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

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) = 4 x - 5 a)

c)

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

b)

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

d)

f (- 3) = 4(- 3) - 5 f (- 3) = - 12 - 5 f (- 3) = - 17

æ1 ö æ1 ö f ç ÷ = 4 ç ÷- 5 è4 ø è4ø æ1 ö f ç ÷= 1- 5 è4 ø æ1 ö f ç ÷= - 4 è4 ø

In problems 2–10 evaluate the given function for f ( x) = x 2 + 1 and g ( x) = x - 4 .

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

2. ( f + g )(3) = f (3) + g (3)

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

1-1 ..

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

Calculus

Chapter 1

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

= 4c 2 + 1 - (2c - 4)

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

g (2c) = 2c - 4

f (2c) = 4c 2 + 1

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

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

æf ö f (0) 5. ç ÷(0) = g (0) èg ø 1 -4 æf ö 1 (0) = çè g ÷ 4 ø =

6.

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

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

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

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

2 ×g (- 2) = 2(- 6) 2 ×g (- 2) = - 12

g (- 2) = - 2 - 4 g (- 2) = - 6

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 12 = 2 f (3) - g (2) =6 f (1)

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

1-2 ..

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

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

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

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

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

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

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

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

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

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

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 ) = 4 1 - 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 t £ .

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

1-3 ..

Calculus 15.

Chapter 1

f ( x) = 3 x + 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 w ³ 3.

17.

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

18. g ( x) =

2 x3 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: Identity function

f ( x) = x

h( x ) = x

2

Square function

3

Cube function

m( x) = x

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) = x 2 - 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 is x 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 is x ³ 0.

1-5 ..

Calculus 4.

Chapter 1

f ( x) = 3 x - 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.

2

5. g ( x) = ( x - 5) - 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) = - x 2 + 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 even root. Since the radicand must be greater than or equal to zero, the domain is x ³ 4.

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) = 3 x - 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 - 3 x - 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 = x 2 shifted 2 units right and 4 units up.

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

y = 1- 3 x

1-7 ..

Calculus

Chapter 1

20. The shape of y = x 2 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, then ax = ay and x y = . a a Quadratic Functions: Standard form of a quadratic: f ( x) = ax 2 + bx + c where a,b,c are real and a ¹ 0.

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

Axis of symmetry:

x=h

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 x - 13 5 x + 7 - 5 x = 9 x - 13 - 5 x 7 = 4 x - 13 7 + 13 = 4 x - 13 + 13 20 = 4 x 20 4 x = 4 4 5= x 1-9 ..

Calculus 2.

3.

Chapter 1

7 y + 3(6 y - 11) = 167 7 y + 18 y - 33 = 167 25 y - 33 = 167 25 y - 33 + 33 = 167 + 33 25 y = 200 25 y 200 = 25 25 y =8 m m +4 = +7 5 2 10m 10m + 4(10) = + 7(10) 5 2 2m + 40 = 5m + 70 40 = 3m + 70 - 30 = 3m - 10 = m

For 4–11 find:

then:

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 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 2

f ( x) = ( x - 1) - 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.

2

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

3, 0).

2

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

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

f (0) = (- 1) - 3 f (0) = - 2 Therefore, the y-intercept is (0, –2). 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 is y ³ - 3. i. Based on the graph, the function is decreasing over the interval (- ¥ ,1). j. Based on the graph, the function is increasing over the interval (1, ¥ ).

1-11 ..

Calculus

5.

Chapter 1 2

f ( x) = ( x - 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.

2

f ( x ) = ( x - 2) + 4 0 = x2 - 4 x + 4 + 4 0 = x2 - 4 x + 8 Solving the above equation by the quadratic equation will result in complex roots, therefore no x-intercepts are present. e. The y-intercepts are found by setting x = 0.

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 is y ³ 4. i. Based on the graph, the function is decreasing over the interval (- ¥ , 2). j. Based on the graph, the function is increasing over the interval (2, ¥ ).

1-12 ..

Calculus 6.

Chapter 1

f ( x) = - x 2 + 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 setting x = 0. f ( x) = - x 2 + 7

f (0) = - 02 + 7 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 is y £ 0. i. Based on the graph, the function is decreasing over the interval (0, ¥ ). j. Based on the graph, the function is increasing over the interval (- ¥ , 0).

1-13 ..

Calculus

7.

Chapter 1 2

f ( x) = - ( x - 1) - 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.

2

f ( x) = - ( x - 1) - 1 0 = - 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. e. The y-intercepts are found by setting x = 0.

2

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

f (0) = - (0 - 1) - 1 2

f (0) = - (- 1) - 1 f (0) = - 2 Therefore, the y-intercept is (0, –2). f. The graph opens downward, therefore the graph has a maximum value which is the y-coordinate of the vertex or –1. g.

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

1-14 ..

Calculus 8.

Chapter 1

f ( x) = x 2 - 4 x 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) = x 2 - 4 x 0 = x2 - 4 x 0 = x( x - 4) x = 0, 4 Therefore, the x-intercepts are (0, 0) and (4, 0). e. The y-intercepts are found by setting x = 0.

f ( x) = x 2 - 4 x f (0) = 02 - 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 is y ³ - 4. i. Based on the graph, the function is decreasing over the interval (- ¥ , 2). j. Based on the graph, the function is increasing over the interval (2, ¥ ).

1-15 ..

Calculus 9.

Chapter 1

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

5, 0).

2

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

f (0) = (0 + 1) - 5 2

f (0) = (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 is y ³ - 5. i. Based on the graph, the function is decreasing over the interval (- ¥ , - 1). j. Based on the graph, the function is increasing over the interval (- 1, ¥ ).

1-16 ..

Calculus 10.

Chapter 1

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

f (0) = 02 + 2(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 y ³ 0. i. Based on the graph, the function is decreasing over the interval (- ¥ , - 1). j. Based on the graph, the function is increasing over the interval (- 1, ¥ ).

1-17 ..

Calculus 11.

Chapter 1

f ( x) = - x 2 + 10 x - 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 + 10 x - 19 0 = - x 2 + 10 x - 19 0 = x 2 - 10 x + 19 Solve using the quadratic equation x = 5± 6 Therefore, the x-intercepts are (5 + 6,0) and (5 e. The y-intercepts are found by setting x = 0.

