Calculus for Business, Economics, Life Sciences, and Social Sciences, 13E Barnett, Ziegler, Byleen S

Type:

Solution Manual

Calculus for Business, Resource: Economics, Life Sciences, and Social Sciences Edition:

13th Edition

Author(s):

Raymond Barnett Michael Ziegler Karl Byleen

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Calculus

Chapter 1

Name ________________________________ Date ______________ Class ____________

Section 1-1 Functions Goal: To evaluate function values and to determine the domain of functions

The domain of the following functions will be the set of real numbers unless it meets one of the following conditions: 1. The function contains a fraction whose denominator has a variable.

The domain of such a function is the set of real numbers EXCEPT the values of the variable that make the denominator zero.

2. The function contains an even root (square root , fourth root 4 , etc.).

The domain of such a function is limited to values of the variable that make the radicand (the part under the radical) greater than or equal to 0.

1. Evaluate the following function at the specified values of the independent variable and simplify the results.

f(x)=4x-5 a)

c)

f (1 = 4(1)- 5 f ) = 4 -5 f (1 = - 1 ) f ((1 x - 1 = 4( x - 1)-5 ) f (x - ) = 4x - 4 - 5 f (x - 1 = 4x - 9 ) 1 )

b)

d)

f (-3)= 4(-3) - 5 f (-3)=- 12 - 5 f (-3)=- 17 æ1ö 1 f ç ÷4ø= 4 çè ÷-ø5 è 4 æ ö æ1ö f ç ÷=1 - 5 è 4ø æ1ö f ç ÷=- 4 è 4ø

In problems 2–10 evaluate the given function for f (x) = x2 +1 and

f (3 = (3)2 +1 f ) = 9 +1 f (3 =10 ) (3 ) 1-1

2. (f+g)(3)

= f (3)+ g(3) =10 + (- 1) ( f + g )(3) = 9

g( x) = x - 4 . g (3 = 3 - 4 g ) = 1(3 )

..

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Calculus

Chapter 1

f (2c) = (2c)2+1 f (2c) = 4c2 +1

g (2c) = 2c - 4

4. ( fg)(-4) = f(-4)g(-4) = (17)(-8) ( fg)(-4) =- 136

f (-4) = (- 4)2 +1 f (-4) =16+1 f (-4) =17

g (-4) = 4 - 4 g (-4) - 8 = -

æfö f(0) 5. çèg÷ø(0) = g(0)

f (0) = (0)2 +1 f (0) = 0 +1 f (0) =1

g (0) = 0 - 4 g (0) = 4-

3. ( f - g )( 2c )= f(2c) - g (2)c

= 4c21+ - (2c- 4) ( f - g )( 2c )= 4c2 -2c + 5

1 -4 æfö 1 çèg÷ø(0) = - 4 =

6.

2 ×g (-2) = 2(- 6) 2 ×g (-2) = 12 -

g (-2) = 2 - 4 g (-2) - 6 = f (4) =-(4)2+1

7. 3 ×f(4)- 2 ×g (-1) = 3(17)-2(-5) = 51 + 10 3 ×f(4)- 2 ×g (-1) = 61

8.

f(3)- g(2) 10- (-2) = f(1) 2 =

f (4) =16+1 f (4) =17

f (3 = (3)2 +1 f ) = 9 +1 f (3 =10 ) (3 )

12 2

f(3)- g(2) =6 f(1)

f (1 = (1)2+1 f ) =1 +1 f (1 = 2 ) (1 )

1-2 ..

g (-1) = 1-4 g (-1) - 5 = g (2) = 2 -4 g (2) = 2-

Calculus

9.

10.

Chapter 1

g ( - 1 + h)- g (-1) h - 5 - (-5) = h h h = h g ( - 1 + h)- g (-1) =1 h

f(2 + h)- f(2) h2 + 4h+5 - (5) = h h h2 + 4h = h f(2 + h)- f(2) = h +4 h

g ( - + h = - 1 + h- 4 g1 +) = h - 5 (- h 1 )

g (-1) = 1-4 g (-1) - 5 = -

f ( + h = (2 + h)2 +1

f (2) = (2)2 +1 f (2) = 4 +1 f (2) = 5

f 2 + ) = h2 + 4 + 4 +1 f ( + h = h2 + h + 5 2

)

4

(

h

h

2 ) In problems 11–18 find the domain of each function.

11. g(x)=

5 x-2

The domain is restricted by the denominator. Since it cannot equal zero, the domain is all real numbers except 2.

12.

f(x)=

2x 3x+7

The domain is restricted by the denominator. Since it cannot equal zero, the domain 7 is all real numbers except - .

3 13.

h(t)=41-2t

The domain is restricted by the radicand since it has an even root. Since the radicand 1 must be greater than or equal to zero, the domain is 2t£.

14.

g(x)=1-2x2 There are no restrictions on the domain, therefore the domain is all real numbers.

1-3 ..

Calculus 15.

Chapter 1

f(x)=3x+4

There are no restrictions on the domain since it has an odd root, therefore the domain is all real numbers.

16. h(w)=

w-3

The domain is restricted by the radicand since it has an even root. Since the radicand must be greater than or equal to zero, the domain is w³3.

17.

f(x)=2x3+5x2-x+17 There are no restrictions on the domain, therefore the domain is all real numbers.

18. g(x)=

2x3 5

There are no restrictions on the domain since there is no variable in the denominator, therefore the domain is all real numbers.

1-4 ..

Calculus

Chapter 1

Name ________________________________ Date ______________ Class ____________

Section 1-2 Elementary Function: Graphs and Transformations Goal: To describe the shapes of graphs based on vertical and horizontal shifts and reflections, stretches, and shrinks

Basic Elementary Functions:

f (x) = x

Identity function

h(x) = x2

Square function

m(x) = x3

Cube function

n( x) = x

Square root function

p( x) = 3 x g( x) = x

Cube root function Absolute value function

In problems 1–14 describe how the graph of each function is related to the graph of one of the six basic functions. State the domain of each function. (Do not use a graphing calculator and do not make a chart.)

1. g(x)=x2-4 The graph is the square function that is shifted down 4 units. There are no restrictions on the domain, therefore, the domain is all real numbers.

2. f(x)=

x+5

The graph is the square root function that is shifted up 5 units. The domain is restricted by the radicand since it has an even root. Since the radicand must be greater than or equal to zero, the domain isx ³ 0.

3.

f(x)=-

x

The graph is the square root function that is reflected over the x–axis. The domain is restricted by the radicand since it has an even root. Since the radicand must be greater than or equal to zero, the domain isx ³ 0.

1-5 ..

Calculus 4.

Chapter 1

f(x)=3x-2

The graph is the cube root function that is shifted 2 units to the right. There are no restrictions on the domain since it has an odd root, therefore the domain is all real number.

5.

g(x)=(x-5)2 -3

The graph is the square function that is shifted to the right 5 units and down 3 units.

There are no restrictions on the domain, therefore, the domain is all real numbers.

6.

f(x)=-x2+1

The graph is the square function that is reflected over the x–axis and shifted up 1 unit.

There are no restrictions on the domain, therefore, the domain is all real numbers.

7. g(x)=2-

x-4

The graph is the square root function that is shifted 4 units to the right, reflected over the x–axis, and shifted 2 units up. The domain is restricted by the radicand since it has an x ³is 4. even root. Since the radicand must be greater than or equal to zero, the domain 8. h(x)=x+5 The graph is the absolute value function that is shifted 5 units to the left. There are no restrictions on the domain, therefore, the domain is all real numbers.

9. g(x)=3x-3 The graph is the cube root function that is shifted 3 units down. There are no restrictions on the domain since it has an odd root, therefore the domain is all real number.

10.

f(x)=x+2-4

The graph is the absolute value function that is shifted 2 units to the left and 4 units down. There are no restrictions on the domain, therefore, the domain is all real numbers.

11. h(x)=-x-3+2 The graph is the absolute value function that is shifted 3 units to the right, reflected over the x–axis, and shifted 2 units up. There are no restrictions on the domain, therefore, the domain is all real numbers.

1-6 ..

Calculus

Chapter 1

12. f(x)=x3+2 The graph is the cube function that is shifted 2 units up. There are no restrictions on the domain, therefore, the domain is all real numbers.

13.

f(x)=-(x+2)3+4

The graph is the cube function that is shifted 2 units to the left, reflected over the x– axis, and then shifted 4 units up. There are no restrictions on the domain, therefore, the domain is all real numbers.

14. h(x)=3-3x-4 The graph is the cube root function that is shifted 4 units to the right, reflected over the x–axis, and then shifted 3 units up. There are no restrictions on the domain since it has an odd root, therefore the domain is all real numbers.

In Problems 15–23 write an equation for a function that has a graph with the given characteristics.

15. The shape of y=x3 shifted 6 units right.

y=(x-6)3

16. The shape of y=

x shifted 4 units down.

y= x-4

17. The shape of y=x reflected over the x-axis and shifted 2 units up. y=-x+2 18. The shape of y=x2 shifted 2 units right and 4 units up. y=(x-2)2+4 19. The shape of y=3x reflected over the x-axis and shifted 1 unit up.

y=1-3x

1-7 ..

Calculus

Chapter 1

20. The shape of y = x2 reflected over the x-axis and shifted 3 units down. y=-x2-3

21. The shape of y = x shifted 4 units left. y= x+4 22. The shape of y = x3 shifted 6 units right and 2 units down.

y=(x-6)3 - 2

23. The shape of y = x shifted 6 units right and 5 units up. y=x-6+5

1-8 ..

Calculus

Chapter 1

Name ________________________________ Date ______________ Class ____________

Section 1-3 Linear and Quadratic Functions Goal: To describe functions that are linear and quadratic in nature

1. If x = y and a is any real number, then x±a=y ± a.

Equality Properties:

2. If x = y and a is any nonzero real number, thenax = ay and y x= . a a Quadratic Functions: Standard form of a quadratic:f (x) = ax2 +bx + c where a,b,c are real anda ¹ 0.

Vertex form of a quadratic: f (x) = a(x- h) 2 + k where a ¹ 0 and (h,k)is the vertex. x =h

Axis of symmetry:

Minimum/Maximum value:

If a>0, then the turning point (or vertex) is a minimum point on the graph and the minimum value would be k.

If a<0, then the turning point (or vertex) is a maximum point on the graph and the maximum value would be k.

In problems 1–3, solve for the variable: 1.

5 x + 7 = 9 - 13 5 x + 7 - 5 x= x - 13- 5x 7 = 9 - 13 7 += x1 - 313+13 2 0= 4 20 = x4x 4 44 5 =x 4 x ..

1-9

Calculus 2.

Chapter 1

7 y + 3(6y- 11) =167 167 7 y + 18y- 33 =167 167 25y- 33 =+ 33 25y - 33+33 = 25 y = 200 25 y 200 = 25 25 y =8

3.

m

m +4 = + 7 5 2 4(10) 10m 10m + 40 = + 7(10) 5 2 40 2m + = 5m +7 -30 = -10 3m 0 = 3m +7 =m 0

For 4–11 find:

a. the domain b. the vertex c. the axis of symmetry d. the x-intercept(s) e. the y-intercept f. the maximum or minimum value of the function then: g. Graph the function. h. State the range. i. State the interval over which the function is decreasing. j. State the interval over which the function is increasing.

1-10 ..

Calculus 4.

Chapter 1

f(x)=(x-1)2 -3 a. The function is a quadratic, therefore the domain is all real numbers. b. The function is in vertex form, therefore the vertex is (1, – 3). c. The axis of symmetry is the x–value of the vertex, therefore the axis of symmetry is x = 1. d. The x-intercepts are found by setting f (x) = 0.

f ( x ) =( x - )21- 3

0=x2-2x+1-3 0=x2-2x-2 Solve using the quadratic equation x =1± 3 Therefore, the x-intercepts are (1+3,0) and (1e. The y-intercepts are found by settingx = 0.

3, 0).

2

f (x )= (x - 1) - 3 2

f ( = 0 (0)- 1) - 3 f ( = 0 (-)1)2 - 3 f (0)= 2 - the y-intercept is (0, –2).

Therefore, f. The graph opens upward, therefore the graph has a minimum value which is the ycoordinate of the vertex or –3. g.

h The graph has a minimum value of –3, therefore the range yis³-3. . Based on the graph, the function is decreasing over the interval (- ¥ ,1). i. Based on the graph, the function is increasing over the interval (1, ¥ ). j.

1-11 ..

Calculus 5.

Chapter 1

f(x)=(x-2)2 +4 a. The function is a quadratic, therefore the domain is all real numbers. b. The function is in vertex form, therefore the vertex is (2, 4). c. The axis of symmetry is the x–value of the vertex, therefore the axis of symmetry is x=2. d. The x-intercepts are found by setting f (x) = 0.

f ( x )= (x - 2)2+ 4 0= x - 4 x +4 + 4 x 0= 2 - 4 +8 Solvingxthe above equation by the quadratic equation will result in complex roots, therefore no x-intercepts are present.

2 are found by setting x=0. e. The y-intercepts 2

f (x )= (x - 2) + 4 2

f ( = 0 (0)- 2) + 4 2 f ( = 0 (-)2) + 4

f (0)= 8 Therefore, the y-intercept is (0, 8). f. The graph opens upward, therefore the graph has a minimum value which is the ycoordinate of the vertex or 4. g.

h The graph has a minimum value of 4, therefore the range yis³ 4. . Based on the graph, the function is decreasing over the interval (- ¥ ,2). i. Based on the graph, the function is increasing over the interval (2, ¥ ). j.

1-12 ..

Calculus 6.

Chapter 1

f(x)=-x2+7 a. The function is a quadratic, therefore the domain is all real numbers. b. The function in vertex form is f(x)=-(x-0)2+7, therefore the vertex is (0, 7). c. The axis of symmetry is the x–value of the vertex, therefore the axis of symmetry is x=0. d. The x-intercepts are found by setting f(x)=0. f ( x) = - x 2 + 7

0=-x2+7 Solve using the quadratic equation x=± 7 Therefore, the x-intercepts are( 7,0) and (- 7 , 0). e. The y-intercepts are found by settingx = 0.

f (x )= x2 + 7 ) +7 f ( -0 02 f (0)= 0 + 7 f (0)= 7 Therefore, the y-intercept is (0, 7). f. The graph opens downward, therefore the graph has a maximum value which is the y-coordinate of the vertex or 7. g.

h The graph has a maximum value of 2, therefore the range yis£ 0. . Based on the graph, the function is decreasing over the interval (0, ¥ ). i. Based on the graph, the function is increasing over the interval (- ¥ ,0). j.

1-13 ..

Calculus 7.

Chapter 1

f(x)=-(x-1)2 -1 a. The function is a quadratic, therefore the domain is all real numbers. b. The function is in vertex form, therefore the vertex is (1, –1). c. The axis of symmetry is the x–value of the vertex, therefore the axis of symmetry is x = 1. d. The x-intercepts are found by setting f (x) = 0.

f ( x )=

0-

(x - 1)2- 1 x2 + 2 x - 1-1

0= x2 + 2 x - 2 Solving the above equation by the quadratic equation will result in complex roots, therefore no x-intercepts are present. The y-intercepts are found by x = 0. e. setting = 2 f (x )= - (x - 1) 12 f ( -0 )(0 - )1 1 2 f ( = 0 )(- 1) - 1 f (0)- 2 Therefore, the y-intercept is (0, –2). = opens downward, therefore the graph has a maximum value which is f. The graph the y-coordinate of the vertex or –1.

-

g.

= -

h The graph has a maximum value of –1, therefore the range yis£-1.

(1, ¥ ). . Based on the graph, the function is decreasing over the interval i. Based on the graph, the function is increasing over the interval (- ¥ ,1). j.

1-14 ..

Calculus 8.

Chapter 1

f(x)=x2-4x a. The function is a quadratic, therefore the domain is all real numbers. b. The function in vertex form is f(x)=(x-2)2-4, therefore the vertex is (2, – 4). c. The axis of symmetry is the x–value of the vertex, therefore the axis of symmetry is x=2. d. The x-intercepts are found by setting f(x)=0. f ( x) = x2

0 = 4x x2 0 = - 4x x(xx = 0,4 4) Therefore, the x-intercepts are(0, 0)and (4,0). e. The y-intercepts are found by settingx =0. f ( x) = x2 - 4 x

f (0)= 0 2 - 4(0) f (0)= 0 +0 f (0)= 0 Therefore, the y-intercept is (0, 0). f. The graph opens upward, therefore the graph has a minimum value which is the ycoordinate of the vertex or – 4. g.

h The graph has a minimum value of – 4, therefore the range is³-4. y (-¥,2). . Based on the graph, the function is decreasing over the interval i. Based on the graph, the function is increasing over the interval (2,¥). j.

1-15 ..

Calculus 9.

Chapter 1

f(x)=x2+2x-4 a. The function is a quadratic, therefore the domain is all real numbers. b. The function in vertex form is f(x)=(x+1)2-5,therefore the vertex is (–1, –5). c. The axis of symmetry is the x–value of the vertex, therefore the axis of symmetry is x=-1. d. The x-intercepts are found by setting f(x)=0.

2

f ( x ) = ( x + )1 - 5

0=x2+2x+1-5 0 = x2 +2x- 4 Solve using the quadratic equation x = - 1± 5 Therefore, the x-intercepts are( -1 + 5,0) and (-1e. The y-intercepts are found by settingx = 0.

5,0).

2

f ( x) = (x+ 1) - 5 2

f ( = 0 (0)+ 1) - 5 f ( = 0 ()2 1)- 5 f (0)= - 4 Therefore, the y-intercept is (0, –4). f. The graph opens upward, therefore the graph has a minimum value which is the ycoordinate of the vertex or –5. g.

h The graph has a minimum value of –5, therefore the range yis³-5. . Based on the graph, the function is decreasing over the interval (-¥,-1). i. Based on the graph, the function is increasing over the interval (- 1, ¥ ). j.

1-16 ..

Calculus 10.

Chapter 1

f(x)=x2+2x+1 a. The function is a quadratic, therefore the domain is all real numbers. b. The function in vertex form is f(x)=(x+1)2+0,therefore the vertex is (–1, 0). c. The axis of symmetry is the x–value of the vertex, therefore the axis of symmetry is x=-1. d. The x-intercepts are found by setting f(x)=0. f ( x) = x22 + x +1

0 = x22 + x +1 0 = (x+1)( x +1) x = -1 (-1,0).

Therefore, there is only one x-intercept which is e. The y-intercepts are found by setting x=0. f (x )= x2 +2 x +1

f (0)= 0 22(0) + +1 f (0)= 0 +0 +1 f (0)= 1 Therefore, the y-intercept is (0, 1). f. The graph opens upward, therefore the graph has a minimum value which is the ycoordinate of the vertex or 0. g.

h The graph has a minimum value of 0, therefore the range is y³ 0. . Based on the graph, the function is decreasing over the interval (-¥,-1). i. Based on the graph, the function is increasing over the interval (-1,¥). j.

1-17 ..

Calculus 11.

Chapter 1

f(x)=-x2+10x-19 a. The function is a quadratic, therefore the domain is all real numbers. b. The function in vertex form is f(x)=-(x-5)2+6, therefore the vertex is (5, 6). c. The axis of symmetry is the x–value of the vertex, therefore the axis of symmetry is x=5. d. The x-intercepts are found by setting f(x)=0. f ( x) = - x 2 + 1 0 x - 1 9

0=-x2+10x-19 0 = x2 - 10x+19 Solve using the quadratic equation x=5± 6 Therefore, the x-intercepts are(5 + 6,0) and (5e. The y-intercepts are found by settingx = 0.

6,0).

f ( x) = - x+2 10 x - 19 f ( = )+ 10(0)- 19 0 - 02 f (0)= 0 +0 - 19 f (0)= - 19 Therefore, the y-intercept is (0, –19). f. The graph opens downward, therefore the graph has a maximum value which is the y-coordinate of the vertex or 6. g.

h The graph has a maximum value of 4, therefore the range is£y6. . Based on the graph, the function is decreasing over the interval (5, ¥ ). i. Based on the graph, the function is increasing over the interval (- ¥ ,5). j.

1-18 ..

Calculus

Chapter 1

12. The revenue and cost functions for a company that manufactures components for washing machines were determined to be:

R(x)=x(200-4x)

and

C(x)=160+20x

where x is the number of components in millions and R(x) and C(x) are in millions of dollars. a) How many components must be sold in order for the company to break even? (Breakeven points are when R(x)=C(x).) (Round answers to nearest million.)

R(x)= C( x) x(200- 4x)=160+20x 200x- 4x2 =160+20x 0 = 4x2 - 180x+160 Solve the equation by the quadratic equation x » 0.9, 44.09 The company would need to sell approximately 1 million or 44 million to break even. b) Find the profit equation. (P(x)=R(x)-C(x))

P(x = R(x)- C(x) ) = x(200- 4x)- (160+20x) P(x = 200x- 4x2 - 160- 20x ) = 4x2 +180x- 160 P(x c) Determine the maximum profit. How many components must be sold in order to achieve ) that maximum profit?

P(x The ) maximum profit occurs at the vertex of the profit function. The x-coordinate is b 180 -180 x=- 2a =- 2(-4) = -8 =22.5. To find the y-coordinate of the vertex, substitute the value into the function as follows: P(x) = 4 x2+180x - 160

P ( 22.5) - 4(22.5)2 +180(22.5)- 160 P ( 22.5)= 2025 + 4050 - 160 P ( 22.5)=1865 The maximum profit of $1,865 million is achieved when 22.5 million components are = sold.

1-19 ..

Calculus

Chapter 1

13. A company keeps records of the total revenue (money taken in) in thousands of dollars from the sale of x units (in thousands) of a product. It determines that total revenue is a function R(x) given by

R(x)=300x-x2 It also keeps records of the total cost of producing x units of the same product. It determines that the total cost is a function C(x) given by

C(x)=40x+1600 a) Find the break-even points for this company. (Round answer to nearest 1000.)

R( x) =C(x) 300 x - x 2 =40x+1600 0 = x2 - 260x+1600 Solve the equation by the quadratic equation x » 6.307, 253.693 The company would need to sell approximately 6,000 or 254,000 to break even. b) Determine at what point profit is at a maximum. What is the maximum profit? How many units must be sold in order to achieve maximum profit?

