change To find the average rate of change, we average the rates calculated for the specified intervals:

Average rate of change 

 32.5  0  23.33

3 55.83 Average rate of change  3  18.61 miles per year 46. Calculate the rate of change from 2015 to 2018: Value in 2018  Value in 2015 2018  2015 $63,179  $55, 775 Rate of change  2018  2015 $7404   2468 dollars per year 3 years Rate of change 

Calculate the rate of change from 2018 to 2020:

Value in 2020  Value in 2018 2020  2018 $67, 521  $63,179 Rate of change  2020  2018 $4342   2171 dollars per year 2 years Rate of change 

Calculate the rate of change from 2017 to 2021: Value in 2021  Value in 2017 2021  2017 $70, 784  $60, 336 Rate of change  2021  2017 $10, 448   2612 dollars per year 4 years Rate of change 

Average Rate of Change To find the average rate of change, we average the rates calculated for the specified intervals:  2468  2171  2612  Average rate of change  3 7251 Average rate of change  3  2417 dollars per year 47. a) From 0 units to 1 unit the average rate of change is 70  0  70 pleasure units per unit. 1 0 From 1 unit to 2 units the average rate of change is 109  70 39   39 pleasure units per unit. 2 1 1 From 2 units to 3 units the average rate of change is 138  109 29   29 pleasure units per unit. 32 1 From 3 units to 4 units the average rate of change is 161  138 23   23 pleasure units per unit. 43 1 b) Answers may vary. As you consume more and more of a good, the additional utility, or amount of pleasure, associated with that good will start to fall. The additional satisfaction from the 1st to the 2nd slice of pizza is greater than the additional satisfaction from the 9th to the 10th slice of pizza.