6,0).

f ( x) = - x 2 + 10 x - 19 f (0) = - 02 + 10(0) - 19 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 y £ 6. i. Based on the graph, the function is decreasing over the interval (5, ¥ ). j. Based on the graph, the function is increasing over the interval (- ¥ ,5).

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 - 4 x)

and

C ( x) = 160 + 20 x

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 - 4 x) = 160 + 20 x 200 x - 4 x 2 = 160 + 20 x 0 = 4 x 2 - 180 x + 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) P( x) = x(200 - 4 x) - (160 + 20 x) P( x) = 200 x - 4 x 2 - 160 - 20 x P( x) = - 4 x 2 + 180 x - 160 c) Determine the maximum profit. How many components must be sold in order to achieve that maximum profit?

The maximum profit occurs at the vertex of the profit function. The x-coordinate is b 180 - 180 x=== = 22.5. To find the y-coordinate of the vertex, substitute the 2a 2(- 4) -8 value into the function as follows: P( x) = - 4 x 2 + 180 x - 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) = 300 x - x 2 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) = 40 x + 1600 a) Find the break-even points for this company. (Round answer to nearest 1000.)

R( x) = C ( x) 300 x - x 2 = 40 x + 1600 0 = x 2 - 260 x + 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) P( x) = 300 x - x 2 - (40 x + 1600) P( x) = - x 2 + 260 x - 1600 The maximum profit occurs at the vertex of the profit function. The x-coordinate is b 260 - 260 x=== = 130. To find the y-coordinate of the vertex, substitute the value 2a 2(- 1) -2 into the function as follows: P( x) = - x 2 + 260 x - 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) = 0.01x 2 - 20 x + 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) = 0.01x 2 - 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=== = 1000. To find the y-coordinate of the vertex, substitute the 2a 2(0.01) 0.02 value into the function as follows: C ( x) = 0.01x 2 - 20 x + 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) = 0.04 x 2 - 20 x + 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) = 0.04 x 2 - 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=== = 250. To find the y-coordinate of the vertex, substitute the 2a 2(0.04) 0.08 value into the function as follows: C ( x) = 0.04 x 2 - 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 + 3.50 x 17.50 x = 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) = an x n + an- 1x n- 1 +

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

Definition: Rational function

n( x ) d ( x) ¹ 0 where n(x) and d(x) are polynomials. The domain is the set of d ( x) all real numbers such that d ( x) ¹ 0. 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 = c is 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 - 7 x - 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 + 4 x 2 - 4 x - 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 - 3x 2 - 10 x + 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 + 4 x 2 - 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) = x 4 - 2 x3 + x 2 + 2 x - 2 = ( x - 1)( x + 1)( x 2 - 2 x + 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 + 5 x 4 - 20 x 2 - x + 15 = ( x + 3)( x - 1)( x + 1)( x 2 + 2 x - 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 x. Therefore, f ( x) = = 2 x = 2 x , as x increases or decreases without bound, the 2 x 22 xx 1 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.

b.

c.

Chapter 1

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

x- 5=0 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). d.

Vertical asymptotes occur at values where the function is not defined. Since 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

3x 3 x. Therefore, f ( x) = = xx5= , as x increases or decreases without bound, the x - 5 x - x 1 - 5x 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.

b.

c.

Chapter 1

3x x- 3 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.

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). 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 f ( x) =

3x

3x 3 x. Therefore, f ( x) = = xx3= , as x increases or decreases without bound, the x - 3 x - x 1 - 3x 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) = a.

b.

c.

Chapter 1

2x - 4 x +3 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 of x is zero.

2x - 4 x+3 2(0) - 4 f (0) = 0+3 -4 f (0) = 3 Therefore, the y–intercept is (0, - 43 ). f ( x) =

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 2x 4 4 2x - 4 x - x 2 - x x. Therefore, f ( x) = = x 3 = , as x increases or decreases without bound, the 3 x +3 + 1 + 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 line y = 2. f.

1-29 ..

Calculus

11.

f ( x) = a.

b.

c.

Chapter 1

4+ x 4- x The function is defined everywhere except when the denominator is zero.

4- x = 0 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). d.

Vertical asymptotes occur at values where the function is not defined. Since 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 x 4 4 + x x + x x +1 x. Therefore, f ( x) = = 4 x = 4 , as x increases or decreases without bound, the 4- x x - x x - 1 f ( 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.

Chapter 1

f ( x) =

1 - 5x 1 + 2x

a.

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

1+ 2x = 0 2x = - 1 1 x=2 1 .

2 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.

The domain is therefore all real numbers except b.

1 - 5x = 0 - 5x = - 1 1 x= 5 Therefore, the x–intercept is ( 15 , 0). c.

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

1 - 5x 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 = - 12 , the vertical asymptote is the line x = - 12 . f ( x) =

e.

Horizontal asymptotes are found by dividing all terms by the highest power of 1 5x 1- 5 1 - 5x x - x x. Therefore, f ( x) = = = x , as x increases or decreases without bound, the 1 + 2 x 1x + 2xx 1x + 2 numerator tends to 0 – 5, or –5 and the denominator tends to 0 + 2, or 2; so f(x) tends to - 52 .

The horizontal asymptote is the line y = - 52 .

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

C ( n) . n

C (n) 20n + 52, 000 = 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 n n , as n increases without highest power of n. Therefore, C (n) = = n 1 n bound, the numerator tends to 20 + 0, or 20 and the denominator is always 1; so C ( n) 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 b) The average cost per hour of operating the machine is given by C (h) = Find the average cost function. C (h) =

C ( h) . h

C (h) 21.75h + 36, 000 = 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 21.75 + 36,000 h h h , as h increases by the highest power of h. Therefore, C (h) = = h 1 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) = x 2 + 8 x + 6000 The average cost per printer at a production level of x printers per day is C ( x) =

C ( x) . x

a) Find the average cost function.