The profit equation is:

P(x = R(x)- C(x) )

= 300x- x 2 - (40x+1600)

P(x = - x2 +260x- 1600 ) The maximum profit occurs at the vertex of the profit function. The x-coordinate is bP(x 260 -260 x=- 2a =- 2(-1) = -2 =130. To find the y-coordinate of the vertex, substitute the value ) into the function as follows: P(x) = x2+ 260x - 1600

P(130)- (130)2 +260(130)- 1600 P(130) = 16, 900 + 33, 800 - 1600 P(130) =15, 300 The maximum profit of $15,300 thousands or $15,300,000 is achieved when 130,000 = units are sold.

1-20 ..

Calculus

Chapter 1

14. The cost, C(x), of building a house is a function of the number of square feet, x, in the house. If the cost function can be approximated by

C(x)=0.01x2-20x+25,000

where 1000 £ x £ 3500

a) What would be the cost of building a 1500 square foot house?

Substitute the value of 1500 into the cost function:

C(x) = 0.01x2- 20 x +25,000 C (1500) = 0.01(1500)2 - 20(1500)+25,000 C (1500)= 0.01(2, 250, 000) - 30, 000 + 25, 000 C (1500)=17, 500 It will cost $17,500 to build a 1500 square foot house. b) Find the minimum cost to build a house. How many square feet would that house have?

The minimum cost occurs at the vertex of the cost function. The x-coordinate is b -20 20 x==- 2(0.01) = =1000. To find the y-coordinate of the vertex, substitute the 2a 0.02 value into the function as follows:

C(x) = 0.01x2- 20x +25000 C (1000)= 0.01(1000)2 - 20(1000)+25,000 C (1000)=10,000- 20,000+25,000 C (1000)=15, 000 The minimum cost of $15,000 is achieved when a 1000 square foot home is built.

1-21 ..

Calculus

Chapter 1

15. The cost of producing computer software is a function of the number of hours worked by the employees. If the cost function can be approximated by

C(x)=0.04x2-20x+6000

where 200 £ x £ 1000

a) What would be the cost if the employees worked 800 hours?

Substitute the value of 800 into the cost function: C(x) = 0.04x2- 20 x +6000

C (800)= 0.04(800)2 - 20(800)+6000 C (800)= 0.04(640, 000) - 16, 000 + 6000 C (800)=15, 600 If the employees work 800 hours, it will cost $15,600 to produce the software. b) Find the number of hours the employees should work in order to minimize the cost. What would the minimum cost be?

The minimum cost occurs at the vertex of the cost function. The x-coordinate is b -20 20 x==- 2(0.04) = 0.08 =250. To find the y-coordinate of the vertex, substitute the 2a value into the function as follows:

C(x) = 0.04x2- 20 x +6000 C(250)= 0.04(250)2 - 20(250) + 6000 C(250) = 2500- 5000+6000 C(250) = 3500 The minimum cost of $3500 is achieved when the employees work 250 hours.

16.

Break-even Analysis. A publisher for a promising new novel figures fixed costs (overhead, advances, promotion, copy, editing, typesetting, and so on) at $67,000 and variable costs (printing, paper, binding, shipping) at $3.50 for each book produced. If the book is sold to distributors for $21 each, how many must be produced and sold for the publisher to break even?

Let x = the number of books produced. Since the break even point is the point when cost is the same as the revenue:

21x = 67,000 +3.50x 17.50x = 67, 000 x = 3828.571429 Therefore, the publisher must produce 3829 books to break even.

1-22

..

Calculus

Chapter 1

Name ________________________________ Date ______________ Class ____________

Section 1-4 Polynomial and Rational Functions Goal: To describe and identify functions that are polynomial and rational in nature

Definition: Polynomial function f(x)=axn+a

xn-1+

+a1x+a0 for n a nonnegative integer, called the degree of the polynomial. The coefficients a0,a1, ,anare real numbers withan ¹ 0. The domain of a polynomial function is the set of all real numbers. n

n-1

Definition: Rational function n(x)

d(x)¹0where n(x) and d(x) are polynomials. The domain is the set of d(x) all real numbers such that d(x)¹0. f(x)=

Vertical Asymptotes: Case 1: Suppose n(x) and d(x) have no real zero in common. If c is a real number such that d(x)=0,then the line x=cis a vertical asymptote of the graph.

Case 2: If n(x) and d(x) have one or more real zeros in common, cancel common linear factors and apply Case 1 to the reduced fraction.

Horizontal Asymptotes: Case 1: If degree n(x) < degree d(x), then y = 0 is the horizontal asymptote.

Case 2: If degree n(x) = degree d(x), then y = a/b is the horizontal asymptote, where a is the leading coefficient of n(x) and b is the leading coefficient of d(x).

Case 3: If degree n(x) > degree d(x), there is no horizontal asymptote.

For 1–6 determine each of the following for the polynomial functions: a. the degree of the polynomial b. the x–intercept(s) of the graph of the polynomial c. the y–intercept of the graph of the polynomial 1-23 ..

Calculus

Chapter 1

1. f(x)=x3-7x-6=(x+2)(x-3)(x+1) a. The degree of the polynomial is the highest exponent, which is 3. b. The x–intercept(s) are found by setting the polynomial equal to zero. The polynomial is already in factored form, therefore if we set the factors equal to zero, the zeros of the polynomial occur at –1, –2, and 3. c. The y-intercept occurs when the x value is zero. If the x value is zero, the only term that will not be zero is the constant term, therefore the y-intercept is (0, –6).

2.

f (x) =x3+4x2-4x-16=(x-2)(x+2)(x+4) a. The degree of the polynomial is the highest exponent, which is 3. b. The x–intercept(s) are found by setting the polynomial equal to zero. The polynomial is already in factored form, therefore if we set the factors equal to zero, the zeros of the polynomial occur at 2, –2, and –4. c. The y-intercept occurs when the x value is zero. If the x value is zero, the only term that will not be zero is the constant term, therefore the y-intercept is (0, –16).

3.

f (x) =x3-3x2-10x+24=(x+3)(x-2)(x-4) a. The degree of the polynomial is the highest exponent, which is 3. b. The x–intercept(s) are found by setting the polynomial equal to zero. The polynomial is already in factored form, therefore if we set the factors equal to zero, the zeros of the polynomial occur at –3, 2, and 4. c. The y-intercept occurs when the x value is zero. If the x value is zero, the only term that will not be zero is the constant term, therefore the y-intercept is (0, 24).

1-24 ..

Calculus 4.

Chapter 1

f(x)=x3+4x2-x-4=(x+4)(x+1)(x-1) a. The degree of the polynomial is the highest exponent, which is 3. b. The x–intercept(s) are found by setting the polynomial equal to zero. The polynomial is already in factored form, therefore if we set the factors equal to zero, the zeros of the polynomial occur at –4, –1, and 1. c. The y-intercept occurs when the x value is zero. If the x value is zero, the only term that will not be zero is the constant term, therefore the y-intercept is (0, –4).

5.

f (x) =x4-2x3+x2+2x-2=(x-1)(x+1)(x2-2x+2) a. The degree of the polynomial is the highest exponent, which is 4. b. The x–intercept(s) are found by setting the polynomial equal to zero. The polynomial is already in factored form. The third factor must be solved using the quadratic equation, therefore, the zeros of the polynomial occur at 1, –1, and 1± 3. c. The y-intercept occurs when the x value is zero. If the x value is zero, the only term that will not be zero is the constant term, therefore the y-intercept is (0, –2).

6.

f (x) =x5+5x4-20x2-x+15=(x+3)(x-1)(x+1)(x2+2x-5) a. The degree of the polynomial is the highest exponent, which is 5. b. The x–intercept(s) are found by setting the polynomial equal to zero. The polynomial is already in factored form. The fourth factor must be solved using the quadratic equation, therefore, the zeros of the polynomial occur at –3, –1, 1, and -1± 6. c. The y-intercept occurs when the x value is zero. If the x value is zero, the only term that will not be zero is the constant term, therefore the y-intercept is (0, –15).

1-25 ..

Calculus

Chapter 1

For the given rational functions in 7–12: a. Find the domain. b. Find any x-intercept(s). c. Find any y-intercept. d. Find any vertical asymptote. e. Find any horizontal asymptote. f. Sketch a graph of y=f(x) for -10£x£10.

7. f(x)=

1

2x a.

The function is defined everywhere except when the denominator is zero. The domain is therefore all real numbers except 0. b.

The x–intercept(s) are found by setting the function equal to 0. Since the function is a rational expression and the numerator cannot be zero, the function value cannot be zero. Therefore, there is no x–intercept. c.

The y–intercept is found when the value of x is zero. Since x = 0 is not in the domain, there is no y–intercept. d.

Vertical asymptotes occur at values where the function is not defined. Since the function is not defined for x = 0, the vertical asymptote is the line x = 0. e.

Horizontal asymptotes are found by dividing all terms by the highest power of 1

1

1 =2x=2x,as x increases or decreases without bound, the x. Therefore, f(x)= 2x 2x 1 2x denominator is always 1 and the numerator tends to 0; so f(x) tends to 0. The horizontal asymptote is the line y = 0. f.

1-26 ..

Calculus

8.

f(x)= a.

Chapter 1

3x x-5 The function is defined everywhere except when the denominator is zero.

x-5=0

b.

c.

x=5 The domain is therefore all real numbers except 5.

The x–intercept(s) are found by setting the function equal to 0. Since the function is a rational expression, the function value is zero when the numerator is zero.

3x=0 x=0 Therefore, the x–intercept is (0, 0).

The y–intercept is found when the value of x is zero.

3x x- 5 3(0) f (0) = 0- 5 0 f (0) = =0 -5 Therefore, the y–intercept is (0, 0).

Vertical asymptotes occur at values where the function is not defined. Since d. the function is not defined for x = 5, the vertical asymptote is the line x = 5. e.

Horizontal asymptotes are found by dividing all terms by the highest power of f (x )=

3x

x. Therefore, f(x)=

3x 3 ,as x increases or decreases without bound, the = x = x- 5 x - 5 1-x 5 x

x

numerator is always 3 and the denominator tends to 1 – 0, or 1; so f(x) tends to 3. The horizontal asymptote is the line y = 3. f.

1-27 ..

Calculus

9.

f(x)= a.

Chapter 1

3x x-3 The function is defined everywhere except when the denominator is zero.

x-3=0

b.

c.

x=3 The domain is therefore all real numbers except 3.

The x–intercept(s) are found by setting the function equal to 0. Since the function is a rational expression, the function value is zero when the numerator is zero.

3x=0 x=0 Therefore, the x–intercept is (0, 0).

The y–intercept is found when the value of x is zero.

3x x- 3 3(0) f (0) = 0- 3 0 f (0) = =0 -3 Therefore, the y–intercept is (0, 0).

Vertical asymptotes occur at values where the function is not defined. Since d. the function is not defined for x = 3, the vertical asymptote is the line x = 3. e.

Horizontal asymptotes are found by dividing all terms by the highest power of f (x )=

3x

x. Therefore, f(x)=

3x 3 ,as x increases or decreases without bound, the = x = x- 3 x - 3 1-x 3 x

x

numerator is always 3 and the denominator is tends to 1 – 0, or 1; so f(x) tends to 3. The horizontal asymptote is the line y=3. f.

1-28 ..

Calculus

10. f(x)=

Chapter 1

2x-4 x+3

a.

b.

c.

The function is defined everywhere except when the denominator is zero.

x+3=0 x=-3 The domain is therefore all real numbers except –3.

The x–intercept(s) are found by setting the function equal to 0. Since the function is a rational expression, the function value is zero when the numerator is zero.

2x- 4 =0 2x = 4 x =2 Therefore, the x–intercept is (2, 0).

The y–intercept is found when the value ofisx zero.

2x - 4 x +3 2(0) - 4 f (0) = 0 +3 -4 f (0) = 3 Therefore, the y–intercept is (0,-4). f (x )=

3

d.

Vertical asymptotes occur at values where the function is not defined. Since the function is not defined for x = –3, the vertical asymptote is the line x = –3. e.

Horizontal asymptotes are found by dividing all terms by the highest power of 2 x - 4 2x - 4x= 2 - 4 =x x,as x increases or decreases without bound, the x. Therefore, f(x)= x+3 x + 3 1+ 3 x x

x

numerator tends to 2 – 0, or 2 and the denominator tends to 1 – 0, or 1; so f(x) tends to 2. The horizontal asymptote is the liney = 2. f.

1-29 ..

Calculus

11.

f(x)=

Chapter 1

4+x 4-x

a.

The function is defined everywhere except when the denominator is zero.

4-x=0

b.

c.

4=x The domain is therefore all real numbers except 4.

The x–intercept(s) are found by setting the function equal to 0. Since the function is a rational expression, the function value is zero when the numerator is zero.

4+x=0 x=-4 Therefore, the x–intercept is (–4, 0).

The y–intercept is found when the value of x is zero.

4 +x 4- x 4 +0 f (0) = 4- 0 4 f (0) = =1 4 Therefore, the y–intercept is (0,1).

Vertical asymptotes occur at values where the function is not defined. Since d. the function is not defined for x = 4, the vertical asymptote is the line x = 4. e.

Horizontal asymptotes are found by dividing all terms by the highest power of 4 4 4 +x =x +x x=x +1 x. Therefore, f(x)= ,as x increases or decreases without bound, the 4 - x 4 - x 4x - 1 f (x )=

x

x

numerator tends to 0 + 1, or 1 and the denominator tends to 0 – 1, or –1; so f(x) tends to –1.

The horizontal asymptote is the line y=-1. f.

1-30 ..

Calculus

12.

f(x)=

Chapter 1

1-5x 1+2x

a.

The function is defined everywhere except when the denominator is zero.

1+2x = 0 2x = - 1 x=- 1 2 1 The domain is therefore all real numbers except2-. b.

The x–intercept(s) are found by setting the function equal to 0. Since the function is a rational expression, when the numerator is zero, the function value is zero.

1-5x = 0 -5x =- 1 x=1 5 Therefore, the x–intercept is (1,0).

5

c.

The y–intercept is found when the value ofisx zero.

1 - 5x f (x )= + 1 2x 1 5(0) f (0) = 1 + 2(0) 1 f (0) = 1 =1 Therefore, the y–intercept is (0,1). d.

Vertical asymptotes occur at values where the function is not defined. Since the function is not defined for x=-1,the vertical asymptote is the line x=-1.

2

2

e.

Horizontal asymptotes are found by dividing all terms by the highest power of 1- 5x 1 - 5 x 1 - 5 x. Therefore, f(x)= 1 + 2 x =x x=x ,as x increases or decreases without bound, the 1 + 2 x 1x + 2 x

x

numerator tends to 0 – 5, or –5 and the denominator tends to 0 + 2, or 2; so f(x) tends - 52to .

The horizontal asymptote is the liney=-5.

2

1-31 ..

Calculus

Chapter 1

f.

13. A video production company is planning to produce a documentary. The producer estimates that it will cost $52,000 to produce the video and $20 per video to copy and distribute the tape. a) Assuming that the total cost to market the video, C(n), is linearly related to the total number, n, of videos produced, write an equation for the cost function.

C(n)=20n+52,000

C(n) b) The average cost per video for an output of n videos is given by C(n)= n .

C(n) 20n+52,000 Find the average cost function. C(n)= = n n c) Sketch a graph of the average cost function for 500 £ n £ 3000 .

The x–axis scale shown is from 500 to 3000. Each tick mark is 500 units. The y–axis scale shown is from 0 to 200. Each tick mark is 50 units.

d) What does the average cost per video tend to as production increases?

To find the value that the function tends to go towards, you will find the horizontal asymptote of the function. Horizontal asymptotes are found by dividing all terms by the 20n 52,000 + 20+ 52,000 n highest power of n. Therefore, C(n)= n = n ,as n increases without n 1 n C(n) bound, the numerator tends to 20 + 0, or 20 and the denominator is always 1; so tends to 20. This means that the average cost per video tends towards $20 each.

1-32 ..

Calculus

Chapter 1

14. A contractor purchases a piece of equipment for $36,000. The equipment requires an average expenditure of $8.25 per hour for fuel and maintenance, and the operator is paid $13.50 per hour to operate the machinery. a) Assuming that the total cost per day, C(h) , is linearly related to the number of hours, h, that the machine is operated, write an equation for the cost function.

C(h)=21.75h+36,000 C(h) b) The average cost per hour of operating the machine is given by C(h)=h .

C(h) 21.75h+36,000 Find the average cost function. C(h)= = h h c) Sketch a graph of the average cost function for 1000£h£8000.

The x–axis scale shown is from 1000 to 8000. Each tick mark is 1000 units.

The y–axis scale shown is from 0 to 100. Each tick mark is 20 units.

d) What cost per hour does the average cost per hour tend to as the number of hours of use increases?

To find the value that the function tends to go towards, you will find the horizontal asymptote of the function. Horizontal asymptotes are found by dividing all terms 21.75n 36,000 36,000 + h 21.75+ h 1 by the highest power of h. Therefore, C(h)= = h ,as h increases h h

without bound, the numerator tends to 21.75 + 0, or 21.75 and the denominator is always 1; so C(h)tends to 21.75. This means that the average cost per hour tends towards $21.75 as the number of hours of use increases.

1-33 ..

Calculus

Chapter 1

15. The daily cost function for producing x printers for home computers was determined to be:

C(x)=x2+8x+6000 C( x) The average cost per printer at a production level of x printers per day is C(x)= . x a) Find the average cost function.

C(x) x2 +8 x +6000 C(x) = x = x b) Sketch a graph of the average cost function for25 £ x £ 150 .

The x–axis scale shown is from 25 to 150. Each tick mark is 25 units. The y– axis scale shown is from 50 to 400. Each tick mark is 50 units.

c) At what production level is the daily average cost at a minimum? What is that minimum value?

Based on the graph above, the minimum value occurs around the third tick mark, which has a value of 75. By substituting values into the average value equation, we would have the following:

C(75) =163 C(76) =162.947 C(77) =162.922 C(78) =162.923 C(79) =162.949 Therefore, the minimum average cost of 162.92 occurs when 77 printers are produced.

1-34 ..

Calculus

Chapter 1

16. The monthly cost function for producing x brake assemblies for a certain type of car is given by: C(x)=3x2+36x+9000 The average cost per brake assembly at a production level of x assemblies per month is C( x) C(x)= x . a) Find the average cost function.

C(x) 3x2 +36x+9000 = x x b) Sketch a graph of the average cost function for 0£x£150.

C(x)=

The x–axis scale shown is from 0 to 150. Each tick mark is 25 units. The y– axis scale shown is from 300 to 500. Each tick mark is 50 units.

c) At what production level is the daily average cost at a minimum? What is that minimum value?

Based on the graph above, the minimum value occurs just beyond the third tick mark, which has a value of 50. By substituting values into the average value equation, we would have the following:

C(53) = 364.811 C(54) = 364.667 C(55) = 364.636 C(56) = 364.714 Therefore, the minimum average cost of 364.63 occurs when 55 brake assemblies are produced.

1-35 ..

Calculus

Chapter 1

1-36 ..

Calculus

Chapter 1

Name ________________________________ Date ______________ Class ____________

Section 1-5 Exponential Functions Goal: To describe and solve functions that are exponential in nature

Rules for Exponents:

a m ×a n = am+nProduct Rule

a 0 =1, a ¹ 0 Zero Exponent Rule

am

n = amn Power Rule (am)

a

am-n n =

Quotient Rule

In problems 1–8, describe in words the transformations that can be used to obtain the graph of g(x) from the graph of f(x).

1. g(x)=4x+3-4; f(x)=4x The function f is shifted 3 units to the left and 4 units down.

2. g(x)=-3x-5; f(x)=3x The function f is reflected over the x–axis and shifted 5 units down.

3. g(x)=2x-4-6; f(x)=2x The function f is shifted 4 units to the right and 6 units down.

4. g(x)=-5x-3+2; f(x)=5x The function f is shifted 3 units to the right, reflected over the x–axis, and shifted 2 units up.

1-37 ..

Calculus

Chapter 1

5. g(x)=10x-2-5; f(x)=10x The function f is shifted 2 units to the right and 5 units down.

6. g(x)=-10x-3; f(x)=10x The function f is reflected over the x–axis and shifted 3 units down.

7. g(x)=ex+1+2; f(x)=ex The function f is shifted 1 unit to the left and 2 units up.

8. g(x)=-ex+5; f(x)=ex The function f is reflected over the x–axis and shifted 5 units up.

In Problems 9–20, solve each equation for x.

2 10. 10 x =102x+8

2 3 105x+4 9. 10x=

x2 = 2 x+ 8

2x - 3 = 5 +4 - 7 =x 7 3 - =x 3 x

x2 - 2 x - 8 = 0 (x- 4)( x + 2 )= 0 x = - 2,4

2

12. 8 x = 88 x

13. ( x + 6)3 = (3x - 8)3

x2 = 8x

x + 6 =3x- 8 1 4 = 2x 7=x

x2 - 8x = 0 x(x- 8) = 0 x = 0,8

1-38 ..

2 11. 65x-4 =6x

5x - 4 =x2 x2 -5x + 4 =0 ( x - 4)( x -1)=0 x =1,4

14. (2x - 7)5 = ( x +1)5

2x - 7 = x + 1 x=8

Calculus

Chapter 1 16. (ex)4=ex2

15. (e3)4 = ex

17. (e2x)x=e15+x

2x 2 =15 + x

4x = x2

3(4) = x 12 = x

x2 - 4x = 0 0 x(x - 4) = 0,4 x=

2 x2 - x-15 = 0 (2x + 5)(x-3) = 0 = - 5 ,3

x

2

2

2

2

20. 9 x ×9=92 x

19. 2 x = 212x×2-32

18. 3x ×34 = 33 x

x + 4 =3 2

x+1= 2 2

x2 =12x- 32 x - 12 x + 3 2= 0 2 - 4)( x-8) = 0 ( x = 4,8 x

3x2- x - 4 = x (3x- 4)( x +1)=0 x = 0 ,- 1 4 3

INTEREST FORMULAS Simple Interest:

2x2- x -1= x (2x+1)(x -1)= 0 x = 0- 21 ,1

A = P (1+ rt)

Compound Interest:

mt æ rö A=Pç1+ m÷ ø è

Continuous Compound Interest:

A=Pert

where P is the amount invested (principal), r (expressed as a decimal) is the annual interest rate,t is time invested (in years), m is the number of times a year the interest is compounded, and A is the amount of money in the account after t years (future value).

1-39 ..

Calculus

Chapter 1

(Round answers for 21 –28 to the nearest dollar) 21. Fred inherited $35,000 from his uncle. He decides to invest his money for 5 years in order to have the greatest down-payment when he buys a house. He can choose from 3 different banks.

Bank A offers 1% compounded monthly.

Bank B offers .5% compounded continuously.

Bank C offers .75% compounded daily.