C ( x) =

C ( x) x 2 + 8 x + 6000 = x x

b) Sketch a graph of the average cost function for 25 £ 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) = 3x 2 + 36 x + 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) =

C ( x) 3x 2 + 36 x + 9000 = x x

b) Sketch a graph of the average cost function for 0 £ x £ 150 .

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 = a m+n Product Rule

a0 = 1, a ¹ 0 Zero Exponent Rule

am

(am ) = amn Power Rule

a

= a m- n n

Quotient Rule

n

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

4. g ( x) = - 5x- 3 + 2; f ( x) = 5 x 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) = 10 x- 2 - 5; f (x)=10 x The function f is shifted 2 units to the right and 5 units down.

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

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

8. g ( x) = - e x + 5; f ( x) = e x 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 = 102 x+8

9. 102 x- 3 = 105 x+4

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

2

12. 8 x = 88 x

x2 = 8x

5x - 4 = x2

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

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

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

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

1-38 ..

2

x2 = 2 x + 8

x + 6 = 3x - 8 14 = 2 x 7=x

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

11. 65 x- 4 = 6 x

2x - 7 = x +1 x =8

Calculus

Chapter 1

15. (e3 )4 = e x

16. (e x )4 = e x

2

17. (e2 x ) x = e15+ x

4 x = x2

3(4) = x 12 = x

2 x 2 = 15 + x

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

18. 3x ×34 = 33 x

2 x 2 - x - 15 = 0 (2 x + 5)( x - 3) = 0 x = - 52 ,3

2

2

20. 9 x ×9 = 92 x

19. 2 x = 212 x ×2- 32

x 2 = 12 x - 32

x + 4 = 3x 2

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

3x 2 - x - 4 = 0 (3x - 4)( x + 1) = 0 x = 43 , - 1

INTEREST FORMULAS Simple Interest:

2

x + 1 = 2 x2 2 x2 - x - 1 = 0 (2 x + 1)( x - 1) = 0 x = - 12 ,1

A = P(1 + rt ) mt

Compound Interest:

æ rö A = P ç1 + ÷ è mø

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 = P ç1 + ÷ è mø

Bank B:

A = Pert

12(5)

A = 35, 000e(0.005)(5)

60

A = 35, 000e0.025 A » $35,886

æ 0.01 ö A = 35, 000 ç1 + ÷ è 12 ø A = 35000 (1.000833) A » $36, 794

mt

Bank C:

æ rö A = P ç1 + ÷ è mø

365(5)

æ 0.0075 ö A = 35, 000 ç1 + ÷ è 365 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 = P ç1 + ÷ è mø

4(18)

æ 0.012 ö A = 10, 000 ç1 + ÷ è 4 ø 72

A = 10, 000 (1.003) 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 = P ç1 + ÷ è mø

365(18)

æ 0.01 ö A = 5000 ç1 + è 365 ÷ ø

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 rö A = P ç1 + ÷ è mø

12(10)

æ 0.018 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 rö A = P ç1 + ÷ è mø

365(5)

æ 0.012 ö 30, 000 = P ç1 + ÷ è 365 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 = P ç1 + ÷ ç1 + ÷ è mø è mø

12(3)

æ 0.015 ö A = 3000 ç1 + ÷ è 12 ø 36

A = 3000 (1.00125)

365(1)

æ 0.022 ö çè1 + ÷ 365 ø

(1.000060)365

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 = P ç1 + ÷ ert è mø 365(6)

æ 0.0125 ö A = 8000 ç1 + ÷ è 365 ø

e(0.02)(4)

2190 0.08

A = 8000 (1.000034)

e

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

log a x + log a y = log a xy log a x - log a y = log a

Exponential Function : f ( x) = a x , a > 0, a ¹ 1

x y

y = log a x

means x = a y

log a x n = n log a x

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

1. log3 x = 4

2. log3 ( x + 1) = 2

x = 34 x = 81

x + 1 = 32 x +1 = 9 x =8

3. log3 38 = 7 + 3x

4. log 2 26 = 4 - 3x

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

2 4 - 3 x = 26 4 - 3x = 6 - 3x = 2 2 x=3

1-43 ..

Calculus

Chapter 1

5. ln( x + 6) = 2

6. log x (2 x + 3) = 2

e2 = x + 6

x2 = 2 x + 3 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. log2 (7 x - 19) = log 2 (2 x + 9)

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

8 - 5 x = 2 x - 13 21 = 7 x 3= x Since the value of 3 creates a negative log, there is no solution.

7 x - 19 = 2 x + 9 5 x = 28 x = 5.6

9. ln x + ln 3 = 3

10. ln x - ln 3 = 1

ln 3x = 3

ln 3x = 1

e3 = 3 x

e1 = 3x

e3 =x 3 6.6952 » x

3e = x 8.1548 » x

11. log( x - 1) - log 4 = 3

12. ln( x - 1) - ln 6 = 2 x- 1

x- 1

log 4 = 3 103 =

ln 6 = 2

x- 1 4

e2 =

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

x- 1 6

6e2 = x - 1 6e2 + 1 = x 45.3343 » x

1-44 ..

Calculus

Chapter 1

13. 23 x = 12

14. 5x- 1 = 17

log 5 x - 1 = log17 ( x - 1) log 5 = log17 x log 5 - log 5 = log17 x log 5 = log17 + log 5 log17 + log 5 x= log 5 x » 2.7604

log 23 x = log12 3x log 2 = log12 log12 x= 3log 2 x » 1.1950

16. 42 x+3 = 5x- 2

15. 7 x- 1 = 8x

log 7 x - 1 = log 8 x ( x - 1) log 7 = x log 8 x log 7 - log 7 = x log 8 x log 7 - x log 8 = log 7 x(log 7 - log 8) = log 7 log 7 x= log 7 - log 8 x » - 14.5727

log 42 x+3 = log 5 x- 2 (2 x + 3) log 4 = ( x - 2) log 5 2 x log 4 + 3log 4 = x log 5 - 2 log 5 2 x log 4 - x log 5 = - 2 log 5 - 3log 4 x(2 log 4 - log 5) = - 2 log 5 - 3log 4 - 2 log 5 - 3log 4 x= 2 log 4 - log 5 x » - 6.3429 18. e x+4 = 14.654