Which bank offers the best plan so Fred can earn the most money from his investment? mt

Bank A:

æ rö A=P1ç + ÷mø è æ 12 ( 5) 0.01 ö ç + A = 35,000 1 ÷ è 12 ø

Bank B:

)60 A = 35000 (1.000833 A » $36, 794

A = Pert 35, A= 000e A= 35, A» 000e $35, 886

( 0.005)( 5) 0.025

mt

Bank C:

æ rö A = Pç1+ ÷ m ø è æ 365(5) 0.0075 ö A = 35,000ç1 + 365 ÷ è ø

)1825 A = 35, 000 (1.000021 A » $36, 337 Therefore, Bank A is the best option.

22. The day your first child is born you invest $10,000 in an account that pays 1.2% interest compounded quarterly. How much will be in the account when the child is 18 years old and ready to start to college? mt

æ rö A = Pç1 + ÷mø è

4(18)

æ 0.012 ö A =10,000 1 4 ÷ ø çè + )72 A =10,0 001.003 ( A » $12, 407

1-40 ..

Calculus

Chapter 1

23. When your second child is born, you are able to invest only $5000 but the account pays 1% interest compounded daily. How much will be in the account when this child is 18 years old and ready to start to college? mt

æ rö A = Pç1 + ÷mø è

365(18)

æ 0.01 ö A = 5000 1 ÷ çè + 365ø

)6570 A = 5000 (1.000027 A » $5986 24. When your third child comes along, money is even tighter and you are able to invest only $1000, but you are able to find a bank that will let you invest the money at 1.75% compounded continuously. How much will be in the account when this third child is 18 years old and ready to start to college?

A = Pert A =1000e( 0.0175)(18) A =1000e0.315 A » $1370 25. Joe Vader plans to start his own business in ten years. How much money would he need to invest today in order to have $25,000 in ten years if Joe’s bank offers a 10-year CD that pays 1.8% interest compounded monthly. mt

æ r ö A = Pç+ 1 ÷ è mø æ 0.018 12 (10 ) ö 25, 000= P ç1 + ÷ è 12 ø 120

25, 000= P (1.0015) P » $20, 885

1-41 ..

Calculus

Chapter 1

26. Bill and Sue plan to buy a home in 5 years. How much would they need to invest today at 1.2% compounded daily in order to have $30,000 in five years? mt

æ r ö A = Pç+ 1 ÷ è mø æ 0.012 365( 5) ö 30, 000= P ç1 + 365 ÷ è ø

)1825 30, 000= P (1.000088 P » $28, 253 27. Suppose you invest $3000 in a four-year certificate of deposit (CD) that pays 1.5%interest compounded monthly the first 3 years and 2.2% compounded daily the last year.

What is the value of the CD at the end of the four years? mt

mt

æ rö æ rö A = Pç1 + ÷mø ç1 + ÷mø è è

12 ( 3) 365(1)

æ 0.022 ö 1 365 ÷ ç+è ø

æ 0.015 ö A = 3000 èç1 + ÷ 12 ø

)36(1.000060)365 A = 3000(1.00125 A » $3208

28. Suppose you invest $8000 in a 10-year certificate of deposit (CD) that pays 1.25% interest compounded daily the first 6 years and 2% compounded continuously the last four years. What is the value of the CD at the end of the 10 years? mt

æ rö A=P1ç + ÷mø ert è 365( 6 )

æ 0.0125 ö A = 8000 1 365 ÷ ø çè +

e( 0.02 )( 4 )

e 0.08 )2190 A = 8000 (1.000034 A » $9341

1-42 ..

Calculus

Chapter 1

Name ________________________________ Date ______________ Class ____________

Section 1-6 Logarithmic Functions Goal: To solve problems that are logarithmic in nature

Properties of Logarithms

lo a x + loga y = loga x y g a x - loga y = loga x y lo loga xn = nloga x

Exponential Function: f(x) = a x , a > 0, a ¹ 1 y=logax means x=ay

g In Problems 1–20 find the value ofx. (Evaluate to four decimal places if necessary.)

1. log3 x = 4

2. log3 ( x+1)= 2

x = 34 x = 81

x +1=3 x +1=2 x= 9 8 6 log 22 = 4 - 3x 4.

3. log3 38 = 7 + 3x

2 4-3x = 2 4 - 3x = 6 - 3x = 6 2 x=- 2 3

37+3x= 38 7 + 3x = 8 3x =1 x=1 3

1-43 ..

Calculus

Chapter 1

6. logx(2x+3)=2

5. ln(x+6)=2

x2 = 2 x+3

e2 = x +6

x2 - 2 x - 3 = 0 (x - 3)( x+1) = 0 x =3 x¹ 1 The log function cannot have a negative base.

e2 - 6 = x 1.3891 » x

7. log 2 (7 -x19) = log 2 (2 + x 9)

8. log3(8 - 5 x=log3(2x-13) )

7 x -19 = 2 x + 9

8 - 5x = 2 -13

5x = 2 8 x =5.6

21 = x 3=7 Since thex value of 3 creates a negative log, there is no solution. x

9. ln x + ln3 = 3

10. ln x-ln3=1

ln3x = 3

lnx =1

e3 = 3x

e31 =3 x

e3 = x 3 6.6952 » x

3e = x 8.1548 » x

11. log( x-1)- log4 = 3

12. ln(x -1)- ln6 = 2

log 4

x-1

103

x- 1

=3

ln 6 = 2 e2 x= - 1

x-1

=4

6

(4)103 = x 1 4(1000) +1 = 4001 = x x

6e2= x 1 6e2 +1 = 45.3343 » x x

1-44 ..

Calculus

Chapter 1

14. 5x-1=17

13. 23x=12

x-1 = log5 log 17 log (x - 1) log 5= 17 log 17 5 log17+log5 x log 5 - log = x log 5 =

log23x = log 3xlog2 = 12 x = log 12 312log 2 x» 1.1950

x = log17+log5 log5 x» 2.7604

15. 7 x- 1 = 8 x

16. 42 x+3 = 5 x- 2

log7x-1 = log8x 7 x log (x - 1) log = x log 7 - log =7 8 x x log7-xlog8 = log 8 x(log7- log8)= 7 x = log log7 7log 7 - log 8 x » - 14.5727 17. 7 x+1 =102x

log42x+3= log5x-2 (2x+3)log4 = (x- 2)log5 2xlog4+3log4 = xlog5- 2log5 2xlog4-xlog5 =- 2log55) 3log4 x(2 log 4 - log =x = - 2log5- 3log4 3log4 2log4-log5 » x 6.3429

18. e x+4 =14.654

log 7 x +1 = log 102 (x+1)log7 = x 2 x xlog7+log7 = log 10 2 log 10 x log7- 2xlog10 =-x log x(log 7 - 2 log10)=- 7 x = log- log7 7 2log10 log7x » 0.7317

ln ex+4 =ln(14.654) x + 4 =ln(14.654) x =ln(14.654)- 4

19. x + 5 =e3

20. e 4 x = e20

x = e3-5 x » 15.0855

x » - 1.3153

4x = 20 x =5 1-45 ..

Calculus

Chapter 1

21. You want to accumulate $20,000 by your son’s eighteenth birthday. How much do you need to invest on the day he is born in an account that will pay 1.4% interest compounded quarterly? (Round your answer to the nearest dollar.) mt æ r ö A = Pç+ 1 ÷ è mø æ 0.014 4(18) ö 20, 000= P ç1 + 4 ÷ è ø

20, 000 = P (1.0035)72 $15, 552» P 22. Using the information in Problem 21, how much would you need to invest if you waited until he is 10 years old to start the fund? mt

æ r ö A = Pç+ 1 ÷ è mø æ 0.014 4(8) ö 20, 000= P ç1 + 4 ÷ è ø 20, 000 = P (1.0035)32 $17, 884» P 23. A bond that sells for $1000 today can be redeemed for $1200 in 10 years. If interest is compounded quarterly, what is the annual interest rate for this investment? (Round your answer to two decimal places when written as a percentage.)

1-46 ..

Calculus

Chapter 1 mt æ rö A = P ç1+÷ m ø è 4(10)

æ rö 1200 = 1000 ç1 + è ÷4ø 40

æ rö 1.2 = ç1+ ÷ ø è 4 1

1.2 40 =1 + 1

1.2 40 - 1=

r 4

r 4

1

4(1.2 40 - 1)= r 1.83% »r 24. A bond that sells for $18,000 today can be redeemed for $20,000 in 6 years. If interest is compounded monthly, what is the annual interest rate for this investment? (Round your answer to two decimal places when written as a percentage.) mt

æ rö A = P ç1+÷ m ø è 12(6)

æ r ö 20, 000=18,000 1 12 ÷ çè + ø 72

r ö æ 1.111 = ç1+ ÷ ø è 12 1

1.111 72 =1 + 1

1.111 72 - 1= 1

r 12

r 12 r

12(1.111 72 - 1)= 1.76% »r 25. What is the minimum number of months required for an investment of $10,000 to grow to at least $20,000 (double in value) if the investment earns 1.8% annual interest rate compounded monthly? What would be the actual value of the investment after that many months?

1-47 ..

Calculus

Chapter 1 mt

æ rö A= Pç1+ ÷ m ø è æ 12t 0.018 ö ç + 20, 000=10,000 1 ÷ è 12 ø 2 = (1.0015)12t

)12t ln2 = ln1.0015 ( ln2 =12t ln(1.0015) ln2 =t 12 ln 1.0015 38.5 » t It will take about(38.5)(12) = 462 months to double.

r ö mt æ ÷ mø ç + A=P1 è 12 ( 38.5)

æ 0.018 ö A =10,000 1 ÷ çè + 12 ø

)462 A =10, 000 (1.0015 A =10, 000(1.998668) A » $19, 987 26. What is the minimum number of months required for an investment of $10,000 to grow to at least $30,000 (triple in value) if the investment earns 1.8% annual interest rate compounded monthly? What would be the actual value of the investment after that many months?

1-48 ..

Calculus

Chapter 1 mt

æ rö A= Pç1+ ÷ m ø è 12t

æ 0.018 ö 30, 000=10,000 1 ÷ çè + 12 ø 3 = (1.0015)12t

)12t ln3 = ln1.0015 ( ln3 =12t ln(1.0015) ln3 =t 12 ln 1.0015 61.1 » t It will take about(61.1)(12) = 733 months to triple. mt æ rö A = Pç1 + ÷ m ø è

12 ( 61.1)

æ 0.018 ö A =10,000 1 ÷ çè + 12 ø

)733 A =10, 000 (1.0015 A =10, 000(3.000192) A » $30, 002

27. Some years ago Ms. Martinez invested $7000 at 2% compounded quarterly. The account now contains $10,000. How long ago did she start the account? (Round your answer UP to the next year.)

1-49 ..

Calculus

Chapter 1 mt

æ rö A = P ç 1 ÷mø + è

4t

æ 0.02 ö 10, 000 = 7000 ç1 + 4 ÷ è ø 10 = 7

(1.005)4t

)4t l 10 = ln1.005 (

7 10 n = 4tln(1.005) 7 l ln 10 7 =t

4 ln 1.005 n 17.88 » t

It took approximately 18 years to have a balance of $10,000.

28. Some years ago Mr. Tang invested $18,000 at 3% compounded monthly. The account now contains $24,000. How long ago did he start the account? (Round your answer UP to the next year.)

mt

æ rö A = Pç1+ ÷ m ø è 12t

æ 0.03 ö 24, 000 =18,000 ç1 + è 12 ÷ ø

)12t

4 = 1.0025 3

(

)12t l 4 = ln1.0025 ( 3

n 4 =12t ln(1.0025) ln4 l

3

=t 3 12 ln 1.0025 n 9.60 » t It took approximately 10 years to have a balance of $24,000.

29. In a certain country the number of people above the poverty level is currently 25 million and growing at a rate of 4% annually. Assuming that the population is growing continuously, the population, P (in millions), t years from now, is determined by the formula: 1-50 ..

Calculus

Chapter 1

P=25e0.04t In how many years will there be 30 million people above the poverty level? 40 million?

(Round your answers to nearest tenth of a year.)

30 million people

40 million people

P = 25e0.04

P = 25e0.04

= 25et 30 0.04 = e0.04t 1.2

= 25et 40 0.04 = e0.04t 1.6

t

t

= lne 0.04t ln 1.2 = 0.04t ln 1.2 ln 1.2 =t 0.04 4.6 » t

= lne 0.04t ln 1.6 = 0.04t ln 1.6 ln 1.6 =t 0.04 11.8 » t

It will take approximately 4.6 years to reach 30 million people and 11.8 years to reach 40 million people.

30. The number of bacteria present in a culture at time t is given by the formula N=20e0.35t, where t is in hours. How many bacteria are present initially (that is when t = 0)? How many are present after 24 hours? How many hours does it take for the bacteria population to double? (Round your answers to nearest whole number.)

1-51 ..

Calculus

Chapter 1

Initially there are N=20e0.35(0)=20e0=20bacteria present.

After 24 hours there will be N=20e0.35(24)=20e8.4=20(4447.066748)=88,941bacteria present.

N = 20e0.35 40 = 20et 2 = 0.35t 0.35 e ln2 = 0.35t t lne ln2 = 0.35t ln2 =t 0.35 1.98 » t The number of bacteria will double after approximately 2 hours.

1-52 ..

Calculus

Chapter 2

Name ________________________________ Date ______________ Class ____________

Section 2-1 Introduction to Limits Goal: To find limits of functions

Definition: Limit We write

limf(x)=L or f(x)®Las x®c

x®c

if the functional value f(x)is close to the single real number L whenever x is close, but not equal, to c (on either side of c).

Definition:

One sided limits

limf(x)=Lis the limit of the function as x approaches the value c from the left.

x®c-

limf(x)=Lis the limit of the function as x approaches the value c from the right.

x®c+

Properties of Limits: 1.

2.

limk=k for any constant k

x® c

limx=c

x® c

3.

lim[f(x)+g(x)]=limf(x)+limg(x) x® c

x® c

4.

5.

6.

x®c

lim[f(x)-g(x)]=limf(x)-limg(x) x® c

x® c

x®c

limkf(x)=klimf(x) for any constant k x® c

x® c

lim[f(x)×g(x)]=limf(x)×limg(x) x® c

x® c

x®c

lim f(x)

7.

8.

æxf ( ) ö c lim ç limg(x)¹0 ÷= lim g(x) provided x®c x®c è g ( x) ø x® x® c lim n f (x) = n limf (x) (the limit value must be positive for n even.) x® c

x® c

2-1 ..

Calculus

Chapter 2

1 - 5 Find each limit if it exists 1.

lim(7x+2)=(7(2)+2)=14+2=16 x®2

2.

3.

lim5x=5(-3)=-15

x®-3

limx(4x+1)=limx×lim(4x+1)=5[4(5)+ x®5

x®5

4.

x®5

1)] = 5(21) =105

lim x+8 æx+8ö x®-1 -1+8 7 lim ç ÷= = -1+2 = 1 =7 limx+2 x® -1x+2ø è x®-1

5.

lim

-5x+1=

4

-5(-3)+1=15+1=16=

x®-3

6 - 8 Find the value of the following limits given that

limf(x)=4 and limg(x)=-3. x®3

6.

x®3

lim7f(x)=7×4=28 x®3

7.

18

lim[3f(x)-2g(x)]=3(4)-2(-3)=12+6= x®3

8.

æ2f(x)ö 2(4) 8 8 lim ç ÷= = =x®3è3g(x) ø 3(- 3) - 9 9

2-2 ..

Calculus

9.

Chapter 2

ìï x2 + 2 if x <1 Let f (x) =í . Find: ïî 4x - 1 if x ³ 1 a)

limf ( x)

x®1-

Since we are looking for the left–hand limit, use the top function.

limf(x)=lim(x2+2)=((1)2+2)=3 x®1-

b)

x®1-

limf ( x)

x®1+

Since we are looking for the right–hand limit, use the bottom function. limf(x)=lim(4x-1)=(4(1)-1)=3

x®1+

c)

x®1+

limf (x Since the left– and right–hand limits are the same value,

x® 1

limf )

=3.

x® 1

(x d)

f(1)

)

The value of 1 is defined in the bottom function, therefore f(1) = 4(1)- 1 = 3.

10.

Let f (x)=í a)

ì5x-6

if x£2

î2x+4

if x>2

. Find:

limf(x)

x®2-

Since we are looking for the left–hand limit, use the top function. limf(x)=lim(5x-6)=5(2)-6=4

x® 2-

b)

x® 2-

limf ( x)

x®2+

Since we are looking for the right–hand limit, use the bottom function. limf(x)=lim(2x+4)=(2(2)+4)=8

x® 2+

x® 2+

2-3 ..

Calculus c)

Chapter 2

limf ( x)

x®2

Since the left– and right–hand limits are not the same value, limf(x) does x®2

not exist. d)

f(2) f(2) = 5(2) - 6 = 4.

The value of 2 is defined in the top function, therefore

11.

æx2+4x-5ö ÷. Find Let f(x)= ç è x- 1 ø a)

limf(x) x®1

Substituting in a value of 1 will result in a zero in the denominator, therefore we must try to remove the problem by first simplifying the function:

f(x)=

x2 +4x- 5 x- 1

=

(x+5)(x- 1) =x+5, x¹1 x- 1

Therefore, limf(x)=lim(x+5)=1+5=6 x®1

b)

c)

x®1

x2 +4x- 5 (-1)2 +4(- 1)- 5 - 8 = = =4 x- 1 - 1- 1 x® -1 -2

limf(x)=lim

x® -1

x 2 + 4 x - 5 ( - 5 ) 2 + 4 ( - 5 ) - 50 = = =0 x® -5 x- 1 - 5- 1 -6

limf(x)=lim

x® -5

2-4 ..

Calculus

12.

Chapter 2

æx- 2 ö ç Let f(x) = èx÷. Find 2ø a)

limf(x)

x®2-

As x approaches 2 from the left side, the value of the numerator will be positive and the value of the denominator will be negative (because the x values are smaller than 2). The limit will therefore, be a value of –1 (because they are opposite signs). b)

limf(x)

x®2+

As x approaches 2 from the right side, the value of the numerator will be positive and the value of the denominator will be positive (because the x values are larger than 2). The limit will therefore, be a value of 1 (because they are the same sign). c)

limf(x) does not exist because the left–hand and right–hand limits are

x®2

different values.

2-5 ..

Calculus

Chapter 2

2-6 ..

Calculus

Chapter 2

Name ________________________________ Date ______________ Class ____________

Section 2-2 Infinite Limits and Limits at Infinity Goal: To find limits of functions as they approach infinity

Limits of Power Functions at Infinity: If p is a positive real number and k is any real number except 0, then 1.

k lim =0 x®-¥xp

3.

limkxp=±¥

2.

k =0 lim x®¥xp

limkxp=±¥ 4.

x®-¥

x®¥

provided that xp is a real number for negative values of x. The limits in 3 and 4 will be either positive or negative infinity, depending on k and p.

Limits of Rational Functions at Infinity:

If f ()x=

a x m +m-1x a m - 1+ m

n- 1

bxn+b n-1x n

+

+a1x+a 0,a +b1x+b0

m ¹0,b n ¹0 ,

There are three cases to consider: amxm amxm then limf(x)=lim and lim f(x)=lim x ® ¥ bnx n x®¥ x®-¥ x ® - ¥ bnxn

1.

If m<n,then limf (x) = lim f (x) = 0.

2.

3.

If m = n, limf(x)=limf(x)=

x® ¥

x® - ¥

am. x®-¥ bn x®¥ If m>n,then the limit will be ¥ or - ¥ , depending on the values of m, n,

am,and bn.

2-7 ..

Calculus

Chapter 2

1 - 3 Find each limit. Use¥ or - ¥ when appropriate.

1.

f(x)= a)

x x+4 lim x® -4-

f(x)

If the value is substituted into the function, we would have a zero in the denominator.

Since there is no way to remove the problem, we will look at what happens around the value of –4 when approached from the left. These values are smaller than –4, therefore the denominator will approach 0 from the negative side (that is, the denominator will always be negative). Since the numerator will always be negative, lim f(x)=¥. x® -4-

b)

lim f(x) x® -4+

If the value is substituted into the function, we would have a zero in the denominator.

Since there is no way to remove the problem, we will look at what happens around the value of –4 when approached from the right. These values are larger than –4, therefore the denominator will approach 0 from the positive side (that is, the denominator will always be positive). Since the numerator will always be negative, lim f(x)=-¥. x® -4+

c)

limf(x) does not exist because the left–hand and right–hand limits are

x®-4

different infinite limits.

2.

f(x)= a)

4x+1 (x- 5)2 limf(x)

x®5-

If the value is substituted into the function, we would have a zero in the denominator.

Since there is no way to remove the problem, we will look at what happens around the value of 5 when approached from the left. These values are smaller than 5, but the denominator value is being squared and will always be positive. Since the value of the numerator is 21 as the x value approaches 5, limf(x)=¥. x®5-

2-8 ..

Calculus b)

Chapter 2 limf(x)

x®5+

If the value is substituted into the function, we would have a zero in the denominator.

Since there is no way to remove the problem, we will look at what happens around the value of 5 when approached from the right. These values are larger than 5, but the denominator value is being squared and will always be positive. Since the value of the numerator is 21 as the x value approaches 5, limf(x)=¥. x®5+

c)

limf(x)

= ¥ because the left– and right–hand limits both approach infinity.

x®5>

3.

f(x)= a)

x2 +3x- 5 x+5 lim f(x)

x® -5-

If the value is substituted into the function, we would have a zero in the denominator.

Since there is no way to remove the problem, we will look at what happens around the value of –5 when approached from the left. These values are smaller than –5, therefore the denominator will approach 0 from the negative side (that is, the denominator will always be negative). Since the numerator is positive around –5, lim f(x)=-¥. x® -5-

b)

lim f(x) x® -5+

If the value is substituted into the function, we would have a zero in the denominator.

Since there is no way to remove the problem, we will look at what happens around the value of –5 when approached from the right. These values are larger than –5, therefore the denominator will approach 0 from the positive side (that is, the denominator will always be positive). Since the numerator is positive around –5, lim f(x)=¥. x® -5+

c)

limf(x) does not exist because the left–hand and right–hand limits are

x®-5

different infinite limits.

2-9 ..

Calculus

Chapter 2

4 - 6 Find each function value and limit. Use ¥ or - ¥ where appropriate.