17. 7 x+1 = 102 x

log 7 x+1 = log102 x ( x + 1) log 7 = 2 x log10 x log 7 + log 7 = 2 x log10 x log 7 - 2 x log10 = - log 7 x(log 7 - 2 log10) = - log 7 - log 7 x= log 7 - 2 log10 x » 0.7317

ln e x +4 = ln(14.654) x + 4 = ln(14.654) x = ln(14.654) - 4 x » - 1.3153

19. x + 5 = e3

20. e4 x = e20

x = e3 - 5 x » 15.0855

4 x = 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 rö A = P ç1 + ÷ è mø

4(18)

æ 0.014 ö 20, 000 = P ç1 + ÷ è 4 ø 72

20, 000 = P (1.0035) $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 rö A = P ç1 + ÷ è mø

4(8)

æ 0.014 ö 20, 000 = P ç1 + ÷ è 4 ø 32

20, 000 = P (1.0035) $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 + ÷ è mø

4(10)

æ rö 1200 = 1000 ç1 + ÷ è 4ø 40

æ rö 1.2 = ç1 + ÷ è 4ø 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 + ÷ è mø

12(6)

æ r ö 20, 000 = 18, 000 ç1 + ÷ è 12 ø 72

æ r ö 1.111 = ç1 + ÷ è 12 ø 1

1.111 72 = 1 + 1

1.111 72 - 1 =

r 12

r 12

1

12(1.111 72 - 1) = r 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 = P ç1 + ÷ è mø

12t

æ 0.018 ö 20, 000 = 10, 000 ç1 + ÷ è 12 ø 12t

2 = (1.0015)

12t

ln 2 = ln (1.0015)

ln 2 = 12t ln(1.0015) ln 2 =t 12 ln1.0015 38.5 » t It will take about (38.5)(12) = 462 months to double. mt

æ rö A = P ç1 + ÷ è mø

12(38.5)

æ 0.018 ö A = 10, 000 ç1 + ÷ è 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 = P ç1 + ÷ è mø

12t

æ 0.018 ö 30, 000 = 10, 000 ç1 + ÷ è 12 ø 12t

3 = (1.0015)

12t

ln 3 = ln (1.0015)

ln 3 = 12t ln(1.0015) ln 3 =t 12 ln1.0015 61.1 » t It will take about (61.1)(12) = 733 months to triple. mt

æ rö A = P ç1 + ÷ è mø

12(61.1)

æ 0.018 ö A = 10, 000 ç1 + ÷ è 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 ç1 + ÷ è mø

4t

æ 0.02 ö 10, 000 = 7000 ç1 + ÷ è 4 ø 10 = 1.005 4t 7

(

)

4t

ln 10 = ln (1.005) 7

ln 10 = 4t ln(1.005) 7 ln 10 7

=t 4 ln1.005 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 = P ç1 + ÷ è mø

12t

æ 0.03 ö 24, 000 = 18, 000 ç1 + ÷ è 12 ø 4 = 1.0025 12t 3 12t ln 43 = ln 1.0025 ln 43 = 12t ln(1.0025)

(

)

(

)

ln 43

=t 12 ln1.0025 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 = 25e0.04t

P = 25e0.04t

30 = 25e0.04t

40 = 25e0.04t

1.2 = e0.04t

1.6 = e0.04t

ln1.2 = ln e0.04t ln1.2 = 0.04t ln1.2 =t 0.04 4.6 » t

ln1.6 = ln e0.04t ln1.6 = 0.04t ln1.6 =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 = 20 bacteria present.

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

N = 20e0.35t

40 = 20e0.35t 2 = e0.35t ln 2 = ln e0.35t ln 2 = 0.35t ln 2 =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

lim f ( x) = L or f ( x) ® L as 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

lim f ( x) = L is the limit of the function as x approaches the value c from the left.

x® c -

lim f ( x) = L is the limit of the function as x approaches the value c from the right.

x® c +

Properties of Limits: 1.

2.

3.

4.

5.

6.

7.

8.

lim k = k for any constant k

x® c

lim x = c

x® c

lim[ f ( x) + g ( x)] = lim f ( x) + lim g ( x)

x® c

x® c

x® c

lim[ f ( x) - g ( x)] = lim f ( x) - lim g ( x)

x® c

x® c

x® c

lim kf ( x) = k lim f ( x) for any constant k

x® c

x® c

lim[ f ( x) ×g ( x)] = lim f ( x) ×lim g ( x)

x® c

x® c

x® c

lim f ( x) æ f ( x) ö x® c lim ç = provided lim g ( x) ¹ 0 ÷ x® c x® c è g ( x) ø lim g ( x)

x® c n lim f ( x) = n lim f ( 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.

2.

3.

4.

lim (7 x + 2) = (7(2) + 2) = 14 + 2 = 16

x® 2

lim 5 x = 5(- 3) = - 15

x® - 3

lim x(4 x + 1) = lim x ×lim (4 x + 1) = 5[4(5) + 1)] = 5(21) = 105

x® 5

x® 5

x® 5

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

5.

lim

x® - 3

- 5 x + 1 = - 5(- 3) + 1 = 15 + 1 = 16 = 4

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

lim f ( x) = 4 and lim g ( x) = - 3.

x® 3

6.

7.

8.

x® 3

lim 7 f ( x) = 7 ×4 = 28

x® 3

lim[3 f ( x) - 2 g ( x)] = 3(4) - 2(- 3) = 12 + 6 = 18

x® 3

æ2 f ( x) ö 2(4) 8 8 = = =lim ç ÷ 9 x® 3 è 3 g ( x) ø 3(- 3) - 9

2-2 ..

Calculus

9.

Chapter 2

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

lim f ( x)

x® 1-

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

lim f ( x) = lim ( x 2 + 2) = ((1)2 + 2) = 3

x® 1-

b)

x® 1-

lim f ( x)

x® 1+

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

x® 1+

c)

x® 1+

lim f ( x) Since the left– and right–hand limits are the same value,

x® 1

lim f ( x) = 3.

x® 1

d)

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

10.