4.

f(x)= a)

4x+1 6x-3 f(20) f(20)=

4(20)+1

=

6(20)- 3 b)

81 »0.6923 117

f(200) 4(200)+1

f(200)=

=

6(200)- 3 c)

801 »0.6692 1197

4 2 m = n, the 6 = 3 because the function is a rational expression and

limf(x)=

x®¥

limit is the ratio of the coefficients.

5.

x-5

f(x)=

3x2+2x+2 a)

f(10)

10- 5

f(10)=

3(10)2 +2(10)+2 b)

5

=

322

f(100)

100- 5

f(100)=

95

=

3(100)2 +2(100)+2 c)

»0.0155

»0.0031

30,222

limf(x)=0 because the function is a rational expression and m < n.

x®¥

2-10 ..

Calculus

6.

f(x)= a)

Chapter 2

x3 + 2x -1

3-8 x f(-10) f(-10)=

b)

f(-100)

f(-100)= c)

(- 10)3 +2(- 10)- 1 - 1021 »-12.3012 = 3- 8(-10) 83

(- 100)3 +2(- 100)- 1 3- 8(-100)

=

- 1,000,201

»-1245.5803

803

limf(x)=-¥ because the function is a rational expression,

x®-¥

m > n, and

when x becomes increasingly negative, the value of the function will be increasingly negative.

7-9

Find the vertical and horizontal asymptotes for the following functions.

7.

f(x)=

5x x+4

Vertical asymptotes are found by setting the denominator equal to zero. Therefore, there would be a vertical asymptote atx + 4 = 0Þ x = - 4.

Horizontal asymptotes are found by finding the limit of the function as it goes to infinity. The limit of this function is 5 becausem = nand the limit is the ratio of the coefficients. Therefore, the horizontal asymptote is y=5.

8.

f(x)=

x2-3 x2+4

Vertical asymptotes are found by setting the denominator equal to zero. Since the denominator cannot be zero, there are no vertical asymptotes.

Horizontal asymptotes are found by finding the limit of the function as it goes to infinity. The limit of this function is 1 becausem = nand the limit is the ratio of the coefficients. Therefore, the horizontal asymptote is y=1.

2-11 ..

Calculus

9.

f ( x) =

Chapter 2

x2 + 4 x - 5 x2 +8 x +15

The function can be reduced as follows: f(x)=

x2 +4x- 5

(x+5)(x- 1) x- 1 , x ¹ - 5 and x ¹ - 3 = = x2 +8x+15 (x+5)(x+3) x+3

Vertical asymptotes are found by setting the denominator equal to zero. Therefore, there would be a vertical asymptote atx + 3 = 0Þ x = - 3.

Horizontal asymptotes are found by finding the limit of the function as it goes to infinity. The limit of this function is 1 because m=nand the limit is the ratio of the coefficients. Therefore, the horizontal asymptote is y=1.

2-12 ..

Calculus

Chapter 2

Name ________________________________ Date ______________ Class ____________

Section 2-3 Continuity Goal: To determine if functions are continuous at specific points and intervals

Definition: Continuity A function f is continuous at the point x = cif 1.

limf(x)exists

x®c

2.

f (c) exists

3.

limf (x) = f (c)

x®c

Continuity Properties: 1.

A constant functionf (x) = k, where k is a constant, is continuous for all x.

2.

3.

4.

For n a positive integer,f (x) = xn is continuous for all x.

A polynomial function is continuous for all x.

A rational function is continuous for all x except those values that make a denominator 0.

For n an odd positive integer greater than 1, nf(x)is continuous wherever f ( x) is continuous.

5.

6.

For n an even positive integer,n f ( x) is continuous wherever f(x)is continuous and nonnegative.

Constructing Sign Charts: 1.

Find all partition numbers. These are all the values that make the function discontinuous or 0.

2.

Plot the numbers found in step 1 on a real–number line, dividing the number line into intervals.

3. f ( x) at each test value to Select a test value in each open interval and evaluate determine whether f ( x) is positive or negative.

4.

Construct a sign chart, using the real–number line in step 2.

2-13 ..

Calculus

Chapter 2

1 - 5 Using the continuity properties, determine where each of the functions are continuous.

1.

f(x)=3x3-4x2+x+7 Since the function is a polynomial function, it is continuous for all x.

2.

f(x)=

x3 +4x2 +3x- 7 x2 - 15x+26

Since the function is a rational function, it is continuous for all x except when the denominator is 0. Therefore, x2-15x+26=0,or ( x - 2)(x -13) = 0, when x=2or x=13.

So the function is continuous for all x except x=2or x=13.

3.

f(x)=

3x-2 x2+3

Since the function is a rational function, it is continuous for all x except when the denominator is 0. The denominator cannot have a value of 0, therefore the function is continuous for all x.

4.

f(x)=3x-10

Since the function is an odd positive root, the function is continuous for all values of x where the radicand is continuous. The radicand is a polynomial function, therefore the function is continuous for all x.

5.

f(x)=

x2-25

Since the function is an even root, the function is continuous for all values of x where x2 - 25³ 0 and the points in the radicand is continuous and nonnegative. Therefore, question are ±5.For all values between –5 and 5, the radicand would be negative, therefore, ( - ¥ , - 5] È [5, ¥ ). the function is continuous on the interval

2-14 ..

Calculus

Chapter 2

6 - 7 Use a sign chart to solve each inequality. Express answers in inequality and interval notation.

4x2-29x+7<0

6.

Find the partition numbers: 4x2-29x+7=0 (4x- 1)(x- 7)=0 1 and 7.

Therefore, the partition numbers are 4

The number line would be broken into three parts involving the intervals (-¥,1),(1,7),and(7,¥). Pick test values. We choose 0, 1, and 8 for the test values. Find 4

4

the function values using these three test values.

f (0) = 4(0) 2 -29(0)+7 f (0) = 7

f (1 = 4(1)2 - 29(1) +7 f (8 = 4(8)2 - 29(8) + 7 f ) = 4 - 29+7 f ) = 256 232 + 7 f (1 =- 18 f (8 = 31 ) ) (1 (8 Only the test value 1 makes the inequality true, so the solution14is< x < 7, or (14 ,7). ) )

2-15 ..

Calculus

7.

Chapter 2

x3 +3x2 x +5

>0

Find the partition numbers:

x3 + 3x2 = 0

x+5=0

x2 ( x +3) = 0 Therefore, the partition numbers are 0, –3, and –5.

The number line would be broken into four parts involving the intervals (-¥,-5),(-5,-3),(-3,0),and(0,¥). Pick test values. We choose –6, –4, –1, and 1 for the test values. Find the function values using these four test values.

f(-6)=

f (- 1) =

(-6) 3 +3(-6) 2 - 108 = =108 -6+5 -1 (-1)3 +3(- 1)2 -1+5

=

2

=0.5

4

f(-4)=

f(1) =

(-4) 3 +3(-4) 2 - 16 = =-16 1 -4+5

(1)3 + 3(1)2 4 = »0.67 1+5 6

The test values of –6, –1, and 1 make the original inequality true, so the solution is x < - 5 or x>-3,or(-¥,-5)È(-3,¥).

2-16 ..

Calculus

Chapter 2

Name ________________________________ Date ______________ Class ____________

Section 2-4 The Derivative Goal: To find the first derivative of a function using the four step process.

Definition: Average Rate of Change

For y=f(x),the average rate of change from x=a to x=a+his f(a+h)- f(a) (a+h)- a

=

f(a+h)- f(a) h

,h¹0

where h is the distance from the initial value of x to the final value of . x Definition: Instantaneous Rate of Change

For y=f(x),the instantaneous rate of change atx=ais f(a+h)h®0 h lim

f(a)

, if the limit exists.

(a, f(a )) and to find This formula is also used to find the slope of a graph at the point the first derivative of a function,f (x).

Procedure: 1.

2.

3.

4.

Finding the first derivative: Find f ( x + h).

Find f ( x + h) - f (x).

Find . f(x+h)- f(x) h f(x+ )h f(x) Find lim h h®0

.

2-17 ..

Calculus

Chapter 2

1 - 3 Use the four step procedure to find f¢ (x) and then find f ¢(1), f ¢(2),andf ¢(3).

1.

f(x)=7x-3 Step 1:

f(x+h)=7(x + h) - 3 = 7x + 7h-3

Step 2:

f(x+h)-f(x)

Step 3:

f(x+h)- f(x) 7h =7 = h h

Step 4:

f(x+h)h®0 h lim

= 7 x + 7h - 3 - (7x- 3) = 7 h

f(x)

= lim 7 = 7 h®0

¢(3) = 7.

f¢(x)=7,f¢(1)=7,f¢(2)=7,and f

2.

f(x)=-3x2+5x-6 Step 1:

f(x+h) = - 3 x + h)2 +5(x + h) - 6

= - ( x2 + 2xh + h2 ) + 5(x + h) - 6 f ( x + h ) = -3x2 - 6xh - 3h2 + 5x + 5h - 6 Step 2:

( f ( x + h - f (x = 3x2 - 6 xh - 3h2 + 5 x + 5h - 6 - (-3x2 + 5 x - 6) f ( x + ) - f ) - 6xh - 3h2 + 5h

h (x = f ( x + h) - f (x) - 6xh- 3h2 + 5h ) ) -= Step 3: h h f ( x + h) - f (x) = - 6 x - 3h+5 h x f ( + h)- f (x) Step 4: lim = lim(-6x 3h + 5) h h®0 h®0 = 6x- 3(0) + 5 f ( x + h)- f (x) - 6x+5 lim h h® 0 Therefore, f ¢(x) = - 6 x + 5 = 2-18 ..

Calculus

3.

Chapter 2

f ¢(1) = 6(1)+ 5 f ¢(1) - 1 f (x) = 2 x = - x- 5

f ¢(2) = - 6(2) + 5 f ¢(2) = - 7

f ¢(3) = 6(3) +5 f ¢(3) - 13 = -

2( x + h) 2 x +2h = ( x+h)- 5 x + h - 5

Step 1:

f ( x + h) =

Step 2:

f ( x + h) - f ( x) =

2 x +2 h 2x x + h -5 x- 5 2 x( x + h-5) (2x+2h )(x -5) = ( x + h - 5)( x - 5) ( x + h - 5)(x- 5) h - 10 f ( x + h) - f ( x) = ( x + h - 5)( x - 5) - 10h f ( x + h) - f ( x) = h ( x + h - 5)( x - 5)h - 10 f ( x + h) - f ( x) = h ( x + h - 5)( x -5)

Step 3:

f ( x + h)- f ( x) -10 = lim h h®0 h®0(x+h-5)(x- 5) -10 = (x+0- 5)(x- 5)

Step 4:

lim

f ( x + h)- f ( x) -10 = 5)2 (xh h®0 lim

Therefore, f ¢(x) =

f ¢(1)=

-10 (x- 5)2

5 - 10 -10 -10 =- 8 = = (1- 5)2 (-4)2 16

f ¢(2) =

10 - 10 - 10 -10 =- 9 = = (2- 5)2 ( 3)2 9

f ¢(3)=

5 - 10 - 10 -10 =- 2 = = (3- 5)2 ( 2)2 4 2-19 ..

Calculus 4.

Chapter 2

The profit, in hundreds of dollars, from the sale of x items is given by

P(x)=2x2-5x+6 a. b. c.

Find the average rate of change of profit from x = 2 to x = 4.

Find the instantaneous rate of change equation using the four–step procedure.

Using the equation found in part b, find the instantaneous rate of change when x = 2 and interpret the results.

Solution: a.

h = 4 - 2 = 2.

First find the function values at 2 and 4. Note that 2 2

P (2) = 2(2) - 5(2) + 6 P (2) = 8-10 + 6 P (2) = 4

P (4) = 2(4) - 5(4) + 6 P (4) = 32-20+6 P (4) =18

Now use the average rate of change formula:

P ( a +h) - P ( a ) P(2+2) - P(2) = h 2 P(4)- P(2) = 2 18-4 = =7 2 b.

P(x + h ) =2( x + h)2 -5( x + h) + 6

Step 1:

= 2( x2 + 2xh + h2 ) - 5(x + h) + 6 P(x + h ) =2 x2 + 4xh + 2h2 - 5x - 5h + 6 Step 2:

Step 3:

P( + h - P(x = 2 x + 4 xh + 2h2 - 5 x - 5h + 6 - (2x2 - 5 x + 6) x +) - ) = 4 2 + 2h2 - 5h x P( h P(x +2h2- 5h P(x + h) - P(x) 4xh x ) ) = h h h P(x + h) - P(x) = 4x + 2h- 5 h

2-20 ..

Calculus

Chapter 2

P(x + h)- P(x) = lim(4x+2h- 5) h h®0 h®0 = 4x+2(0)- 5

Step 4:

lim

lim

P(x + h)- P(x) h

h®0

=4x-5

Therefore, P¢(x)=4x-5 P¢(2)=4(2)-5=3 means that when the second item was sold, your profit c. increased by $300.

5.

The distance of a particle from some fixed point is given by

s(t)=t2+5t+2 where t is time measured in seconds. a. b. c.

Find the average velocity from t = 4 to t = 6.

Find the instantaneous rate of change equation using the four–step procedure.

Using the equation found in part b, find the instantaneous rate of change when t = 4 and interpret the results.

Solution: a.

First find the function values at 4 and 6: 2 s (4) = (4) 5(4) + 2 20+2 + s (4) =16 + s (4) = 38

2 s (6) = (6) 5(6) + 2 s (6) = ++ + s (6) = 6836 30 2

h = 6 - 4 = 2.

Now use the average rate of change formula. Note that

s(t + h)- s (t) s(4+2)- s(4) = 2 h s(6)- s(4) = 2 68- 38 = =15 2

2-21 ..

Calculus b.

Chapter 2

s (t + h) = (t + h)2 + 5(t + h)2+

Step 1:

s (t + h) Step 2:

=2 +t2th +h2 + 5t + 5h 2+

s( + h)- s( t) = t2 + 2th + h2 +5t + 5h 2+(t -

2

+ 5t + 2)

t + h)- s( t) = 2 th+h2 +5h Step 3:

Step 4:

s( s(t + h) - s (t) 2th +h2+ 5h t = h h s(t + h) - s (t) +5 =2t + h h s (t + h)- s (t) = lim(2t +h + 5) h h®0 h®0 = 2t + 0 + 5 lim

s(t + h)- s (t) 2t 5 = + h h®0 Therefore, s¢(t)=2t+5 lim

c. s¢(4)=2(4)+5=9 units per second.

means that after 2 seconds the particle is traveling at 9

2-22 ..

Calculus

Chapter 2

Name ________________________________ Date ______________ Class ____________

Section 2-5 Basic Differentiation Properties Goal: To find the first derivatives using the basic properties

Notation: If y = f ( x), then f ¢(x), y¢, d all represent the derivative of f at x. y d x

Theorems: 1.

If y = f ( x )= C,then f ¢()x = 0 (Constant Function Rule) If y=f (x )= xn ,where n is a real number, then f¢(x)=nxn-1 (Power Rule)

2.

3.

4.

If y=f (x) = (kux ), then f¢(x)=ku¢(x) (Constant Multiple Property) If y=f (x) = ux ) ±v(x),then f¢(x)=u¢(x)±v¢(x)(Sum and Difference Property) (

1 - 6 Find the indicated derivatives.

1.

y¢ for y=x3 Use theorem 2 to find the derivative:

2.

d for y=1 y x7 d x

y¢=3x2

Convert the problem to the form xn and then use theorem 2. dy =-7x-8=-

y=1=x-7 x7

dx

7 x8

2-23 ..

Calculus 3.

Chapter 2

d(2u2-3u+8)

du

Use theorem 4 to break the original function into pieces, then use a combination of theorems 1–3. d (2u2-3u+8) du

d 2u2 - d3u+d8 = du du du

= 4u-3 + 0 = 4u-3

4.

f¢(x)if f(x)=x3.2+7x2.1-3x+7

Use theorem 4 to break the original function into pieces, then use a combination of theorems 1–3.

( x2.1 )- 3 ¢( f¢(x) = f¢ ( x3.2 ) + 7 f¢ f x)+f¢(7) = 3.2 x2.2 + 7(2.1)x1.1 - 3 + 0 = 3.2 x2.2 + 14.7x1.1 - 3 5.

d(1+6

x-5x+8)

dx x4

Use theorem 4 to break the original function into pieces, then use a combination of theorems 1–3. d 1 + d 6 x - d 5x + d 8 x-5x+8) = dx 4 dx dx dx

d (1+6 x4 dx

6.

x 1 d x- 4 + d = 6 x 2 - d 5x + d 8 dx dx dx dx - 5 6(1) - 12 = 4x + 2 x - 5 + 0 - 4 + 3 -5 x5 x

= 3

4

f¢(x)if f(x)=x5+4x5-2x -

-1

5+5

Use theorem 4 to break the original function into pieces, then use a combination of theorems 1 - 3.

4

3

f¢(x) = f ¢(x 5 ) + 4 f ¢ (x5) - 2f¢(x 1

2

-1

5)+f¢(5)

-6

= 4 x- 5 + 12 x- 5 + 52 x 5 5 5

2-24 ..

Calculus 7.

Chapter 2

Given the function f(x)=7x2-9x+2 a. b. c. d.

Find f¢(x).

Find the slope of the graph of f at x=3.

Find the equation of the tangent line at x=3.

Find the value of x where the tangent is horizontal.

Solution: a.

f ( x) = 7 x-2 9 x + 2 f ¢( x) = 7¢() f x2 - 9 f ¢(x) + f ¢(2) f ¢( x) =14-9 x

b.

Substitute the value into the equation in part a to find the slope.

f ¢( x) =14 x -9 f ¢(3) =14(3) - 9 m = 33 c.

Substitute the given value into the original function to find the y value of the point and then use that point with the slope found in part b. The point is (3, 38).

y - y1= m(x- x1) y - (38) = 33(x - 3) y - 38 = 33 - 99 y= x - 61 d.

33 The tangent is horizontal when the first derivative has a value of 0. x f ¢( x) =14 - 9 0 =x - 9 9 =14 0 . »6xx 4 14 x

2-25 ..

Calculus

Chapter 2

8 - 9 Use the following information for both problems: If an object moves along the y axis (marked in feet) so that its position at time (inx seconds) is given by the indicated function, find: a. b. c.

8.

v = f '(x) The instantaneous velocity function The velocity when x=0andx=4 The time(s) when v=0

f(x)=3x2-12x-8 Solution: a.

f ( x) = 3 x2 - 12 x -8 f ¢( x) = 3 f ¢(x2 )- 12f ¢(x) - f ¢(8) f ¢( x) = 612x v = 612x -

b.

c.

= 6x-12 f ¢()x = 6x- 12 v( x) = 6(0) - 12 v(0) =v(0) 12

f ¢()x = 6x-12 v( x) = 6x- 12 v(4) = 6(4)- 12 v(4) =12

v( x) = 6 - 12 0 = x - 12 12 = 6 2 =x 6 x

2-26 ..

Calculus 9.

Chapter 2

2 f (x) = x3 - 21 2 x +30x

Solution: a.

f (x ) = x 3 - 2 1 x 2 + 3 0 x 2

f ¢(x) = f ¢(x3) - 21 f ¢(x2)+ 30f ¢(x) 2

f¢()=3x2-21x+30 x v =3x2 - 21x+30 b.

f ¢()x = 3x2-21x+30 v( x) = 3x2 - 21x+30

f ¢()=3x2-21x+30 x v( x) =3x2 - 21x+30 v(0) =3(0)2 - 21(0)+30 v(0)=0-0+30 v(0)=30

c.

v(4) = 3(4)2 - 21(4) + v(4) = 30 v(4) = 48-84+30 -6

v()=3x2-21x+30 x 0 =3x2 - 21x+30 0 =3(x2 - 7x+10) 0=3(x- 5)(x- 2) x =2,5

10.

Find f¢(x)if f(x)=(5x-6)2 First simplify the function by using the FOIL to expand it.

f(x)=(5x-6)2=(5x-6)(5x-6)=25x2-60x+36 Now use theorem 4 to break the original function into pieces, then use a combination of theorems 1–3.

f ( x) = 25 x2 - 60x +36 f ¢( x) = 25 f ¢(x 2)- 60f¢( x) + f ¢(36) f ¢( x) = 50 x - 60

2-27 ..

Calculus 11.

Chapter 2

Find f¢(x)if f(x)=9x-5.

x

First simplify the function.

f(x)=9x-5=9x-5=9-5x-1 x

x

x

Now use theorem 4 to break the original function into pieces, then use a combination of theorems 1 - 3.

f ( x) = 9 - 5 x - 1 f ¢( x) = f ¢(9)- 5 f ¢(x- 1 ) f ¢( x) = 0 + 5x- 2 5 f ¢( x) = x2

2-28 ..

Calculus

Chapter 2

Name ________________________________ Date ______________ Class ____________

Section 2-6 Differentials Goal: To use differentials to solve problems

Definition: Differentials

If y=f(x)defines a differentiable function, then the differential dy or df is defined as the product of f¢(x)and dx, where dx=Dx. Symbolically, dy=f¢(x)dx or

df=f¢(x)dx

where dx=Dx. Recall that Dy=f(x+Dx)-f(x)

1.

D

,and y given x1 = 2 and x2 = 5.

Given the functiony = 3 x4 ,find Dx, Dy D x When x1 = 2, y1 = 3(2)4 = 48. When x2 = 5, y2 = 3( 5) 4=1875.

2.

D Dx 5 - 2 Dy 1875- 48 = 1827 = 609 3 y D = 3 = 1827 x Dx Dy = = ,and yD given x1 = 3 and x2 = 6.

Given the functiony = 3 x4 ,find Dx, Dy D x When x1 = 3, y1 = 3( 3) 4= 243. When x2 = 6, y2 = 3(6)4 = 3888.

Dx 6 - 3 = 3 Dx =

Dy 3888- 243 = 3645 Dy =

2-29 ..

D = 3645 3 =1215 y D x

Calculus 3.

Chapter 2

Given the functiony = 3x3 -12x 2 + 4 x +16, find dy.

y = 3x3 - 12 x2 + 4x +16 dy = 9x2 - 24 x + 4 dx 2 dy

= (9 x - 24 x + 4)dx

4.

x3

Given the functiony = x5 (2- 12 ), find dy.

x3) y = x5 (2 - 12 1x8 y = 2x5 - 12 dy = 8x7 10x4 - 12 dx dy = (10x4 - 2 x 7dx ) 3

5.

Given the functiony = 25- 7 x2 - x3 ,find dy and Dy given x = 4 and dx =Dx= 0.1.

y = 25 - 7 x2 - x3

d = ( -14x- 3x2)dx

dy = - 14 x -3 x2 dx

y = [ -14(4)- 3(4)2](0.1)

dy = (- 14 x -3 x2)dx

d = ( -56- 48)(0.1) y = ( -104)(0.1)

Dy= f(x+Dx)-f(x) Dy= f(4+0.1)- f(4) f Dy= (4.1) - f (4)

d = - 10.4 y d

y d Dy= (25- 7(4.1)2 - (4.1)3)- (25y 7(4)2 - (4)3) Dy=- 161.591- (- 151) Dy=- 10.591

2-30 ..

Calculus

Chapter 2

6.

A company will sell N units of a product after spend x thousand dollars in advertising, as given by N=120x-x2 10£x£60 Approximately what increase in sales will result by increasing the advertising budget from $15,000 to $17,000? From $25,000 to $27,000?

N =120-x x2

d = (12 - 2 xdx ) d = (12 - 2 xdx ) dN =120 - 2 x N = 0 - 2(15))(2) N = 0 - 2( 25))( 2) dx d =180 (12 d =140 (12 dN = (120- 2 )xdx N 0 N 0 Therefore, increasing dthe advertising budget from d$15,000 to $17, 000 will result in an increase of 180 units and an will only result in a 140 N increase from $25,000 to $27,000 N unit increase.

7.

The average pulse rate y (in beats per minute) of a healthy person x inches tall is given approximately by y=

590 x

30£x£75

Approximately how will the pulse rate change for a change in height from 49 inches to 52 inches?

y=

590 -1 =590x 2 x - 12

3

y = 590 x

d = ( -295 x 2)dx

-3 dy =- 295 x 2 dx

y = ( -295(49)

dy

- 32

= (- 295 x

-

3

2)(3)

d = ( - 295)(3)

)dx

y =-

343 885 » - 2.58 343

d Therefore, if a person grows from 49 to 52 inches, their pulse rate would decrease by y approximately 2.5 beats per minute. d y

2-31 ..

Calculus

Chapter 2

2-32 ..

Calculus

Chapter 2

Name ________________________________ Date ______________ Class ____________

Section 2-7 Marginal Analysis in Business And Economics Goal: To solve problems involving marginal functions in business and economics

Definition: Marginal Cost, Revenue, and Profit If x is the number of units of a product produced in some time interval, then total cost =C(x) total revenue =R(x) total profit =R(x)-C(x)

marginal cost =C(x) ¢ marginal revenue =R(x) ¢ marginal profit =R¢(x)-C (x) ¢

and and and

Definition: Marginal Average Cost, Revenue, and Profit If x is the number of units of a product produced in some time interval, then

Cost per unit: average cost = C=

C(x) x

and marginal average cost =C¢()=d x dx C(x)

x R¢()= d R(x) and marginal average revenue = dx x P(x) dP(x) x and marginal average profit =P ¢()= Profit per unit: average profit = P= R(x)

Rev. per unit: average revenue =R=

dx

x

1 - 10 Find the indicated function if cost and revenue are given by

C(x)=6000-40x+0.006x2 and 1.

R(x)=10,000x-200x2

Marginal cost function

C(x)=6000-40x+0.006x2 C¢(x)=-40+0.012x 2.

Average cost function C(x)= C(x) =6000-40x+0.006x2=6000-40+0.006x x

x

x

2-33 ..

Calculus 3.

Chapter 2

Marginal average cost function

C(x) = 6000 - 40 + 0.006x x

C(x) = 6000x-1- 40+0.006x C¢(x) = 6000x-2 +0.006 C¢(x) 4.

5.

3000 + 0.003 x2

= Marginal revenue function R(x) =10, x - 200 x2 R¢(x)=000 - 400 x 10, Average000 revenue function 2

R(x) = R(x) = 10,000 xx-200x = 10, 000 -200x x 6.

Marginal average revenue function

R( x)=10,000- 200x R¢( x)=-200 7.

Profit function

P(x) = R(x)- C(x) = (10,000x- 200x2)- (6000- 40x+ 0.006 x2 ) P(x) =- 200.006x2 +10,040x- 6000 8.

9.

Marginal profit function

P(x) = 200.006x2 +10, 040 x- 6000 P¢(x)- 400.012x +10, 040 = Average profit function 2

- 6000 P(x) = P(x) = - 200.006x +10,040x = - 200.006 x+ 10, 040 -6000 x x x

2-34 ..

Calculus 10.

Chapter 2

Marginal average profit function

¢ = R¢( x) - C¢( x) P(x) ¢ = ( -200)6000 + 0.006) P(x) - (x2 - 200.006 ¢ 6000 P(x) = x2

11.

Consider the revenue (in dollars) of a stereo system given by

R(x)=1000+1000x x

a. b.

Find the exact revenue from the sale of the 101st stereo.

Use marginal revenue to approximate the revenue from the sale of the 101st stereo.

Solution: a.

To find the exact revenue, find the revenue from the 101st and 100th and subtract their values:

R(x) = 1000 + 1000x x + 1000(101) R(101)= 1000 101

R(101)=101,009.90

R(x) = 1000 + 1000x x R(100)= 1000 + 1000(100) 100

R(100)=100, 010

R(101)-R(100)=101009.90-100010=999.90 Therefore, the actual revenue from the 101st stereo was $999.90. b.

Find the marginal revenue formula and then substitute in 101.

R(x) = 1000 +1000 x x

R¢(x) =

1000 + 1000 x2

R(x) =1000x-1+1000x

R¢(101)-

1000 + 1000

R¢(x)= 1000x-2 +1000 R¢(x)-

1000 + 1000 x2

(101)2

R¢(x) = 0.098+1000 R¢(x) = 999.90 -

= Therefore, the approximate revenue from the = 101st stereo was $999.90.

2-35 ..

Calculus 12.

Chapter 2

The total cost (in dollars) of manufacturing x units of a product is:

C(x)=10,000+15x a. b. c.

Find the average cost per unit if 300 units are produced.

Find the marginal average cost at a production level of 300 units and interpret the results.

Use the results in parts a and b to estimate the average cost per unit if 301 units are produced.

Solution: a.

The cost of producing 300 units is C(300) = 10, 000 +15(300)= 14, 500.

Therefore the average cost is C(300)=14500=$48.33.

300

b.

C(x)

=10000x-1+15.

The average cost function is C(x)= x = 10,000+15x x Therefore, the marginal average cost function for producing the 300th unit would be:

C(x) =10000 x- 1 +15 C¢(x)=- 10000 x- 2 C(x) = -

10000 x2

C(x) =

10000 x2

C (300)-

10000 (300)2

C (300)= 0.11

This means that the average cost is decreasing by $0.11 per unit produced. c.

= The average cost per unit for the 301st unit would be $48.33 – $0.11 = $48.22.

-

2-36 ..

Calculus 13.

Chapter 2

The total profit (in dollars) from the sale of x units of a product is:

P(x)=30x-0.03x2+200 a. b.

Solution: a.

Find the exact profit from the 201st unit sold.

Find the marginal profit from selling the 201st unit.

To find the exact profit, find the profit from the 201st and 200th and subtract their values:

x2+ P(x) = 30 x0.03 200

x2+ P(x) = 30 x0.03 200

P ( 201) = 30(201) - 0.03(201)2 + 200 P ( 201)= 5017.97

P ( 200) = 30(200) - 0.03(200) 2 + 200 P ( 200)= 5000

P(201)-P(200)=5017.97-5000=17.97 Therefore, the actual profit from the 201st unit was $17.97. b.

Find the marginal profit formula and then substitute in 201.

P(x) = 30 x - 0.03x2+200 P¢(x)= 30 0.06 x -

P¢(x) = 30 - 0.06 x P¢(201) = 30 - 0.06(201) P¢(201) =17.94

Therefore, the approximate profit from the 201st stereo was $17.94.

2-37 ..

Calculus

Chapter 2

14.

The total cost and revenue (in dollars) for the production and sale of x units are given, respectively, by:

C(x)=32x+36,000 and a. b. c.

R(x)=300x-0.03x2

Find the profit function P(x).

Determine the actual cost, revenue, and profit from making and selling 101 units.

Determine the marginal cost, revenue, and profit from making and selling the 101st unit.

Solution: a.

P(x = R(x)- C(x) = (300x- 0.03x2) - (32x+36,000)

)

P(x =- 0.03x2 +268x- 36, 000 b.

)

C(x) = 32x+36,000 C(100)= 32(100) + 36, 000 C(100)= 39, 200

C(x) = 32x+36,000 P(x C(101)= 32(101)+36,000 ) C(101) = 39, 232 R(x) = 300x- 0.03x2

R(x) = 300x- 0.03x2

R(101)= 300(101) - 0.03(101)2 R(101)= 29, 993.97

R(100)= 300(100) - 0.03(100)2 R(100)= 29, 700

P(x) = R(x)- C(x) P(101)= R(101) - C(101) P(101)= 29, 993.97 - 39, 232 P(101)=- 9238.03

P(x) = R(x)-C(x) P(100)= R(100) - C(100) P(100)= 29, 700 - 39, 200 P(100)=- 9500

Actual Cost

Actual Revenue

Actual Profit

C(101)- C(100) 39, 232- 39, 200 32

R(101)-R(100) 29,993.97- 29,700 293.97

P(101)-P(100) - 9238.03 - (- 9500) 261.97

2-38 ..

Calculus c.

Chapter 2 Marginal functions would be:

R(x) = 300x - 0.03x2 R¢(x)= 300 - 0.06x

C(x) = 32 x+36,000 C¢(x)= 32

P¢ = R(x)-C¢(x) ¢ (x) = (300- 0.06x)- (32) P¢ = 0.06x+268 (x) Marginal values for the 101st unit are: P¢ (x) C¢(x) = 32 R¢(x) = 300-0.06x C¢(101)= 32 R¢(101)= 300 - 0.06(101) R¢(101)= 293.94 P¢(x) = 0.06x+268 P¢(101)- 0.06(101) + 268 P¢(101)= 261.94 -

2-39 ..

Calculus

Chapter 2

2-40 ..

Calculus

Chapter 3

Name ________________________________ Date ______________ Class ____________

Section 3-1 The Constant e and Continuous Compound Interest Goal: To work with the constante and solve continuous compound problems

Definition: The number e

æ e=limç1+ n®¥è Theorem 1:

n

1ö ÷ nø

or

1

e=lim(1+s)s s®¥

Continuous Compound Interest Formula

If a principal P is invested at an annual rate r (expressed as a decimal) compounded continuously, then the amount A in the account at the end of t years is given by

A=Pert

1.

A to the nearest cent.

Use a calculator to evaluate

A=500e0.03t for t = 3, 6, 9 A = 500e0.03t500e

A = 500e0.03t

A = 500e0.03t500e

A = 500e0.03(3) 0.09 A = 500(1.09417) A = 547.09

A = 500e0.03( 6 )

A = 500e0.03(9) 0.27 A = 500(1.30996 ) A= 654.98 A=

A = 500e0.18 A = 500(1.19722) A = 598.61

A=

3-1 ..

Calculus 2.

Chapter 3

Solve for t to two decimal places.

4=e0.07t 4 = e0.07t ln4 = lne 0.07t ln4 = 0.07t ln4 =t 0.07 19.80 = t 3.

Solve for r to two decimal places.

3=e4r 3 = e4r ln3 = lne 4r ln3 = 4r ln3 = r 4 0.2747 = r 27.47 = r 4.

Bank A offers a 5 year CD that earns 1.75% compounded continuously. a.)

If $5,000 is invested in this CD, how much will it be worth in 5 years?

A = Pert 0.0175)( 5) A = 5000e(5000e 0.0875 A = 5000(1.09144 ) A= 5457.21 A=

The account will be worth $5457.21 in 5 years.

3-2 ..

Calculus b.)

Chapter 3 How long will it take for the account to be worth $10,000?

10, 000= 5000e0.0175t 2 = e0.0175t ln2 = lne 0.0175t ln2 = 0.0175t ln2 =t 0.0175 39.61 = t It will take approximately 39.61 years to double in value.

5.

A note will be worth $25,000 at maturity 10 years from now. How much should you be willing to pay for the note now if money is worth 2.6% compounded continuously?

A = Pert ( 0.026 )(10 )

25, 000 = Pe 0.26 25, 000 = Pe 25, 000 = P (1.29693) 19, 276.29= P

You should be willing to invest $19,276.29 today.

6.

A woman invests $15,000 in an account that pays 3.8% compounded continuously. If the woman needs $20,000, how long will it take to accumulate that much money?

A = Pe rt 20, 000 =15, 000e ( 0.038) t 4 = e( 0.038) t 3 l 4 = ln e( 0.038) t 3 n 4 = 0.038t 3 lnl 4 3 =t

0.038 n 7.57 = t

It will take approximately 7.57 years for the woman to have the money she needs.

3-3 ..

Calculus

Chapter 3

3-4 ..

Calculus

Chapter 3

Name ________________________________ Date ______________ Class ____________

Section 3-2 Derivatives of Exponential and Logarithmic Functions Goal: To find the first derivative of functions that are exponential or logarithmic in nature.

Definition: Derivatives of Exponential and Logarithmic Functions

For b>0,b¹1,

d x dx bx = b lnb d dx 1 æ1ö logb x = ç÷ lnb è x ø

d dx ex=ex d dx lnx= 1 x

1 - 5 Find the first derivative for the given function.

1.

f(x)=4ex+5x4-12x+46

f (x) = 4e x + 5 x4 - 12 x + 4 6 f ¢(x) = 4e x + 20 x3 - 12

2.

f (x) = - 7e x + 5lnx +12x - 35 f ( x) = 7e x + 5lnx+12x - 35 f ¢(x) -

5 7e x + +12 x

= -

3-5 ..

Calculus 3.

Chapter 3

f(x)=4x3+2x2-5x+lnx5+ex

f (x = 4 x3 + 2 x2 - 5 x + ln x5 + e x f ) = 4 x3 + 2 x2 - 5 x +5ln x + e x (x =12 x2 + 4 x - 5 5 e x f ¢(x) ) + + x

4.

y =10x - e x+3 - 5 x y =10 x -e3-5x d =10 x ln10- 5 y d x

5.

y = log7 x + 3x5 - 4 x

y = log7 x+3x5 - 4x 1 æ1ö dy = ÷+15 x4 - 4 x ln4 ç xø dx ln 7 è

3-6 ..

Calculus

Chapter 3

Given the function f(x)=5ex+7,find the equation of the line tangent to the graph of 6. the function at x=0.

First find the first derivative and substitute the given x value. This will give the slope of the tangent line to the graph of the function.

f (x) = 5e x + 7

f ¢( x) = 5ex

f ¢(x) = 5e x

f ¢(0) = 5e f ¢(0) = 5

0

Now find the ycoordinate of the point by substituting the given value into the original function:

f (x = 5e x + 7 f ) = 5e0 + 7 f (0 =12 ) (0) Finally, use the slope and point to find the equation of the tangent line.

y - y1 = m(x- x1) y - 12 = 5( - 0) y - 12 = x y= 5x +12 5x

3-7 ..

Calculus

Chapter 3

Given the function f(x)=lnx5+2,find the equation of the line tangent to the graph 7. of the function atx = e.

First find the first derivative and substitute the given x value. This will give the slope of the tangent line to the graph of the function.

f ( x) = ln x5 + 2 f ( x) = 5ln x + 2

f ¢(x) = 5 f ¢(e) = x

5 f ¢(x) = x

f ¢(e) = 5e5- 1

Now find the ycoordinate of the pointeby substituting the given value into the original function:

f ( x) =lnx5+ 2 f ( e ) =lne5+ 2 f ( e) = 5 + 2 f ( e) = 7 Finally, use the slope and point to find the equation of the tangent line.

y - y1 = m ( x- x 1) 5e-1(xy - 7= e) 5e-1xy - 7= 5 5ey= 1x+2 y=1.839x+2

3-8 ..

Calculus

Chapter 3

Name ________________________________ Date ______________ Class ____________

Section 3-3 Derivatives of Products and Quotients Goal: To find the first derivative of functions that are written as products or quotients.

Theorem 1: Product Rule

x S¢(x) exist, then If y =f(x)= F ()x S x()and if F ¢()and f ¢()= x FxS x SxF¢(x) () ¢()+ () Theorem 2: Quotient Rule If y =f(x)=

T(x) B(x)

and if T¢(x)and B(x) ¢ exist and B(x)¹ 0 , then

f¢(x)=

x )- T(x )B¢(x ) B(x )T ¢(

[B( x)]2

1 - 10 Find the first derivative and simplify.

1.

f(x)=4x5(2x2-13x+2)

Since the problem is a product, find the functions and derivatives and then follow the product rule above.

F(x) = 4 x5

S(x) = 2 x2 - 13x + 2 S¢(x) = 4 x - 13

F¢(x)= 20 x4

f ¢ x) = F(x)S¢(x)+S(x)F¢(x) f ( x) = (4x5)(4x- 13)+(2x2 - 13 x + 2)( 20x4 ) f ¢ x) =16x6 - 52x5 +40x6 - 260x5 + 20 x4 f ( x) = 56x6 - 312x5 +40x4 ¢ ( ¢ (

3-9 ..

Calculus 2.

Chapter 3

f(x)=(4x2+7)(2x2-9)

Since the problem is a product, find the functions and derivatives and then follow the product rule above.

F(x)=4x2+7

S(x)=2x2-9

F¢(x)=8x

S¢(x)=4x

f ¢ x) = F(x)S¢( x) + (S) x F (¢ x) f ( x) = (4x2 + 7)(4x)+(2 x2- 9)(8 x ) f ¢ x) = 16x3+ 28x+16 -x3 72 x f ( x) = 32x3 - 44 x ¢ 3.

( 13x f(x)= ¢ 3x2 - 2

( Since the problem is a quotient, find the functions and derivatives and then follow the quotient rule above.

B(x)=3x2 -2 B¢(x)=6x

T(x)=13x T¢(x)=13 f ¢(x) =

f ¢(x) =

f ¢(x) =

f ¢(x) =

B( x) ¢(T x )- T ( x ) B¢ ( x ) 2 [B(x)]

(3x2 - 2)(13) - (13x)(6x) 2

é(3x2 - 2)ù ë û 39x2 -26 - 78x2 2

é(3x2 - 2)ù ë û 39x2 - 26 2

é(3x2 - 2)ù ë û

3-10 ..

Calculus

4.

Chapter 3 2x-5

f(x)=

3x2 +5 Since the problem is a quotient, find the functions and derivatives and then follow the quotient rule above.

B(x)=3x2+5

T(x)=2x-5 T¢(x)=2

f ¢(x) =

f ¢(x) =

f ¢(x) =

f ¢(x) =

B( x) ¢( T

B¢(x)=6x )x- T( x B)¢( x )

[B( x-)]2

(3x2 + 5)(2) (2 x - 5)(6x) 2

é(3x2 + 5)ù ë û 6x2 +10 - 12x2 + 30x 2

é(3x2 + 5) ù ë û 2 - 6x + 30x +10 2

é(3x2 +5)ù ë û 5.

f(x)=(3ex)(3lnx)

Since the problem is a product, find the functions and derivatives and then follow the product rule above.

F(x)=3ex

S(x)=3lnx 3 S¢(x)= x

F¢(x)=3ex

x f ¢ x) = F(x) S¢()+ S()x F¢(x) 3 f ( x) = (3ex)( )+(3ln )(3 x xe ) x 9ex ¢ +9 exlnx f ¢(x) = x (

3-11 ..

Calculus 6.

Chapter 3

y=(2x3-5x2)(x2-2x)

Since the problem is a product, find the functions and derivatives and then follow the product rule above.

F(x) = 2 x3 - 5 x2

S(x) = x2 - 2 x x S(x ¢ ) =2 - 2

F¢(x)= 6 x2 - 10 x f ¢ x) = F(x)S¢(x)+S(x)F¢(x)

2 f ( x) = (2x3 - 5x2)(2x- 2)+(x - 2x)(6x2 - 10 x) 2 6x4 - 10 3 f ¢ x) = 4x4 - 4x3 - 10x3 +10x + x - 12 x3 + 20x2

f ( x) =10x4 - 36x3 +30x2 ¢ ( x2-3x+4 y= 7. ¢ 4x2 - 3 ( Since the problem is a quotient, find the functions and derivatives and then follow the quotient rule above.

T(x)=x2-3x+4

B(x)=4x2-3

T¢(x)=2x- 3

B¢(x)=8x

f ¢(x) =

f ¢(x) =

f ¢(x) =

f ¢(x) =

B( x) ¢( T

)x- T( x B)¢( x )

[B( x)]2 (4x2 -3)(2 x - 3) - ( x2 - 3x + 4)(8x) 2

é(4 x2 - 3)ù ë û 8x3 - 12 x 2- 6 x + 9 - 8 x3 + 2 4 x 2 - 3 2 x 2

é(4 x2 - 3)ù ë û 2 12 x - 38x + 9 2

é(4 x2 - 3)ù û ë

3-12 ..

Calculus 8.

Chapter 3

f(x)=5xlog7x

Since the problem is a product, find the functions and derivatives and then follow the product rule above.

F(x) = 5x

S(x)=log 7 x

F¢(x)= 5xln5

S¢(x)=

1 1 ç æ ln7÷ø è xö

f ¢ x) = F ( x) S¢()+S(x)F¢(x) x 1 ö x)(5x ln5) f ( x) = (5x)æ ø+(log7 çè x l n 7 ÷ ¢ 5x f ¢(x) = + (5x)(ln5)(log7 x) x ln7 (

9.

f(x)=

84x x2+4

Since the problem is a quotient, find the functions and derivatives and then follow the quotient rule above.

1

B(x) = x2 + 4 B(x) ¢ = 2x

T(x) = 8 x 4 3

T¢(x)= 2 x - 4 f ¢(x) =

f ¢(x) =

B(x)T¢(x )- TxB ( ) ¢(x) ( 2 [B x)] ( x 2 +4)(2x- 34) - (8x 14)(2x) 2

5

f ¢(x) =

é( x2 + 4)ù ë û - 34

2 x 4+8 x

5

- 16 x 4 2

é( x2 + 4)ù ë û 5 3 2 ( - x7 2 + 4 ) 4 +8 x 4 f ¢(x) = = 3 2 14x2 é(x + 4)ù x 4 ( x2 +4)2 ë û 3-13 ..

Calculus

10.

f(x)=

Chapter 3

(2x3 - 5)(3x2 +1) x2+7

The main problem is a quotient, but in the numerator, there is a product. To find the first derivative of the numerator, follow the product rule.