ì 5 x - 6 if x £ 2 . Find: Let f ( x) = í î 2 x + 4 if x > 2

a)

lim f ( x)

x ® 2-

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

x ® 2-

b)

x ® 2-

lim f ( x)

x ® 2+

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

x ® 2+

x ® 2+

2-3 ..

Calculus c)

Chapter 2

lim f ( x)

x® 2

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

not exist. d)

f (2) The value of 2 is defined in the top function, therefore f (2) = 5(2) - 6 = 4.

11.

æx 2 + 4 x - 5 ö Let f ( x) = ç ÷. Find è x- 1 ø a)

lim f ( 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) =

x 2 + 4 x - 5 ( x + 5)( x - 1) = = x + 5, x ¹ 1 x- 1 x- 1

Therefore, lim f ( x) = lim( x + 5) = 1 + 5 = 6 x® 1

b)

c)

x® 1

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

lim f ( x) = lim

x® - 1

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

2-4 ..

Calculus

12.

Chapter 2

æx - 2 ö Let f ( x) = ç . Find è x- 2 ÷ ø a)

lim f ( 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)

lim f ( 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)

lim f ( 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.

3.

lim

k

x® - ¥ x

p

=0

2.

lim kx p = ±¥

4.

x® - ¥

lim

k

x® ¥ x p

=0

lim kx p = ±¥

x® ¥

provided that x p 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) =

am x m + am- 1x m- 1 + n

bn x + bn- 1x

n- 1

+

+ a1x + a0 + b1x + b0

, am ¹ 0, bn ¹ 0,

a xm a xm then lim f ( x) = lim m n and lim f ( x) = lim m n x® - ¥ x® ¥ x® - ¥ bn x x® ¥ bn x There are three cases to consider: 1.

2.

3. am , and bn .

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

x® - ¥

a If m = n, lim f ( x) = lim f ( x) = m . bn x® ¥ x® - ¥ If m > n, then the limit will be ¥ or - ¥ , depending on the values of m, n,

2-7 ..

Calculus

Chapter 2

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

1.

f ( x) = a)

x x+4 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 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)

lim f ( 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 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, 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, lim f ( x) = ¥ . x® 5-

2-8 ..

Calculus b)

Chapter 2 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, 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, lim f ( x) = ¥ . x® 5+

c)

3.

f ( x) = a)

lim f ( x) = ¥ because the left– and right–hand limits both approach infinity.

x® 5>

x 2 + 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)

lim f ( 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) =

4x +1 6x - 3

a)

f (20) 4(20) + 1 81 = » 0.6923 6(20) - 3 117

f (20) = b)

f (200) f (200) =

4(200) + 1 801 = » 0.6692 6(200) - 3 1197

4 2 = because the function is a rational expression and m = n, the 6 3 x® ¥ limit is the ratio of the coefficients. c)

5.

f ( x) =

lim f ( x) =

x- 5 2

3x + 2 x + 2 a)

f (10)

f (10) = b)

10 - 5 3(10)2 + 2(10) + 2

=

5 » 0.0155 322

f (100)

100 - 5

f (100) =

2

=

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

95 » 0.0031 30, 222

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

x® ¥

2-10 ..

Calculus

6.

Chapter 2

f ( x) =

x3 + 2 x - 1 3 - 8x

a)

f (- 10) f (- 10) =

b)

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

f (- 100) f (- 100) =

c)

(- 100)3 + 2(- 100) - 1 - 1, 000, 201 = » - 1245.5803 3 - 8(- 100) 803

lim f ( x) = - ¥ because the function is a rational expression, m > n, and

x® - ¥

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 at x + 4 = 0 Þ 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 because m = n and 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 because m = n and 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 x 2 + 8 x + 15

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

x2 + 4 x - 5 2

=

x + 8 x + 15

( x + 5)( x - 1) x - 1 = , x ¹ - 5 and x ¹ - 3 ( x + 5)( x + 3) x + 3

Vertical asymptotes are found by setting the denominator equal to zero. Therefore, there would be a vertical asymptote at x + 3 = 0 Þ 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 = n and 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 = c if 1.

lim f ( x) exists

x® c

2.

f (c) exists

3.

lim f ( x) = f (c)

x® c

Continuity Properties: 1.

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

2.

3.

4.

For n a positive integer, f ( x) = x n 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, n f ( 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.

Select a test value in each open interval and evaluate f ( x) at each test value to 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 ) = 3 x3 - 4 x 2 + x + 7 Since the function is a polynomial function, it is continuous for all x.

2.

f ( x) =

x3 + 4 x 2 + 3 x - 7 x 2 - 15 x + 26

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

So the function is continuous for all x except x = 2 or 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) = 3 x - 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) = x 2 - 25

Since the function is an even root, the function is continuous for all values of x where the radicand is continuous and nonnegative. Therefore, x 2 - 25 ³ 0 and the points in question are ±5. For all values between –5 and 5, the radicand would be negative, therefore, the function is continuous on the interval (- ¥ , - 5] È [5, ¥ ).

2-14 ..

Calculus

Chapter 2

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

6.

4 x 2 - 29 x + 7 < 0

Find the partition numbers: 4 x 2 - 29 x + 7 = 0 (4 x - 1)( x - 7) = 0

Therefore, the partition numbers are 14 and 7.

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

(- ¥ , 14 ), ( 14 ,7), and (7, ¥ ).

the function values using these three test values.

f (1) = 4(1)2 - 29(1) + 7 f (1) = 4 - 29 + 7 f (1) = - 18

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

f (8) = 4(8)2 - 29(8) + 7 f (8) = 256 - 232 + 7 f (8) = 31

Only the test value 1 makes the inequality true, so the solution is 14 < x < 7, or ( 14 ,7).

2-15 ..

Calculus

7.

Chapter 2

x3 + 3 x 2 >0 x +5 Find the partition numbers:

x3 + 3 x 2 = 0

x+5 = 0

x 2 ( 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.

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

(- 1)3 + 3(- 1)2 2 f (- 1) = = = 0.5 - 1+ 5 4

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

(1)3 + 3(1)2 4 f (1) = = » 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 + h is

f ( a + h) - f ( a ) f (a + h ) - f (a ) = ,h¹ 0 ( a + h) - a h 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 at x = a is

f ( a + h) - f ( a ) , if the limit exists. h h® 0 lim

This formula is also used to find the slope of a graph at the point (a, f (a)) and to find 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). f ( x + h) - f ( x ) Find . 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), and f ¢(3).