S(x) = 3 x2+1 S¢(x) = 6 x

F(x) = 2 x3 -5 F¢(x)= 6 x2

T ¢ x) = F(x)S¢( x) + S (x)F ¢ x( ) T ( x) = (2x3 - 5)(6 x) + (3x2+1 ) (2 )6 x T ¢ x) = 12x4 - 30 x +18x4 +6x2 T ( x) = 30x4 + 6 x2 - 30 x ¢ Now that we have the derivative for the numerator, follow the quotient rule. ( ¢ = (2 x3 - 5)(3x2 +1) T(x)

B(x) = x2 + 7 B(x) ¢ = 2x

( = 30 x4 + 6 x2 - 30x T¢(x)

f ¢(x) =

f ¢(x) =

f ¢(x) =

f ¢(x) =

B(x)T¢(x)-T(x)B¢(x) 2 [B(x)]

(x2 +7)(30x4 +6x2 - 30x)- (2x3 - 5)(3x2 +1)(2x) 2

é( x2 +7)ù û ë 30x6 + 210x4 + 6 x4 + 42x2 - 30x3 - 210x-12x 5 - 4 x4 + 30x3 +10x 2

é(x2 +7)ù ë û 5 212 x 4 42x2 - 200x 6 30x -12x + + 2

é(x2 +7)ù ë û

3-14 ..

Calculus

Chapter 3

Name ________________________________ Date ______________ Class ____________

Section 3-4 The Chain Rule Goal: To find the first derivative of functions that are written as composite functions.

Definition: Composite Function A function m is a composite of functions f and g if

m(x)=f[g(x)] The domain of m is the set of all numbers x such that x is in the domain g g ( x) is and in the domain of f.

Theorem 1: General Power Rule n

y = f ( x) = [u(x) ] , then If u( x) is a differentiable function, n is any real number, and n-1 u¢(x) f¢(x)=n[u(x)]

Theorem 2: Chain Rule

If y=f(u)and u=g(x)define the composite function y=m(x)=f d dydu = y dudx d x 1.

provided

d d and exist y u d d u x

( )]if f(u) = u3 and g( x) = 3x2 - 7 x +17.

Find m(x) = f [gx m(x = f [g(x)] = f é3x2 - 7x+17ù û ë m(x = (3x2 - 7x+17)3

)

) m(x )

3-15 ..

[g ( x)], then

Calculus 2.

Chapter 3

Find m(x)=f[g(x)]if f(u)=lnuand g(x)=4x2+7x-35. m(x = f [ g ( x)] ù = fé4 x2 + 7 x - 35û ë m(x = ln(4x2 + 7 x - 35) )

) Given y=(x4+3x2-3x+5)3,write this composite function in the form 3. m(x u=g(x). y=f(u)and ) y=u3

g(x)=x4+3x2-3x+5

Given y=e-5x3+4x-23,write this composite function in the form y=f(u)and 4. u=g(x). y=eu

g(x)=-5x3+4x-23

5–10 Find the first derivative for the given functions.

5.

f(x)=(3x4-5x+13)3 We must break the function into y and g(x) and then find each derivative.

u = g( x) = 3x4 - 5 x +13

y = u3

d = g ¢ x() =12 x3 - 5 d = 3u2 u d y x d Now, use the chain rule. u

dydu dudx f ( x) = 3u2(12x3 - 5) f ¢ x) =

f x) = 3(3x4- 5 x +13)2(12x3 - 5) ¢ f x) = 4(6 x2- 7)(2x3 - 7x+3)3 ( ¢ ( ¢ (

3-16 ..

Calculus 6.

Chapter 3

f(x)=4ln(9x-7) We must break the function into y and g(x) and then find each derivative.

y = 4ln u u = g( x) = 9 x - 7 d = ¢ ( d 4 g x) = 9 u = d y u x d Now, use the chain rule. u

dydu dudx 4 f ¢(x) = u (9) 36 f ¢(x) = 9 x - 7 f ¢(x) =

7.

f(x)=3e5x-2 We must break the function into and y g(x) and then find each derivative.

y = 3eu

u = g( x) = 5x - 2

d = ¢ ( d g x) = 5 u = 3eu d y x d Now, use the chain rule. u

f ¢ x) =

dydu dudx

f ( x) = 3eu (5) f x) =15eu ¢ f x) =15e5x- 2 ( ¢ ( ¢ (

3-17 ..

Calculus

8.

Chapter 3

y=6(x2-3x+12)6 We must break the function into y and g(x) and then find each derivative.

u = g( x) = x 2- 3x + 1 2

y = 6u6

d = g ¢ x() = 2x - 3 d = 36u5 u y d x d Now, use the chain rule. u

dydu dudx f ( x) = 36u5(2 x - 3) f ¢ x) =

f x) = 36( 2 x- 3 x +12)5(2x- 3) ¢ f x) = 36(2 x - 3)( x2 - 3x+12)5 ( 9.

¢ 1 ( = 3 2 x2 - 3x = (2x2 - 3x) 3 f (x) ¢ We must break the function into and y g(x) and then find each derivative. ( 1

u = g( x) = 2 x2 - 3x

y =u 3

d u d x

dy 1 - 2 3 = u du 3

= g ¢ x() = 4 x - 3

Now, use the chain rule.

dydu dudx 1 -2 f ¢(x) = 3 u 3 (4x- 3) 4 -2 x - 3 x2 f ¢(x) = (2 -3x) 3 3 4x - 3 f ¢(x) = 2 3(2x2 - 3x) 3 4x - 3 f ¢(x) = 33 (2x 2 - 3x)2 f ¢(x) =

3-18 ..

Calculus

10.

Chapter 3

y=(ln(3x+7))3 We must break the function into y and g(x) and then find each derivative.

y = u3

u = g( x) = ln(3x+7)

d du = g ¢( x) = 3 3x+7 = 3u2 dx y d Since g ( x) is still a composite function, we must break it into pieces again. u g = ln v

v = h( x) = 3 x + 7

dv = d 1 dx h¢( x) = 3 = g v d Now, use the chain rule to find the first derivative of. g v

dgdv dvdx 1 g ¢(x) = v (3) 3 g ¢(x) = 3 x + 7 g ¢(x) =

Now, use the chain rule again.

dydu dudx 2æ 3 ö f ¢(x) = 3u ç ÷ è 3x + 7 ø 9(ln(3x +7))2 f ¢(x) = 3x + 7 f ¢(x) =

3-19 ..

Calculus

Chapter 3

3-20 ..

Calculus

Chapter 3

Name ________________________________ Date ______________ Class ____________

Section 3-5 Implicit Differentiation Goal: To find the first derivative of functions that are not explicit in nature.

Procedure:

1. Find the derivative of the function using the basic derivative rules using y¢or dy dx for the derivative of the variable y.

2. Remove parentheses, if necessary.

3. Group all terms with y¢on one side of the equation and group all terms without y¢on the other side of the equation.

4. Factor out y¢for all terms that have ay¢.

5. Solve for y¢by dividing.

1. Consider the equation

3x2 +2y-19 = 0 a.) Solve the equation fory and then find the first derivative.

y = 32 x2 + 19 2

3x2 + 2 y -19 = 0 2 y =- 3 x2 +19 y =- 3 x2 + 19 2 2

y'-

3

= x -

b.) Use implicit differentiation to find the first derivative.

3x2 + 2 y -19= 0 6 x + 2 y¢+0= 0 2y¢=- 6x y¢=- 3x

3-21 ..

Calculus

Chapter 3

2. Use implicit differentiation to find the first derivative and evaluate it at the given point.

3y-6x3+8=0; (3,-2)

3 y - 6 x3 + 8 = 0

y¢ 6 x2

3 y¢- 18 x2 + 0 = 0

y= 6(3)2 y¢ 54

3 y ¢ =18 x2 y ¢ = 6 x2

= ¢ = 3. Use implicit differentiation to find the first derivative and evaluate it at the given point.

2xy2+3y-2x2=0; (1,4)

2xy2 + 3y - 2x2 = 0

y¢=

(2x)(2 yy¢)+y2 (2)+3y¢- 4 = 0

- 2(4)2 + 4(1) y¢= 4(1)(4)+3

4 xyy¢+2y2 + 3 y¢- x = 0 4 ¢=- 2 y2 + 4 x 4 xyy¢+3y y¢(4 xy +x3) =- 2 y2 + 4 x y¢=

- 2 y2 + 4x 4xy + 3

3-22 ..

- 2y2 +4x 4xy+3

y¢=-

28 19

Calculus

Chapter 3

4. Use implicit differentiation to find the first derivative and evaluate it at the given point.

5lny=ex+x2-y2; (0,3)

5ln y = e x + x2 - y2 5 x 2x y y¢= e + - 2yy¢

y¢=

e0 + 2(0) 5 + 2(3) 3 1

y¢= 23

5 y y¢+ 2 yy¢= ex + 2x æ ö x 5 +2x y¢ç + 2 y ÷=e èy ø

3

3 y¢= 23

ex + 2 x y¢= 5 y +2 y 5. Use implicit differentiation to find the first derivative and evaluate it at the given point. x2y3+xy2-6y+4lnx-6=0; (-2,3)

x2y3 + xy2 - 6y + 4 l nx - 6 = 0 x2 (3 y2y¢)+ y3(2x)+ x(2yy¢)+ y2(1) - 6 y¢+ 4x - 0 = 0 4

3x2y2y¢+ 2xy3 + 2xyy¢+ y2 - 6 y¢+ x = 0 3x y2 y¢+22 xyy¢- 6 y ¢=- 2 x 3 - y - 4 y¢(3x2 y2 + 2 xy - 6)= - 2 y 3 - 2 - x y¢= - 2xy3 - y2 - 4x y¢= 3x2y2 +2xy - 6 - 2(- 2)(3)3 - (3)2 - -42 y¢= 3(-2)2(3) 2 + 2(- 2)(3) - 6 y¢=

101 90

3-23 ..

- 2xxy3 - yy2 - 44 3x2y y2 + 22xy -xx6

Calculus

Chapter 3

6. Use implicit differentiation to find the first derivative and evaluate it at the given point.

2y3-16+5x2+13=0; (3,2)

2y3 - 16 +5 x 2+13=0 1

2 0 (2 y3 - 16) 2 + x 5+13= 1 2

1

+0= 0 (2y3 - 16)- 2 (6 y2 y ') x+10 1

3y 2(2y 3 - 16)- 2 y ' = - 10x 1 '= -10x - 10x(2 y3 -16) 2 y = 1 3 y2 3 y2 (2 y3 -16) - 2

y¢=

- 10(3)(2(2)3 -16)

1

2

3(2)2

0 12 y= 0 y¢

¢ 7. Use implicit differentiation to find the first derivative and evaluate it at the given point. = (4+y)4+6y=x3-2x+7; (1,0)

(4 + y)4+ 6 y = x3 - 2x+ 7 = 4(4 + y)3 y ¢ + 6 y ¢3x2 y¢(4(4 + y ) 3 + 6 )2 = 3x2 y¢=- 23x2 - 2 4(4+ y)3 +6

3(1)2 - 2 y¢= 4(4+0)3 +6 y¢=

1 262

3-24 ..

Calculus

Chapter 3

Name ________________________________ Date ______________ Class ____________

Section 3-6 Related Rates Goal: To solve problems that involve related rates

Procedure: Solving Related Rates Problems 1. Sketch a figure, if possible.

2. Identify all relevant variables, including those whose rates are given and those whose rates are to be found.

3. Express all given rates and rates to be found as derivatives.

4. Find an equation connecting the variables identified in step 2.

5. Implicitly differentiate the equation found in step 4, using the chain rule where appropriate, and substitute in all given values.

6. Solve for the derivative that will give the unknown rate.

1. Assume that x = x( t) and y = y(). t Find

3x2 + 9xy- 2 y3 =15,

dy given dt

dx = 4, x = 2, y =1. dt

Find the first derivative using implicit differentiation:

x2 + 8xy - y3 =12 dx dy dx dy 6x dt + 9x dt + y(9 ) - 6 y2 =0 dt dt

dy Now substitute all known values into the first derivative and solve for dt . dx dy dx dy 6 x dt + 9 x dt + y(9 )- 6y2 = 0 dt dt dy dy 6( 2)( 4)+ 9(2) + (1)(9)(4)-6(1)2 =0 dt dt dy 84 + 12 = 0 dt dy = -7 dt 3-25 ..

Calculus

Chapter 3

2. A boat is being pulled toward a dock by a rope. If the rope is being pulled in at 3 feet per second and the rope is 4 feet above the dock, how fast is the distance between the dock and the boat decreasing when it is 20 feet from the dock?

Based on the information, sketching the rope, the dock, and the boat gives a right triangle. We have 3 variables that we will call a, b, and c (typical sides of a right triangle).

Let the variable a be the rope height from the dock, the variable b be the distance between the boat and the dock, and the variable c be the rope length. Therefore, da db dc =-3. Use the Pythagorean Theorem to set up =0,b=20, =?,c»20.396, a=4, dt dt dt the relation between the variables and use implicit differentiation to solve.

a 2 + b2 = c 2 da db dc 2a +2b = 2c dt dt dt +2(20) db 2( 4)( 0) dt = 2(20.396)(- 3) db 40 dt = - 122.376 db dt = - 3.06 Therefore, the boats distance is decreasing by approximately 3.06 feet per second.

3-26 ..

Calculus

Chapter 3

3. A 20-foot ladder is placed against a vertical wall. Suppose that the top of the ladder slides down the wall at a constant rate of 4 feet per second. How fast is the bottom of the ladder sliding away from the wall when the top of the ladder is 12 feet above the ground?

Based on the information, sketching the ladder, the wall, and the ground gives a right triangle. We have 3 variables that we will call a, b, and c (typical sides of a right triangle).

Let the variable a be the height of the ladder against the wall, the variable b be the distance between the wall the base of the ladder, and the variable c be the ladder length. Therefore, da db dc =0. Use the Pythagorean Theorem to set up the a=12, =-4,b=16, =?,c=20, dt dt dt relation between the variables and use implicit differentiation to solve.

a 2 + b2 = c 2 da + db dc 2a 2b = 2c dt dt dt db 2(12)(-4) + 2(16) dt = 2( 20)( 0) db - 96 +32 dt = 0 db 32 dt = 96 db dt = 3 Therefore, the bottom of the ladder is sliding away from the wall at 3 feet per second.

3-27 ..

Calculus

Chapter 3

4. The radius of a spherical balloon is increasing at a constant rate of 5 inches per minute. How fast is the volume changing when the radius is 15 inches?

As helium is pumped into a balloon, the radius and the volume of the balloon change.

Therefore, we have two variables in the problem. Let the variable V be the volume and the dr dV variable r be the radius. Therefore, r=15, =5,V=4500, =?. Use the volume of a dt dt sphere formula to set up the relation between the variables and use implicit differentiation to solve.

4 V = 3 r 3 dV 2dr = 4r dt dt dV 4(15)2(5) = dt dV = 4500»14,137 dt Therefore, the volume of the balloon is increasing be approximately 14,137 cubic inches per minute.

3-28 ..

Calculus

Chapter 3

Name ________________________________ Date ______________ Class ____________

Section 3-7 Elasticity of Demand Goal: To determine the elasticity of demand for functions

Definition: Relative and Percentage Rates of Change

The relative rate of change of a function f(x)is

f ¢(x) (x) . f

f ¢(x) The percentage rate of change is 100´ . f(x) Theorem 1 : Elasticity of Demand If price and demand are related byx = f ( p), then the elasticity of demand is given by: pf ¢( p) E(p) = - f(p) a. If 0 <E(p) <1,then demand is inelastic. b. If E(p) 1, = then demand is unit. c. If E(p) 1, > then demand is elastic.

1. Given f (x) = 42x - 0.3x2 ,find the relative rate of change. f ( x) = 42x - 0.3x2 f '( x) = 42 - 0.6 x

f '(x) f (x)

=

42- 0.6x 42x- 0.3x2

3-29 ..

Calculus

Chapter 3

2. Given f(x)=7x2-2lnx,find the relative rate of change when x=4. (Round to three decimals if necessary.)

f (x) = 7 x2 - 2lnx f ¢(x) =14 x -

2 x

f (4) = 7(4)2-2ln(4) f (4) =109.227

f ¢(4) =14(4)- 24 f ¢(4) = 55.5

f ¢(x) 55.5 = = 0.508 f (x) 109.227

3. Given f(x)=600-7x,find the relative rate of change when decimals if necessary.)

x = 60. (Round to three

f (x) = 600 - 7 x f ¢(x) =- 7 f (60) = 600- 7(60) f (60) =180

f ¢(60) = - 7

f ¢(x) - 7 = - 0.039 = f (x) 180

4. Given f(x)=4000-6x2,find the percentage rate of change when

x = 30. (Round to

the nearest tenth if necessary.)

f (x) = 4000 -6x2 f ¢(x) =- 12x f (30) = 4000 - 3(30)2 f (30) =1300

f ¢(30) = 12(30) f ¢(30) - 360 = -

3-30 ..

100´

f ¢(x) -360 =100´ 1300 f (x) =100 ´ - 0.277 = 27.7%

Calculus

Chapter 3

5. Given x=f(p)=12,000-300p,use the price–demand equation to find E(p),the elasticity of demand.

f ( p =12, 000-300p f ¢( ) =- 300 p )

pf ¢( p) f(p) p(-300) =12,000- 300 p

E(p) = -

300p 12,000- 300 p p E(p) = 40- p =

6. Given x=f(p)=850-3p-0.04p2,use the price–demand equation to find E(p),the elasticity of demand to determine whether the demand is elastic, inelastic, or has unit elasticity at the indicated values of p. a. p=15 b. p=35 c. p=55

f ( p = 850- 3p- 0.04p 2 f ¢() =- 3- 0.08p p )

E(p) = =E(p) =

E(15) =

E(35) =

E(55) =

3(15)+0.08(15)2 850- 3(15)- 0.04(15)2 3(35) + 0.08(35)2 850- 3(35)- 0.04(35)2 3(55) + 0.08(55)2 850- 3(55)- 0.04(55)2

850- 3p- 0.04 p2

3p+0.08p2 850- 3p- 0.04 p2

63 = 796 = 0.079; demand is inelastic.

=

203

=0.292; demand is inelastic.

696 =

407

=0.722; demand is inelastic.

564

3-31 ..

pf ¢( p) f(p) p(-3- 0.08 )p

Calculus

Chapter 3

7. Given x=f(p)=2400-4p2,use the price–demand equation to find the values of p for which demand is elastic and the values for which demand is inelastic. Assume that price and demand are both positive.

f ( p = 2, 400-4p2 f ¢() =- 8 p p )

pf ¢(p) f p () (p- 8 p) =2400 - 4 p2

E(p) = -

= E(p) =

8p2 2400- 4p2 2 p2 600 - p2

For the demand to be elastic,E(p) >1.

2 p2 >1 600- p2 2 p2 > 600- p2 3 p2 > 600 p2 > 200 p >10 2 For the demand to be inelastic, 0<Ep) ( <1, therefore, the demand is inelastic on the

interval (0,10 2)and elastic on the interval(10 2,10 6).

3-32 ..

Calculus

Chapter 3

1600-2p2,use the price–demand equation to find the values of 8. Given x=f(p)= p for which demand is elastic and the values for which demand is inelastic. Assume that price and demand are both positive.

1

(1600-) p2p2 2 2) f ¢( p) = - 2 p (1600 f ( p) =

2

1

f ¢( p) =

E(p) = -

2

- 2p 1600 - 2p2 pf ¢(p) 1 = - pf (p) × f (p) f(p)

æ 1 - 2p ¢ ö = - pç ÷ 2 èç 1600 - 2 p2 ÷ 1600 - 2 p ø 2p2 = 1600- 2p2

E(p) =

p2 800- p2

For the demand to be elastic,E(p) >1.

p2 800- p2

>1

p2 > 800- p2 2p2 > 800 p2 > 400 p > 20 For the demand to be inelastic,0 < E(p) <1, therefore, the demand is inelastic on the interval (0, 20)and elastic on the interval(20,20 2).

3-33 ..

Calculus

Chapter 3

3-34 ..

Calculus

Chapter 4

Name ________________________________ Date ______________ Class ____________

Section 4-1 First Derivatives and Graphs Goal: To use the first derivative to analyze graphs

Theorem 1 :

Increasing and Decreasing Functions

For the interval (a,b), if f'(x)>0,then f(x)is increasing and the graph of f rises. If f'(x)<0,then f(x)is decreasing and the graph of f falls.

Definition:

Critical Values

The values of x in the domain of f where f'(x)=0or where f'(x)does not exist are called the critical values of f.

Procedure:

First Derivative Test for Local Extrema

Let c be a critical value of f. Construct a sign chart for f'(x)close to and on either side of c. Then f(c)will be a. A local minimum if f'(x)changes from negative to positive. b. A local maximum if f'(x)changes from positive to negative. c. Neither a local minimum nor a local maximum if f'(x)does not change signs.

1. Given the function f (x) = x3 - 27x +12 find: a.

f'(x)

f'(x)=3x2-27

4-1 ..

Calculus

Chapter 4

b. The critical values of the function

f '()= x 3x2 - 27

0 = 3x2 - 27 2 7 =3x2 9 = x2 ±3=x c. The partition numbers for the first derivative The partition numbers are when the first derivative equals 0 or undefined, therefore, the partition numbers are 3 and –3.

2. Given the function f(x)= a.

19 find: 2 x +6

f '(x)

f ( x) =

19

2x+6 1 =19(2x+6)f ( x) f '( x) =-19(2x+6)- 2 (2) or -

38 (2 x+6) 2

b. The critical values of the function

f'(x)=-

38 (2x+6)2

f¢(x) is always negative and cannot equal zero. f¢(x) is undefined atx = - 3.

However, x = - 3 is not in the domain of f, therefore there is no critical value. c. The partition numbers for the first derivative Partition numbers are found by setting the first derivative equal to zero (which can’t happen) and where it is undefined. It is undefined at x=-3,so the partition number is –3.

4-2 ..

Calculus

Chapter 4

3 – 5 For the functions given, find: a. The intervals on which the function is increasing b. The intervals on which the function is decreasing c. The local extrema 3.

f(x)=2x2-16x+33 First find the derivative and the critical values as follows:

f '( x) = 4 - 16 0 = x - 16 16 = 4 4 =x 4 Choose xa number on each side of the critical value to find the value of the first derivative: x Test number: 3

Test number: 5

f '(3 = 4 x -16 f '(5 = 4 x -16 f ) = 4(3) - 16 f ) = 4(5) - 16 f '(3 =12 - 16 f '(5 = 20 - 16 f ) =- 4 f ) =4 '(3 '(5 a. The test number 5 has a first derivative that is a positive value, therefore, the ) ) function is increasing on the interval (4,¥). '(5derivative that is a negative value, therefore, the b. '(3The test number 3 has a first ) function is decreasing on the ) interval (-¥,4). c. Because the function changes from a decreasing function to an increasing function at the critical value 4, the point (4,1) is a local minimum.