1.

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

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

Step 2:

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

Step 3:

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

Step 4:

f ( x + h) - f ( x ) = lim 7 = 7 h h® 0 h® 0 lim

f ¢( x) = 7, f ¢(1) = 7, f ¢(2) = 7, and f ¢(3) = 7.

2.

f ( x) = - 3 x 2 + 5 x - 6 Step 1:

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

Step 2:

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

Step 3:

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

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

lim

2-18 ..

Calculus

3.

Chapter 2 f ¢(2) = - 6(2) + 5 f ¢(2) = - 7

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

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

Step 4:

Therefore, f ¢( x) =

f ¢(1) =

- 10 (1 - 5)

f ¢(2) =

2

- 10 (2 - 5)

f ¢(3) =

f ¢(3) = - 6(3) + 5 f ¢(3) = - 13

2

- 10 (3 - 5)

2

- 10 ( x - 5)2 - 10

=

(- 4) =

2

- 10 (- 3)

=

2

- 10 (- 2)

2

=

- 10 5 =16 8

=

- 10 10 =9 9

=

- 10 5 =4 2 2-19

..

Calculus 4.

Chapter 2

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

P( x) = 2 x 2 - 5 x + 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.

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

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

P(2) = 2(2)2 - 5(2) + 6 P(2) = 8 - 10 + 6 P(2) = 4 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( x 2 + 2 xh + h2 ) - 5( x + h) + 6 P( x + h) = 2 x 2 + 4 xh + 2h 2 - 5 x - 5h + 6 Step 2:

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

Step 3:

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

2-20 ..

Calculus

Chapter 2

P ( x + h) - P ( x ) = lim (4 x + 2h - 5) h h® 0 h® 0 = 4 x + 2(0) - 5 P ( x + h) - P ( x ) lim = 4x - 5 h h® 0 Therefore, P¢( x) = 4 x - 5 Step 4:

lim

c.

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

5.

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

s(t ) = t 2 + 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:

s(4) = (4) 2 + 5(4) + 2 s(4) = 16 + 20 + 2 s(4) = 38

s(6) = (6) 2 + 5(6) + 2 s(6) = 36 + 30 + 2 s(6) = 68

Now use the average rate of change formula. Note that h = 6 - 4 = 2.

s (t + h) - s (t ) s (4 + 2) - s (4) = h 2 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) = t 2 + 2th + h2 + 5t + 5h + 2 Step 2:

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

Step 3:

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

s(t + h) - s(t ) = lim (2t + h + 5) h h® 0 h® 0 = 2t + 0 + 5 s(t + h) - s(t ) lim = 2t + 5 h h® 0 Therefore, s ¢(t ) = 2t + 5 Step 4:

lim

c. s ¢(4) = 2(4) + 5 = 9 means that after 2 seconds the particle is traveling at 9 units per second.

2-22 ..

Calculus

Chapter 2

Name ________________________________ Date ______________ Class ____________

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

dy

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

Theorems: 1.

If y = f ( x) = C , then f ¢( x) = 0 (Constant Function Rule)

2.

3.

4.

If y = f ( x) = x n , where n is a real number, then f ¢( x) = nx n- 1 (Power Rule) If y = f ( x) = ku ( x), then f ¢( x) = ku ¢( x) (Constant Multiple Property) If y = f ( x) = u ( x) ± v( x), then f ¢( x) = u ¢( x) ± v ¢( x) (Sum and Difference Property)

1 - 6 Find the indicated derivatives.

1.

y¢ for y = x3

y ¢= 3x 2

Use theorem 2 to find the derivative:

2.

dy for y = 17 dx x

Convert the problem to the form x n and then use theorem 2. dy = - 7 x - 8 = - 78 dx x

y = 17 = x - 7 x

2-23 ..

Calculus 3.

Chapter 2

d (2u 2 - 3u + 8) du

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

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

4.

f ¢( x) if f ( x) = x3.2 + 7 x 2.1 - 3x + 7

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

f ¢( x) = f ¢( x3.2 ) + 7 f ¢( x 2.1 ) - 3 f ¢( x) + f ¢(7) = 3.2 x 2.2 + 7(2.1) x1.1 - 3 + 0 = 3.2 x 2.2 + 14.7 x1.1 - 3 5.

d ( 1 +6 dx x 4

x - 5 x + 8)

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

d 1 + d 6 x - d 5x + d 8 x - 5 x + 8) = dx 4 dx dx dx x

1

d x- 4 + d 6 x 2 - d 5x + d 8 = dx dx dx dx

= - 4 x - 5 + 6( 12 ) x

- 12

- 5+0

= - 45 + 3 - 5 x

x

6.

3

4

f ¢( x) if f ( x) = x 5 + 4 x 5 - 2 x

- 15

+5

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

3

4

f ¢( x) = f ¢( x 5 ) + 4 f ¢( x 5 ) - 2 f ¢( x = 54 x

- 15

+ 12 x 5

- 25

+ 52 x

- 15

) + f ¢(5)

- 65

2-24 ..

Calculus 7.

Chapter 2

Given the function f ( x) = 7 x 2 - 9 x + 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 ¢( x 2 ) - 9 f ¢( x) + f ¢(2) f ¢( x) = 14 x - 9

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 = 33x - 99 y = 33x - 61 d.

The tangent is horizontal when the first derivative has a value of 0.

f ¢( x) = 14 x - 9 0 = 14 x - 9 9 = 14 x 0.64 » 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 x (in seconds) is given by the indicated function, find: a. b. c.

8.

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

f ( x) = 3x 2 - 12 x - 8 Solution: a.

f ( x) = 3x 2 - 12 x - 8 f ¢( x) = 3 f ¢( x 2 ) - 12 f ¢( x) - f ¢(8) f ¢( x) = 6 x - 12 v = 6 x - 12

b.

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

c.

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

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

2-26 ..

Calculus 9.