4-3 ..

Calculus 4.

Chapter 4

f(x)=x3-27x+12 First find the derivative and the critical values as follows:

f '()= x 3x2 - 27

0 = 3x2 - 27 2 7 =3x2 9 = x2 ±3=x The partition numbers (also critical values of f) arex=-3 and x=3. Choose a number within each interval to find the value of the first derivative: Test number: –4

Test number: 0

Test number: 4

f '(0 = 3(0)2 - 27 f '(4) = 3(4)2 - 27 f '(- = 3(-4) 2 - 27 f ) = 3(0) - 27 f '(4) = 3(16) - 27 f 4) = 3(16) - 27 f '(0 = 0 - 27 f '(4) = 48- 27 f '(- = 48- 27 f ) = - 27 f '(4) = 21 f 4) = 21 '(0 '(a first derivative that is a positive value, a. 4)The test numbers –4 and 4 have ) (- ¥ , - 3) and (3, ¥ ). therefore, the function is increasing on the intervals '(0 '(b. The test number 0 has a first derivative that is a negative value, therefore, the ) interval(-3,3).

4)function is decreasing on the c. Because the function changes from an increasing function to a decreasing (- 3, 66) is a local maximum. Because function at the critical value of –3 the point the function changes from a decreasing function to an increasing function at the critical value 3, the point (3,- 42) is a local minimum.

4-4 ..

Calculus 5.

Chapter 4

f(x)=x4-4x3+3 First find the derivative and the critical values as follows:

f '( x) = 4 3 -

0 = x 3 12x2 - 3) 0 = 4 2 (x0,3 = x

12x2

4 x = 0 and x = 3. Choose a number The partition numbers (also critical values of f) are x within each interval to find the value of the first derivative: x Test number: –1 Test number: 2 Test number: 4

f '(f 1) f '(f 1) '(1) '(1)

= 4(- 1) 3 - 12( -1) 2 = 4(- 1)- 12(1) = - 4- 12 = - 16

f '(2) = 4(2)3 - 12(2)2 f '(2) = 4(8) - 12(4) f '(2) = 32- 48 f '(2) = - 16

f '(4) = 4(4)3 - 12(4)2 f '(4) = 4(64) - 12(16) f '(4) = 256 - 192 f '(4) = 64

a. The test number 4 has a first derivative that is a positive value, therefore, the function is increasing on the interval (3,¥). b. The test numbers –1 and 2 have a first derivative that is a negative value, therefore, the function is decreasing on the intervals (-¥,0)and(0,3). c. The critical value 0 does not give a local extrema because the first derivative does not change sign atx = 0. Because the function changes from a decreasing function to an increasing function at the critical value 3, the point (3,-24) is a local minimum.

4-5 ..

Calculus 6–7

Chapter 4

For the functions given, find: a. The intervals on which the function is increasing b. The intervals on which the function is decreasing c. Sketch the graph of the function

6.

f(x)=2x3+3x2-12x First find the derivative and the critical values as follows:

f '( x) = 6x2 + x - 12

0 = 6x2 6 x - 12 0 = 6( 2++ x - 2) 0 = x +6 2)( x - 1) x - 2,1 = 6( x x = - 2 and x = 1. Choose a The partition numbers (also critical values of f) are number within each interval to find the value of the first derivative: Test number: –3

f '(f 3) f '(f 3) '(3) '(3)

Test number: 0

Test number: 2

f '(0 = 6(0)2 +6(0)- 12 f '(2) = 6(2)2 +6(2)- 12 f ) = 6(0) + 6(0) - 12 f '(2) = 6(4)+6(2)- 12 f '(0 = 0 - 0 - 1 2 f '(2) = 24+12- 12 f ) = - 12 f '(2) = 24 '(0 a. The test numbers –3) and 2 have a first derivative that is a positive value, therefore, the function is increasing on the intervals ( - ¥ , - 2) and (1,¥ ). '(0 b. The test number 0 has a first derivative that is a negative value, therefore, the ) on the interval(- 2,1). function is decreasing = 6(-3) 2 +6(-3) 12 = 6(9) + 6(- 3) - 12 = 54-18-12 = 24

c.

4-6 ..

Calculus 7.

Chapter 4

f(x)=x3-12x+4 First find the derivative and the critical values as follows:

f '()= x 3x2 - 12

0 = 3x2 - 12 1 2 =3x2 4 = x2 ±2=x The partition numbers (also critical values of f) arex=-2 and x=2.Choose a number within each interval to find the value of the first derivative: Test number: –3

f '(f 3) f '(f 3) '(3) '(3)

Test number: 0

Test number: 3

f '(3 = 3(3)2 - 12 f '(0 = 3(0)2 - 12 f ) = 3(9) - 12 f ) = 3(0) - 12 f '(3 = 27 - 12 f '(0 = 0 - 12 f ) =15 f ) = - 12 '(3 '(0 a. The test numbers –3)and 3 have a first derivative)that is a positive value, therefore, the function is increasing on the intervals ( - ¥ , - 2) and (2, ¥ ). '(3 '(0 b. The test number 0 has a first derivative that is a negative value, therefore, the ) ) on the interval(- 2, 2). function is decreasing c. = 3(-3) 2 - 12 = 3(9) - 12 = 27 - 12 =15

4-7 ..

Calculus

Chapter 4

4-8 ..

Calculus

Chapter 4

Name ________________________________ Date ______________ Class ____________

Section 4-2 Second Derivatives and Graphs Goal: To use the second derivative to analyze graphs

Notation: Second Derivative

For y=f(x),the second derivative of f, provided that it exists, is f''(x)=

d

f'(x).

dx

Definition: Concavity

f '(x) is The graph of a function f is concave upward on the interval (a, b) if f '(x) is decreasing on increasing on (a b) and is concave downward on the interval (a, b) if (a, b).

Summary:

Concavity

For the interval (a,b), iff ''(x) > 0, then f '(x) is increasing and the graph of f is concave upward. If f ''( x )< 0, then f '(x) is decreasing and the graph of f is concave downward.

Theorem:

Inflection Point

x = c, then either If y = f (x) is continuous on (a, b) and has an inflection point at f ''(c) = 0 or f ''(c) does not exist.

Procedure:

Graphing Strategy (first version)

Step 1 Analyze f (x). Find the domain and the intercepts Step 2 Analyze f '( x ).Find the partition numbers for, and the critical values of, f '( x ),and determine local extrema.

Step 3 Analyze f ''( x ).Find the partition numbers forf ''( x ),and determine concavity.

Step 4 Sketch the graph of the function f.

4-9 ..

Calculus

Chapter 4

1. Given the function f(x)=3x4-13x2+13x+1,find f''(x).

f '(x) =12 x3 - 26 x+13 ) 36 x2 - 26 f ''( x =

12

2. Given the function f (x) = 2 x5 + 7 x 5 +16, find f ''( x ).

7

5 f '(x) =10 x4 + 84 5x 2

f ''( x = ) 40 x3 + 588 x 5 25

3. Given the function f(x)=2x3-12x2+26x-2,find the x and y coordinates of all inflection points.

Inflection points are found by setting the second derivative equal to zero. f '( x) = 6 x2 - 24 x+26 f ''(x) =12 x - 24

2 + 2 6x - 2 f (2) = 2 x3 - 1x2 f ''(x) =12 - 24 0 =x - 24 f (2) = 2(2)3 - 12(2)2 + 2 6 ( 2 ) 224 =12 f (2) =16- 48+52- 2 2 =xx f (2) =18 12 x Now check the value of the second derivative on both sides of 2.

f''(1)=12(1)-24 f ''(1)=-12

f''(3)=12(3)-24 f ''(3)=12

Because the second derivative changes from negative to positive, the point (2, 18) is an inflection point.

4-10 ..

Calculus

Chapter 4

4 – 6 For the functions given, find: a. The intervals on which the function is concave upward b. The intervals on which the function is concave downward c. The inflection points 4.

f(x)=x4-54x2 First find the second derivative and the partition numbers as follows:

f '(x) = 4x3 - 108 x ) 12 2 - 108 f ''( x = 0 =x

2

- 108

108 =12 2 9 =xx2 x ±3 =12 x The partition numbers arex = - 3 and x = 3. Choose a number on each side of the partition numbers. Find the value of the second derivative at these test numbers Test number: –4

Test number: 0

Test number: 4

f ''(0 =12(0)2 - 108 f ''(4 =12(4) 2 - 108 f ''(- =12(-4) 2 - 108 f ) =12(0) - 108 f ) =12(16) - 108 f 4) =12(16) - 108 f ''(0 = 0 - 108 f ''(4 =192 - 108 f ''(- =192 - 108 f ) = - 108 f ) = 84 f 4) = 84 ''(0 ''(4 ''(a second derivative that is a positive value, a. 4)The test numbers –4 and 4 have ) ) therefore the function is concave upward on the intervals ( - ¥ , - 3) and(3, ¥ ). ''(0 ''(4 ''(b. The test number 0 has a second derivative that is a negative value, therefore the ) on the interval ) 4)function is concave downward (-3,3). c. The partition numbers –3 and 3 are both inflection points because the function (- 3, -405) and changes concavity at each number. The inflection points are (3,- 405).

4-11 ..

Calculus 5.

Chapter 4

f(x)=2x3-6x2+12x-9 First find the second derivative and the partition numbers as follows:

f '(x) = 6x2 - 12 x+12 ) 12 - 12 f ''( x = 0 =x - 12 12 =12 1 =xx 12 The partition number is x=1. Choose a number on each side. Find the value of the x at these test numbers second derivative Test number: 0

Test number: 2

f ''(0 =12(0) - 12 f ''( 2 =12(2) - 12 f ) = 0 - 12 f ) ''( = 24- 12 f ''(0 = - 12 f 2 ) =12 ) ''( 2 The test number 2 has a second derivative that is a positive value, therefore the a. ''(0 ) (1,¥ ). function is concave upward on the interval ) b. The test number0 has a second derivative that is a negative value, therefore the function is concave downward on the interval (- ¥ ,1). c. The partition value 1 is an inflection point because the function changes concavity (1, - 1). at this number. The inflection point is

4-12 ..

Calculus 6.

Chapter 4

f(x)=9e2x-e3x First find the second derivative and the partition numbers as follows:

f '(x) =18e2 x - 3e3 x ) 36e 2 x - 9e3 x f ''( x = 0 =36e 2 x - 9e3 x 3 =36e 2 x 9ex ex =4 x =ln4

The partition number is x=ln4. Choose a number on each side.Find the value of the second derivative at these test numbers

Test number: 0 2(0) 0

f ''(0 = 36e

Test number: 2

- 9e3(0)

f ''(2 = 36e2(2)- 9e3(2)

f ) = 36e - 9e0 f ) = 36e4 - 9e6 f ''(0 = 36- 9 f ''(2 =1965.53- 3630.86 f = 27 f = 1665.33 ) ) ''(0 ''(2 a. The test number 0 has a second derivative that is a positive value, therefore the ) ) function is concave upward on the interval ( - ¥ , ln 4). ''(0 ''(2 b. The test number 2 has a second derivative that is a negative value, therefore the (ln 4, ¥ ). ) function is concave downward ) on the interval c. The partition number ln4 is an inflection point because the function changes concavity at this number. The inflection point (ln is 4,80).

4-13 ..

Calculus

Chapter 4

7–9 Apply the four-step graphing procedure and sketch the graph of the following functions.

7.

f(x)=(x+2)(x2-x-4)

Step 1: Analyze f(x). The domain of the function is all real numbers. To find the intercepts, first find f (0). Next, find the solutions off (x) = 0.

f (0) = (0+ 2)((0)2 -0-4) - 4) f (0) = (2)( f (0) = - 8

0 = (2)( x+

x2- x - 4)

x2 - x - 4 = 0 x= 1 ± 17 2

x + 2 =0or x = -2

Therefore, the x intercepts are1(+17 , 0), (1- 17 ,0) and (–2, 0), and they intercept is 2

2

(0, –8).

Step 2: Analyze the first derivative.

f '(x) = (x + 2)(2x -1)+ ( x2 - x - 4)(1) f '(x) = 2x2 +3 x - 2 + x2 - x - 4 0 = 3x2 + 2 x - 6 - 1 ± 2 5 =x 3 The partition numbers are x=-1+25 and x=-1-25. Choose a number within each 3

3

interval. Find the value of the first derivative at these test numbers Test number: –2

Test number: 0

Test number: 2

f '(2) = 3(2)2 + 2(2) - 6 f '(- = 3(-2) 2 +2( -2) - 6 f '(0 = 3(0) 2 + 2(0) - 6 f ) = 3(0) - 0 - 6 f '(2) = 3(4) + 4 - 6 f 2) = 3(4)- 4- 6 f '(0 =- 6 f '(2) =10 f '(- = 2 ) 2) Therefore, the function is increasing on the intervals (-¥,-1-25)and ( - 1+32 5 ,¥), '(0 '(3 ) 2) decreasing on the interval (-1-25,-1+25), and has a local maximum at x=-1-25 3

3

3

local minimum atx=-1+25.

3

4-14 ..

and a

Calculus

Chapter 4

Step 3: Analyze the second derivative

f '(x) = 3 2 + 2 x - 6 ) x +2 f ''( x = 0 = 6 +2 - 2 =x - 31 = 6x x 6 The partition number is x=-1. Choose a number on each side of the partition 3 x number. Find the value of the second derivative at these numbers.

Test number: –1

Test number: 0

f ''(- = 6 x + 2 f ''(0 = 6 x + 2 f 1) = 6(-1) + 2 f ) = 6(0) + 2 f ''(- = 4 f ''(0 = 2 1) ) Therefore, the function is concave ''(''(0 upward on the interval13 ,(-¥ ), concave 1) on the interval (-¥,-1), and) has an inflection point atx = - 1 . downward 3

3

Step 4: Sketch the function

4-15 ..

Calculus 8.

Chapter 4

f(x)=5x4-10x3

Step 1: Analyze f (x). The domain of the function is all real numbers. To find the intercepts, first find f (0). Next, find the solution off (x) = 0.

f (0) = 5(0)4 -10(0)3 f (0) = 0 - 0 f (0) = 0

0= 0 5x4 - 10x3 x = 5x3(x - 2) = 0,2

Therefore, the x intercepts are (0, 0) and (2, 0), and theintercept y is (0, 0).

Step 2: Analyze the first derivative.

f '( x) = 20x3 - 30x2

0 =10x2 (2x- 3) 3 0, 2 = x The partition numbers are x=0 and x=3. Choose a number within each interval.

2

Find the value of the first derivative at these test numbers.

Test number: –1

Test number: 1

Test number: 2

f '(- = 20(- 3 - 30(-1) 2 f '(1 = 20(1 3 - 30(1)2 f '(2) = 20( 2)3 - 30( 2)2 f 1) = 1) f ) =) f '(2) = 20(8) 30( 4) - 30(1) - 30(1) f '(- =-20(f '(1 =-20(1 f '(2) = 40 10 50 1) 1) ) ) Therefore, the function is increasing on the interval (3,¥ )and decreasing on the intervals '('(1 2 The function )has a local minimum atx=3.

(-¥1),0) and (0,3).

2 2

4-16 ..

Calculus

Chapter 4

Step 3: Analyze the second derivative

f '(x) = 20x - 30x2 )3 f ''( x = - 60x 0 = 60x - 60x 2 0 = 60x(x - 1) 0,1 = x60x The partition numbers are x=0 and x=1. Choose a number on each side of the 2 partition number. Find the value of the second derivative at these test numbers.

Test number: –1

Test number: 0.5

Test number: 2

f ''(- = 60(-1) 2 - 60( -1) f ''(0.5 = 60(0.5)2 - 60(0.5) f ''(2 = 60( 2 - 60( 2) f 1) = 60(1) + 60 f) = 60(0.25) - 30 f ) = 2) - 120 f ''(- =120 f ''(0.5 = - 15 f ''(2 =120 60( 1) ) ) 4) ,0) and (1,¥ )and Therefore, the function is concave upward on the intervals (-¥ ''(''(0.5 ''(2 =0. concave downward on the interval (0, 1), and has an inflection pointxat 1) ) ) Step 4: Sketch the function

4-17 ..

Calculus 9.

Chapter 4

f(x)=8ex-e2x

Step 1: Analyze f (x). The domain of the function is all real numbers. To find the intercepts, first find f (0). Next, find the solutions off (x) = 0.

0= 0 8ex - e 2x x = ex(8- e x) = ln8

f (0) = 8e0 - e2(0) f (0) = 8(1)- 1 f (0) = 7

Therefore, the x intercept is(ln 8, 0),and the yintercept is (0, 7).

Step 2: Analyze the first derivative.

f '( x) = 8e x - 2e2x

0 = 8e x - 2 e2x 0 = 2e x ( 4 -e x ) ln4 = x The partition number is x=ln4. Choose a number within each interval. Find the value of the first derivative at these test numbers.

Test number: 0

Test number: 2

f '(0) = 8e0 - 2e2(0) f '(0) = 8(1)- 2(1) f '(0) = 4

f '(2) = 8e2- 2e2(2) f '(2) = 8 ( 7 .2(54.60) 3 9 ) f '(2) =-5 0 . 0 8

-

Therefore, the function is increasing on the interval (- ¥ , ln 4)and decreasing on the interval The4,function has a local maximum at x= ln4. ¥ ). (ln

4-18 ..

Calculus

Chapter 4

Step 3: Analyze the second derivative

f '(x) = 8e x - 2e2x ) 8e x - 4 e2x f ''( x = 0 = 8e x - 4 e2x 0 = 4e x ( 2 -e x ) ln2 = x

The partition number is x=ln2. Choose a number on each side of the partition number. Find the value of the second derivative at these test numbers.

Test number: 0

Test number: 1

f ' ' = 8e1- 4e2(1) f ''(0 = 8e0 - 4e2(0) f (1) = 8 ( 2 .4(7.39) 7 2 ) f ) = 8(1)- 4(1) f ' ' =- 7.81 f ''(0 = 4 (1) ) Therefore, the function is concave upward on the interval ( - ¥ , ln 2), concave ' ' ''(0 downward on the interval(ln 2,¥ ), and has an inflection point atx = ln2.

(1) ) Step 4: Sketch the function

4-19 ..

Calculus

Chapter 4

10. A company estimates that it will sell N(x)units of a product after spending x thousand dollars on advertising, as given by

N(x)=-0.25x4+23x3-360x2+45,000

25£x£54

When is the rate of change of sales increasing and when is it decreasing? What is the point of diminishing returns and the maximum rate of change of sales?

To find when the rate of change is increasing and decreasing, find the second derivative and use the second derivative test.

N '( x) = x3 + 69 x2 - 720 x N ''(x) -

3x2 +138x - 720

0 = 3( 2 - 46x + 240) 0 - x - 40)( x- 6) 6,40 = = x3( x The partition numbers are x=6 and x=40. Choose a number on each side of the = within the given domain. Find the value of the second derivative at these partition numbers test numbers. Test number: 39

Test number: 41

f ''(39 = 3(39)2 +138(39) - 720 f ''(41 = 3(41)2 +138(41) - 720 + f) - 4563 + 5382- 720 f) - 5043 5658- 720 f ''(39 = 99 f ''(41 = 105 ) ) a. The test number 39 has a second derivative that is a positive value, therefore the ''(39 ''(41 = function is concave upward on the interval (25,40) and so the sales are also ) increasing over this interval. ) b. The test number 41 has a second derivative that is a negative value, therefore the function is concave downward on the interval (40,54) and so the sales are also decreasing over this interval. c. The partition number 40 is the point of diminishing returns because it is an inflection point. The maximum rate of change is N'(40)=17,600.

4-20 ..

Calculus

Chapter 4

Name ________________________________ Date ______________ Class ____________

Section 4-3 L’Hôpital’s Rule Goal: To find limits of functions that are in the form zero over zero or infinity over infinity

Theorems : 1.

L’Hôpital’s Rule for 0/0 Indeterminate forms (Version 1) For c a real number, if limf(x)= 0 and lim g(x) = 0, then x®c

x®c

f(x) f '(x) lim =lim x®c g(x) x®c g'(x) ±¥. providing that the second limit exists or is

2.

(Version 2) The first version of L’Hôpital’s rule remains valid if the symbol x®cis replaced everywhere it occurs with one of the following symbols: x®c+

3.

x®c-

x®¥

x®-¥

Versions 1 and 2 of L’Hôpital’s rule for the indeterminate form 0/0 are also valid if the limit of f and the limit of g are both infinite.

Use L’Hôpital’s rule to find the limit in Problems 1 and 2.

1.

x3 - 64 3x2 3(4)2 48 lim =lim = = =4 x ® 4 x 2 + 4 x - 3 2x ® 4 2 x + 4 2 ( 4 ) + 4 12

2.

3x4 +17 12x3 12 lim =lim =lim =0 x® ¥ 6x5 +25x® ¥ 30x4 x® ¥ 30x

4-21 ..

Calculus

Chapter 4

3.

Explain why L’Hôpital’s rule does not apply to the following problem. If the limit exists, find it by other means.

3x2 - 7 lim x® 2 2x3 - 13 L’Hôpital’s rule does not apply because when the value is substituted into the function, the result is not 0/0.

3x2 - 7 3(2)2 - 7 5 lim = = x® 2 2x3 - 132(2)3 - 13 3

4.

Use L’Hôpital’s rule to find the following limit and then verify that limit by a second method. x2 - 10x+21 2x- 10 2(7)- 10 4 1 lim =lim = = = x ® 7 x 2 - 2 x - 3 5x ® 7 2 x - 2 2(7)- 2 12 3

lim x®7

x2 - 10 x + 21

( x 7)(x- 3) = lim ( x®7 - 7)(x+5) x2 - 2 x - 35 xx- 3 = lim x +5 x®7 4 1 = 7 - 3 = 12 = 3 7+5

Find each limit in Problems 5–8. Note that L’Hôpital’s rule does not apply to every problem and some problems will require more than one application of L’Hôpital’s rule.

5.

lim

x3+7x-6

x® -1 x2

- x

If the value –1 is substituted into the function, the result will be 0/0. Therefore, we use L’Hôpital’s rule.

lim

x® -1

x3 +7x- 6 x2 - x

3x2 +7 3(- 1)2 +7 10 10 =lim = = =x® -1 2x- 1 2(- 1)- 1 -3 3

4-22 ..