Chapter 2

f ( x) = x3 - 21 x 2 + 30 x 2 Solution: a.

f ( x) = x3 - 21 x 2 + 30 x 2 f ¢( x) = f ¢( x3 ) - 21 f ¢( x 2 ) + 30 f ¢( x) 2 f ¢( x) = 3x 2 - 21x + 30 v = 3x 2 - 21x + 30

b.

c.

f ¢( x) = 3 x 2 - 21x + 30

f ¢( x) = 3 x 2 - 21x + 30

v( x) = 3x 2 - 21x + 30

v( x) = 3x 2 - 21x + 30

v(0) = 3(0) 2 - 21(0) + 30 v(0) = 0 - 0 + 30 v(0) = 30

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

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

10.

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

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

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

2-27 ..

Calculus 11.

Chapter 2

Find f ¢( x) if f ( x) = 9 xx- 5 .

First simplify the function.

f ( x) = 9 xx- 5 = 9xx - 5x = 9 - 5 x - 1 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 + 5 x - 2 f ¢( x) = 52 x

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.

Dy

Given the function y = 3x 4 , find Dx, Dy, and Dx given x1 = 2 and x2 = 5.

When x1 = 2, y1 = 3(2) 4 = 48. When x2 = 5, y2 = 3(5)4 = 1875.

Dy = 1875 - 48 Dy = 1827

Dx = 5 - 2 Dx = 3

2.

Dy 1827 = 3 = 609 Dx

Dy

Given the function y = 3x 4 , find Dx, Dy, and Dx given x1 = 3 and x2 = 6.

When x1 = 3, y1 = 3(3) 4 = 243. When x2 = 6, y2 = 3(6)4 = 3888.

Dy = 3888 - 243 Dy = 3645

Dx = 6 - 3 Dx = 3

2-29 ..

Dy 3645 = 3 = 1215 Dx

Calculus 3.

Chapter 2

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

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

dy = (9 x - 24 x + 4)dx 4.

3

x ), find dy.

Given the function y = x5 (2 - 12 3

x ) y = x5 (2 - 12 1 x8 y = 2 x5 - 12 dy 8 x7 = 10 x 4 - 12 dx dy = (10 x 4 - 23 x 7 )dx

5.

Given the function y = 25 - 7 x 2 - x3 , find dy and Dy given x = 4 and dx = Dx = 0.1.

y = 25 - 7 x 2 - x3

dy = (- 14 x - 3 x 2 )dx

dy = - 14 x - 3 x 2 dx 2

dy = [- 14(4) - 3(4) 2 ](0.1) dy = (- 56 - 48)(0.1) dy = (- 104)(0.1) dy = - 10.4

dy = (- 14 x - 3x )dx

Dy = f ( x + Dx) - f ( x) Dy = f (4 + 0.1) - f (4) Dy = f (4.1) - f (4) Dy = (25 - 7(4.1) 2 - (4.1)3 ) - (25 - 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 = 120 x - x 2 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 - x 2

dN = (120 - 2 x)dx dN = (120 - 2(15))(2) dN = 180

dN = 120 - 2 x dx

dN = (120 - 2 x)dx

dN = (120 - 2 x)dx dN = (120 - 2(25))(2) dN = 140

Therefore, increasing the advertising budget from $15,000 to $17, 000 will result in an increase of 180 units and an increase from $25,000 to $27,000 will only result in a 140 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 = 590 x 2 x

y = 590 x

- 12

dy = (- 295 x

dy - 32 = 295 x dx

dy = (- 295 x

- 32

- 32

dy = (- 295(49)

)dx

- 32

)(3)

295 )(3) dy = (- 343

)dx

dy = - 885 » - 2.58 343 Therefore, if a person grows from 49 to 52 inches, their pulse rate would decrease by approximately 2.5 beats per minute.

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)

and and and

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

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

Cost per unit: average cost = C = x

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

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

dx

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

1.

C ( x) = 6000 - 40 x + 0.006 x 2 and Marginal cost function

R( x) = 10, 000 x - 200 x 2

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

Average cost function 2

C ( x)

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

2-33 ..

Calculus 3.

Chapter 2

Marginal average cost function

C ( x) = 6000 - 40 + 0.006 x x C ( x) = 6000 x - 1 - 40 + 0.006 x C ¢( x) = - 6000 x - 2 + 0.006 C ¢( x) = - 3000 2 + 0.003 x

4.

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

5.

Average revenue function

R( x) = 6.

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

Marginal average revenue function

R( x) = 10, 000 - 200 x R¢( x) = - 200 7.

Profit function

P( x) = R( x) - C ( x) = (10, 000 x - 200 x 2 ) - (6000 - 40 x + 0.006 x 2 ) P( x) = - 200.006 x 2 + 10, 040 x - 6000 8.

Marginal profit function

P( x) = - 200.006 x 2 + 10, 040 x - 6000 P¢( x) = - 400.012 x + 10, 040 9.

Average profit function

P( x) =

P ( x ) - 200.006 x 2 +10,040 x - 6000 = = - 200.006 x + 10,040 - 6000 x x x

2-34 ..

Calculus 10.

Chapter 2

Marginal average profit function

P¢( x) = R¢( x) - C ¢( x) P¢( x) = (- 200) - (- 6000 2 + 0.006) x

P¢( x) = 6000 2 - 200.006 x

11.

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

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

R( x) = 1000 + 1000 x x

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

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

R(101) = 101, 009.90

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 x

-1

+ 1000 x

R ¢( x) = - 1000 x - 2 + 1000 R ¢( x) = - 1000 + 1000 x2

R ¢( x) = - 1000 2 + 1000

x R ¢(101) = - 10002 + 1000 (101)

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 + 15 x 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)

The average cost function is C ( x) = x = 10,000x +15 x = 10000 x - 1 + 15.

Therefore, the marginal average cost function for producing the 300th unit would be:

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

C ( x) = - 10000 2

x C (300) = - 100002 (300)

-2

C ( x) = - 10000 2

C (300) = - 0.11

x

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) = 30 x - 0.03x 2 + 200 a. b.

Find the exact profit from the 201st unit sold.

Find the marginal profit from selling the 201st unit.