Calculus

6.

Chapter 4

x4-6x-3 lim x®2x3+2x-6

The value can be substituted into the function and not result in 0/0. Therefore, use simple substitution.

7.

x 4 - 6 x - 3 ( 2 ) 4 - 6 ( 2 ) - 31 lim = = x® 2 x3 +2x- 6(2)3 +2(2)- 6 6 7x3+4x-5 lim x® ¥ 9x4 - 6x+7

If the value ¥ is substituted into the function, the result will be ¥/¥. Therefore, we use L’Hôpital’s rule.

7 x3 + 4 x - 5 21x2 + 4 lim lim = x® ¥ 9 4 ¥ 3 x - 6 x + 7 x® lim 36xx - 6 42 = x2 x®¥ 108 42 = lim x®¥ 216 x =0 8.

x+3 lim x®-3x2-2x-15

If the value –3 is substituted into the function, the result will be 0/0. Therefore, we use L’Hôpital’s rule. x+3 1 1 1 lim =lim = =x ® - 3 x 2 - 2 x - 1x5® - 3 2 x - 2 2 ( - 3 ) - 2 8 Alternative Solution:

x+3 x +3 = lim lim x® - 3 x2- 2 x - 15 x®- 3(x+3)( x - 5)

1 = lim x®-3x-5 1 1 = =8 - 3-5 4-23 ..

Calculus

Chapter 4

4-24 ..

Calculus

Chapter 4

Name ________________________________ Date ______________ Class ____________

Section 4-4 Curve-Sketching Techniques Goal: To graph functions using the final version of the graphing strategy.

Procedure :

Graphing Strategy (Final Version)

Step 1 Analyze f(x). a.

Find the domain. b.

Find the intercepts. c.

Find the asymptotes.

Step 2 Analyze f'(x).Find the partition numbers for, and the critical values of, f'(x). Determine the intervals on which f is increasing and decreasing, and find the local extrema.

Step 3 Analyze f''(x).Find the partition numbers for f''(x). Determine the intervals on which the graph of f is concave upward or downward, and find the inflection points.

Step 4 Sketch the graph of the function.

In problems 1–3, summarize the pertinent information obtained by applying the graphing strategy and sketch the graph.

1.

x-4 f(x)= Step 1: x + 4

Analyze the function.

a.

The domain is all real numbers except wherex + 4 = 0Þ x = - 4.

The x intercepts are found by setting the function (or the numerator) equal to b. zero and the y intercept is found by setting the value of x equal to zero.

f(0)=

x-4=0 x=4

0- 4 -4 = =-1 0+4 4

Therefore, the x intercept is (4, 0), and the y intercept is (0, –1). c.

The vertical asymptote will be at the point of discontinuity xor= - 4.

The horizontal asymptote is found by taking the limit of the function as it goes to infinity. Since the exponents are the same in the numerator and denominator, it is found by calculating the ratio of the coefficients. Therefore, the horizontal asymptoteyis=1.

4-25 ..

Calculus Step 2:

Chapter 4 Analyze the first derivative.

f '( x) =

( x +4)(1) - (x - 4)(1) (x+4)2

f '( x) =

8 ( x +4)2

There is only one partition number, which is the value that makes the denominator zero, or x = - 4. Pick a test number on each side of the partition number and find the value of the first derivative at these test numbers.

Test number: –5

Test number: –3

8

f'(-5) =

2

(- 5 + 4 ) 8 f '(- 5) = =8 (- 1)2

f'(-3) =

8

(- 3+4) 2 8 f '(- 3) = =8 (1)2

( - ¥ , - 4) and (-4, ¥ ).

Therefore, the function is increasing on the intervals Step 3:

Analyze the second derivative

f '(x) = 8(x + 4)- 2 ) 16(x f ''( x = + 4)- 3 = - 16 ''( x ) f (x + 4)3 There is only one partition number, which is the value that makes the denominator zero, or x = - 4. Pick a test number on each side of the partition number and find the value of the second derivative at these test numbers.

Test number: –5 - 16 f''(-5) = (- 5 + 4)3 - 16 = f ''(- 5) = 16 (- 1)3

Test number: –3 - 16 f''(-3) = (- 3 + 4)3 16 f ''(- 3) = = - 16 (1)3

Therefore, the function is concave upward on the interval ( - ¥ , - 4) and concave downward on the interval(-4, ¥ ). However, the function does not have an inflection point because f (-4) is undefined.

4-26 ..

Calculus

Chapter 4

Step 4:

2.

f(x)=

Sketch the graph

-4x x-3

Step 1:

Analyze the function.

a.

The domain is all real numbers except wherex - 3 = 0 Þ x = 3.

The x intercepts are found by setting the function (or the numerator) equal to b. zero and the y intercept is found by setting the value of x equal to zero. f(0)=

-4x=0 x=0

-4(0) 0 = =0 (0- 3) -3

Therefore, the x intercept is (0, 0), and the y intercept is (0, 0).

The vertical asymptote will be at the point of discontinuity xor= 3.

The horizontal asymptote is found by taking the limit of the function as it goes to infinity. Since the exponent in the numerator is the same as the denominator, the horizontal asymptote is y=-4=-4.

1 c.

Step 2:

Analyze the first derivative.

- 4x x- 3 ( x - 3)(-4)- (-4x)(1) f '(x) = (x- 3)2 f (x) =

f '( x) =

12 ( x - 3)2

There is only one partition number; the value that makes the denominator zero x = 3. Pick a test number in each interval. Find the value of the first derivative at these test numbers. is

4-27 ..

Calculus

Chapter 4

Test number: 2

f'(2)=

Test number: 4

12

f'(4)=

(2- 3) 2 12 f '(2) = =12 1

12

(4- 3) 2 12 f '(4) = =12 1

Therefore, the function is increasing on the intervals - ¥( ,3) and (3, ¥ ). Because the function does not decrease, there is no local extrema.

Step 3:

Analyze the second derivative

f'(x)=

12 (x- 3)2

f ''(x) =

(x- 3)2(0) - (12)(2(x - 3 ) (1) (x - 3)4

f ''(x) =

- 24 (x - 3)3

There is only one partition number; the value that makes the denominator zero is x=3. Pick a test number in each interval. Find the value of the second derivative at these test numbers.

Test number: 2

f''(2) =

- 24

(2 - 3)3 - 24 f ''( 2 = = -1 f ) ''( 24

Test number: 4

f''(4) =

- 24

(4 - 3)3 - 24 f ''( 4 = 1 = f ) ''( 24

2) 4) Therefore, the function is concave upward on the interval (-¥,3)and concave downward on the interval (3,¥). However, the function does not have an inflection point because f(3) is undefined.

4-28 ..

Calculus

Chapter 4

Step 4:

3.

Sketch the graph

f(x)=e-3x

2

Step 1:

Analyze the function.

a.

The domain is all real numbers. b.

There are no x intercepts and the y intercept is found by setting the value of x equal to zero.

f(0)=e-3(0)2 =e0=1 Therefore, there is no x intercept, and the y intercept is (0, 1). c.

There is no vertical asymptote because the domain is all real numbers.

The horizontal asymptote is found by taking the limit of the function as it goes to infinity. Therefore, the horizontal asymptote is y=0.

Step 2:

Analyze the first derivative.

f'(x)=-6xe-3x2 There is only one partition number; and also a value that makes the function zero or x=0. Pick a test number on each side of the critical value and find the value of the first derivative at these test numbers.

Test number: –1

Test number: 1 2

f '(-

= - 6(- 1)e-3(-1)2

f '(1 = 6(1)e- 3(1)

f 1) f '(-

= 6e - 3 = 0.299

f ) - 6e - 3 f '(1 = 0.299

1) '(1)

) '(1 = ) 4-29 ..

Calculus

Chapter 4

Therefore, the function is increasing on the interval (-¥,0)and decreasing on the interval (0,¥), and has a local maximum at x=0.

Step 3: Analyze the second derivative -3x2

f ¢( x) = 6xe f ¢¢(x) -

2

2

6x(-6 xe- 3 x ) + e - 3 x (-6) 2

f ¢¢(x) = 6e-3x (6x2 - 1) There are two partition numbers. Set the second derivative equal to zero to find the partition numbers x»±0.41. Pick a test number in each interval of the partition number and find the value of the second derivative at these test numbers.

Test number: –1

Test number: 0 2

2

f ''(-

= 6e- 3(- 1) (6(-1)2 -1)

f ''(0 = 6e - 3(0) (6(0) 2-1)

f 1) f ''(-

= 6e- 3 (5) =1.49

f ) = (6)e0 ( - 1) f ''(0 = 6 ) ''(0 )

1) Test number: 1 ''(e-3(1)2 1) (6(1)2 -1) f ' ' =6 f (1) = 6e- 3 (5) f ' ' =1.49

(1) (- ¥ , - 0.41)and Therefore, the function is concave upward on the intervals ' ' (-0.41,¥). The function is concave downward on the interval( -0.41,0.41). The function has (1) inflection points at x=-0.41 and x=0.41.

Step 4:

Sketch the graph

4-30 ..

Calculus

Chapter 4

Name ________________________________ Date ______________ Class ____________

Section 4-5 Absolute Maxima and Minima Goal: To find absolute maxima or minima on open and closed intervals

Definition: Absolute Maxima and Minima

If f(c)³f(x)for all x in the domain of f, then f(c)is called the absolute maximum.

If f(c)£f(x)for all x in the domain of f, then f(c)is called the absolute minimum.

Theorems: 1.

A function f that is continuous on a closed interval [a, b] has both an absolute maximum and an absolute minimum on that interval.

2. endpoints.

Absolute extrema (if they exist) must always occur at critical values or at

3.

Second Derivative Test Let f be continuous on an interval I with only one critical value c in I. If f'(c)=0and f''(c)>0,then f(c)is the absolute minimum of f on I. If f'(c)=0and f''(c)<0,then f(c)is the absolute maximum of f on I.

Procedure: Finding absolute extrema on closed intervals 1.

2.

3.

4.

5.

Check to make certain that f is continuous over [a, b].

Find the critical values in the interval (a, b).

Evaluate f at the endpoints a and b and at the critical values found in step 2.

The absolute maximum is the largest value found in step 3.

The absolute minimum is the smallest value found in step 3.

4-31 ..

Calculus

Chapter 4

1.

Find the absolute maximum and the absolute minimum for the given function on the given interval.

f(x)=3x2-12x+17 [-1,8] First find the critical value(s) for the function.

f (x) = 3 2 - 12 x+17 f '(x) = x - 12 0 = 6 - 12 12 = x 2 =6 x Calculate the function value at the endpoints and the critical value.

6 f (-1) =x3(- 1) 2 - 12( -1) +17 f (2) = 3(2)2 - 12(2) +17 f (8 = 3(8) 2- 12(8)+17 f (2) =12- 24+17 12+17 f (-1) =x3 + f ) =192 - 96 + 17

f (2) = 5

f (-1) = 32

f (8 =113 ) Therefore, the absolute maximum is at the point (8, 113) and the absolute minimum is (8 at the point (2, 5). ) In Problems 2–3, find the absolute maximum and the absolute minimum, if either exists, for each function.

2.

f(x)=5x3-6x4

First find the critical value(s) for the function.

f (x) = 5x3 - 6x4 f '(x) =15x2 - 24x3 0 = 3x2(5- 8x)

3x = 0 2 =0 xx = 0

5 - 8x = 0 - 8x = - 5

x= 5 8

2 Then use the second-derivative test at the critical values.

f¢¢(x)=30x-72x2

f ¢¢(0) = 30(0) -72(0)2 f ¢¢(0) = 0

f ¢ 58 ) = 30( 58 )-72(5)2 8 f ¢( 5 ) = - 75 8

8

¢ ..

¢(

4-32

Calculus

Chapter 4

.

The second-derivative test fails at 0 and shows that there is a local maximum85 at

Calculate the function value at the critical values.

f (0) = 5(0)3 -6(0)4 f (0) = 0-0 f (0) = 0

- 6(5)4 f ( 85 ) = 5(5)3 8 8 f ( 85 ) = 5(125)- 6( 625 )

512 4096 5 f ( ) = 625 - 3750 4096 8 512 f ( 5 ) = 625 8 2048

625 ) and the local minimum is at the ( 5 , 2048 Therefore, the local maximum is at the point 8

point (0, 0). Also, f(5)=625 is the absolute maximum; there is no absolute minimum.

8

f(x)=

3.

2048

x2-4 x2+1

First find the critical value(s) for the function.

f(x)=

f '( x) = f '(x) =

x2-4 x2+1 (x2 +1)(2x)- ( x2 - 4)(2x) (x2 +1)2

10 x ( x2 +1)2

0 =10 x 0=x To determine if this is a maximum or a minimum, we find the second derivative and use the second-derivative test.

4-33 ..

Calculus

Chapter 4

f '( x) = f ''(x) =

10 x x 4 +2 x2 + 1 (x4 +2 x2 +1)(10)- (10x)(4x3 + 4x) (x4 +2x2 +1)2

f ''(x) =

- 30x4 - 20x2 +10 (x4 +2x2 +1)2

f ''(0) =

- 30(0)4 - 20(0)2 +10 ((0)4 +2(0)2 +1)2

f ''(0) =

10

=10>0

(1)2 Therefore, (0,–4) is a local minimum point because the second derivative is positive.

Also, f(0)=-4 is the absolute minimum; there is no absolute maximum.

In Problems 4–6, find the indicated extremum of each function on the given interval.

7 4.

Absolute minimum value on (0,¥)for f(x)=6+2x+ x .

First find the critical value(s) for the function.

f (x) = 6 2 + 7 x - 1 x -2

f '(x) +

7 f '(x) = 2 7 2 xx 7 0=2- 2 =2 x

7

=2

x2 2x2 = 7 x2 = 7 2

x= 7 2 x » 1.87

There is only one critical value. To determine if this is a minimum, we will find the second derivative and use the second-derivative test.

f '(x) = 2 - 7 x- 2

f ''(1.87 =14(1.87)- 3 f) » 2.14 ) 14 x- 3 f ''( x = ''(1.87 The second derivative of the critical value is positive, therefore, the absolute ) minimum is f(x)=f(7)»13.48.

2

4-34 ..

Calculus 5.

Chapter 4

Absolute minimum value on [0,¥)for f(x)=10x2-8x+26.

First find the critical value(s) for the function.

f (x) =10x2 - 8x + 26 f '(x) = 20x - 8 0 = 20x - 8 -20 x =- 8 x = 0.4 There is only one critical value. To determine if this is a minimum, we will find the second derivative and use the second-derivative test.

f '(x) = 20 x - 8 ) 20 f ''( x = f ''(0.4) = 20

The second derivative of the critical value is positive, therefore, the absolute minimum is f(x)=f(0.4)=24.4.

4-35 ..

Calculus 6.

Chapter 4

Absolute maximum value on (0,¥)for f(x)=3x6-18x4.

First find the critical value(s) for the function.

f (x) = 3x6 - 18x4 f '(x) =18 x5 - 72x3 0 =18 x3 (x2 - 4) 18 x3 = 0 or x2 - 4 = 0 x2 = 4 x =0 x = ±2 Only x=2 is in the interval (0,¥). To determine if the critical value x = 2 is a maximum, we find the second derivative and use the second-derivative test.

f '(x) =18x5 - 72x 3 ) 90x4 - 216x 2 f ''( x = f ''(2) = 90(2)4 - 216(2) 2 f ''(2) =1440- 864 f ''(2) = 576 The second derivative of the critical value is positive, therefore, the absolute maximum is f(x)=f(2)=-96.

4-36 ..

Calculus

Chapter 4

Name ________________________________ Date ______________ Class ____________

Section 4-6 Optimization Goal: To solve application problems that involve maximum and minimum values

Procedure : 1.

2.

3.

4.

Strategy for Solving Optimization Problems Introduce variables, look for relationships among the variables, and construct a mathematical model of the form Maximize (or minimize) f(x)on the interval I. Find the critical values of f(x).

Use the procedure developed in Section 5–5 to find the absolute maximum (or minimum) value off(x)on the interval I and the value(s) of x where this occurs.

Use the solution to the mathematical model to answer all the questions asked in the problem.

Solve the following problems using the four-step procedure outlined above.

1. Find two numbers whose sum is 36 and whose product is maximum.

Step 1:

Maximize A = xy

subject to x + y = 24

Step 2:

x+ y = 24 x = 24- y

A(y = xy ) = - y) y (24 A(y = 24y - y2 )

Find the first derivative:

A(y A ( y) = 24y - y2 ' )(= A ) y 24 - 2 y 0 = 24 - 2 y 2 y = 24 y =12 x = 24- y = 24 - 12 =12

4-37 ..

Calculus

Chapter 4

Step 3: Use the second-derivative test to determine whether the critical value produces a maximum value.

A'(y)=24-2y A''(y)=-2 Since the second derivative is always negative, the value –2 gives a maximum.

Step 4: The numbers 12 and 12 have a sum 24 and gives the maximum product 144.

2.

A fence is to be built to enclose a rectangular area of 900 square feet. The fence along three sides is to be made of material that costs $14 per foot. The material for the fourth side costs $42 per foot. Find the dimensions of the rectangle that will allow for the most economical fence to be built.

Step 1:

Minimize C = 28x +56y

Step 2:

xy = 900 x=

subject to xy = 900

C(y = 2856 x+ y = 28(900 y- 1 )+56y

900

)

y

C(y = 25,200 y- 1 +56y

x = 900 y- 1 Find the first derivative:

) C(y C (y) = 25, 200y- 1 +56y ) y- 2 + 56 C y' )(=- 25, 0 =- 200 25, 200y- 2 = 56 25,

y- 2 + 56

56 y = 25, 200200 2 = 450 yy = 3 50 » 21.21 2 900 » 424.43 x = 900 y = 3 50

4-38 ..

Calculus

Chapter 4

Step 3: Use the second-derivative test to determine whether the critical value produces a maximum value.

C '( y) = - 25,200y-2 +56 C ''( y) = 50, 400 y - 3 C''(3 50) = 50,400(3 50)-3 C''(3 50) » 5.28

Since the second-derivative has a positive value, the critical value50 3 produces a minimum value.

Step 4: A rectangular field that has a width of 21.21 feet (at a cost of $14.00 per foot) and a length of 42.43 feet (at a cost of $42.00 per foot) will minimize the cost of the fence, which is approximately $2,970.

3.

A commercial apple grower estimates from past records that if 40 trees are planted per acre, then each tree will yield an average of 60 pounds of apples per season. If, for each additional tree planted per acre (up to 30), the average yield is reduced by 1 pounds, how many trees should be planted per acre to obtain the maximum yield per acre? What is the maximum yield?

Step 1:

Maximize P(x) = (40 + x)(60 - x)

Step 2:

P(x = (40 + x )(60 - x) )

subject to x £ 30

= 2400 + 20x - x2

P(x Find the first derivative: )

P( x) = 2400+20 - x2 P ' x() = x 20- 2x 0 = 20- 2x 20 2x = x =10

Step 3: Use the second-derivative test to determine whether the critical value produces a maximum value.

P'(x)=20-2x P''(x)=-2

Since the second derivative is always negative, the critical value gives an absolute maximum.

4-39 ..

Calculus

Chapter 4

Step 4: The variable x is the number of additional trees that should be planted.

Therefore, 10 additional trees should be planted per acre for a total of 50 trees per acre. The production will be reduced by 10 pounds per tree for a production level of 50 pounds per tree and a total production of 2500 pounds per acre.

4.

A 400-room hotel in New York City is filled to capacity every night at $100 a room.

For each $2 increase in rent, 4 fewer rooms are rented. If each rented room costs $12 to service per day, how much should the management charge for each room to maximize gross profit? What is the maximum gross profit?

Step 1:

Maximize P(x) = R(x)- C(x)

Step 2:

P(x = (400 - 4 x)(100 + 2x) -12(400-4x) )

subject to x£100

= 35, 200 + 448x - 8x2

P(x Find the first derivative: )

P x( ) = 35, 200 + - 8 x2 448P 'x)(= 448x 0 = 16x 448- 16x 16x = 448 28 x=

Step 3: Use the second-derivative test to determine whether the critical value produces a maximum value.

P'(x)=448-16x P''(x)=-16

Since the second derivative is always negative, the critical value gives an absolute maximum.

Step 4: The variable x is the number of $2 increases that the management should add to the price of the room. Therefore, management should increase the price of the room by $56, which becomes $156 per room. This increase result’s in a decrease of 112 rooms being rented, or 288 rooms. The profit will be 288 rooms rented for $156 per room minus the $12 per room spent to clean them. The maximum profit would be $41,472.

4-40 ..

Calculus

Chapter 5

Name ________________________________ Date ______________ Class ____________

Section 5-1 Antiderivatives and Indefinite Integrals Goal: To find antiderivatives and indefinite integrals of functions using the formulas and properties

Theorem 1 Antiderivatives If the derivative’s of two functions are equal on an open interval (a, b), then the functions differ by at most a constant. Symbolically, if F and G are differentiable functions on the interval (a, b) and F'(x)=G'(x)for all x in (a, b), then F(x)=G(x)+kfor some constant k.

Formulas and Properties of Indefinite Integrals For C and k both a constant

xn+1 n n + 1 +C, 1. òxdx=

n¹-1

2. òexdx=ex+C 1 3. ò x dx=lnx+C,

x¹0

4. òkf(x)dx=kòf(x)dx 5. ò[f(x)±g(x)]dx=òf(x)dx±òg(x)dx

In Problems 1–3, find each indefinite integral and check by differentiating.

1. ò12xdx

y = ò12 xdx 12 x2 +C y= 2 y = 6 x2 + C

Check: y = 6 x2 +C d =12 x + 0 y d =12 x x d y 5-1 d ..

x

Calculus

Chapter 5 1

2. ò9x2dx 1

3

Check: y = 6x2 + C

y = ò9 x 2 dx 3

y=

9x 2 3

1

dy 3 6 x 2 +0 = × dx 2 1 1 dy = 9x2 dx

+C

2 3

2 9x 2 y= × +C 3 1 3

y = 6x2 + C

3. ò2exdx Check: y = 2e x + C d = 2e x + 0 y d = 2ex x d y d

y = ò2exdx y = 2ex+C

xd 4. Find all the antiderivatives for = 3z- 1 + 7. y d dy 3z-1 z = +7 dz dy = (3z-1+ 7)dz

dy = (3 + 7)dz z

ò z + 7) dz òdy = (3 y = 3 l n z+ 7 z + C

5-2 ..