Solution: a.

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

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

P( x) = 30 x - 0.03x 2 + 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 - 0.03x 2 + 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) = 32 x + 36, 000 and a. b. c.

R( x) = 300 x - 0.03x 2

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) P( x) = (300 x - 0.03x 2 ) - (32 x + 36, 000) P( x) = - 0.03x 2 + 268 x - 36, 000

b.

C ( x) = 32 x + 36, 000 C (101) = 32(101) + 36, 000 C (101) = 39, 232

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

R( x) = 300 x - 0.03x 2

R( x) = 300 x - 0.03 x 2

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) = 300 x - 0.03x 2 R¢( x) = 300 - 0.06 x

C ( x) = 32 x + 36,000 C ¢( x) = 32 P ¢( x) = R ¢( x) - C ¢( x) P ¢( x) = (300 - 0.06 x) - (32) P ¢( x) = - 0.06 x + 268

Marginal values for the 101st unit are: C ¢( x) = 32 C ¢(101) = 32

R ¢( x) = 300 - 0.06 x R ¢(101) = 300 - 0.06(101) R ¢(101) = 293.94

P ¢( x) = - 0.06 x + 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 constant e and solve continuous compound problems

Definition: The number e n

æ 1ö e = lim ç1 + ÷ nø n® ¥ è Theorem 1:

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.

Use a calculator to evaluate A to the nearest cent.

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

A = 500e0.03t

A = 500e0.03t

A = 500e0.03(3)

A = 500e0.03(6)

A = 500e0.03(9)

A = 500e0.09 A = 500(1.09417) A = 547.09

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

A = 500e0.27 A = 500(1.30996) A = 654.98

3-1 ..

Calculus 2.

Chapter 3

Solve for t to two decimal places.

4 = e0.07t 4 = e0.07t ln 4 = ln e0.07t ln 4 = 0.07t ln 4 =t 0.07 19.80 = t 3.

Solve for r to two decimal places.

3 = e4r 3 = e4r ln 3 = ln e4r ln 3 = 4r ln 3 =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 A = 5000e(0.0175)(5) A = 5000e0.0875 A = 5000(1.09144) A = 5457.21 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 ln 2 = ln e0.0175t ln 2 = 0.0175t ln 2 =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 25, 000 = Pe(0.026)(10) 25, 000 = Pe0.26 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 = Pert 20, 000 = 15, 000e(0.038)t 4 = e(0.038)t 3 ln 43 = ln e(0.038)t ln 43 = 0.038t ln 43 =t

0.038 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, b ¹ 1,

d x e = ex dx

d x b = b x ln b dx

d 1 ln x = dx x

d 1 æ1 ö logb x = ç ÷ dx ln b è x ø

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

1.

f ( x) = 4e x + 5 x 4 - 12 x + 46

f ( x) = 4e x + 5 x 4 - 12 x + 46 f ¢( x) = 4e x + 20 x3 - 12

2.

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

3-5 ..

Calculus 3.

Chapter 3

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

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

4.

y = 10 x - e x+3 - 5 x y = 10 x - e3 - 5 x dy = 10 x ln10 - 5 dx

5.

y = log 7 x + 3x5 - 4 x

y = log 7 x + 3x5 - 4 x dy 1 æ1 ö 4 x = ç ÷+ 15 x - 4 ln 4 dx ln 7 è x ø

3-6 ..

Calculus

Chapter 3

6.

Given the function f ( x) = 5e x + 7, find the equation of the line tangent to the graph of 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) = 5e x

f ¢( x) = 5e x

f ¢(0) = 5e0 f ¢(0) = 5

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

f ( x) = 5e x + 7 f (0) = 5e0 + 7 f (0) = 12

Finally, use the slope and point to find the equation of the tangent line.

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

3-7 ..

Calculus

Chapter 3

7.

Given the function f ( x) = ln x5 + 2, find the equation of the line tangent to the graph of the function at x = 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.

5 f ¢( x) = f ( x) = ln x5 + 2 x f ( x) = 5ln x + 2 5 f ¢(e) = 5 e f ¢( x) = x f ¢(e) = 5e - 1 Now find the y coordinate of the point by substituting the given value into the original function:

f ( x) = ln x5 + 2 f (e) = ln e5 + 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 - x1 ) y - 7 = 5e- 1 ( x - e) y - 7 = 5e- 1x - 5 y = 5e- 1x + 2 y = 1.839 x + 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 If y = f ( x) = F ( x)S ( x) and if F ¢( x) and S ¢( x) exist, then

f ¢( x) = F ( x) S ¢( x) + S ( x) F ¢( x) Theorem 2: Quotient Rule If y = f ( x) =

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

B( x)T ¢( x) - T ( x) B ¢( x)

[ B( x)]2

1 - 10 Find the first derivative and simplify.

1.

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

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

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

F ( x ) = 4 x5 F ¢( x) = 20 x 4

f ¢( x) = F ( x) S ¢( x) + S ( x) F ¢( x) f ¢( x) = (4 x5 )(4 x - 13) + (2 x 2 - 13 x + 2)(20 x 4 ) f ¢( x) = 16 x 6 - 52 x5 + 40 x 6 - 260 x5 + 20 x 4 f ¢( x) = 56 x 6 - 312 x5 + 40 x 4 3-9 ..

Calculus 2.

Chapter 3

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

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

F ( x) = 4 x 2 + 7 F ¢( x) = 8 x

S ( x) = 2 x 2 - 9 S ¢( x) = 4 x

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

3.

f ( x) =

13x 3x 2 - 2

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

B( x) = 3x 2 - 2 B ¢( x) = 6 x

T ( x) = 13x T ¢( x) = 13 f ¢( x) = f ¢( x) =

f ¢( x) =

f ¢( x) =

B( x)T ¢( x) - T ( x) B ¢( x)

[ B( x)]2 (3x 2 - 2)(13) - (13 x)(6 x) 2

é(3x 2 - 2)ù ë û 39 x 2 - 26 - 78 x 2 2

é(3x 2 - 2) ù ë û - 39 x 2 - 26 2

é(3x 2 - 2) ù ë û

3-10 ..

Calculus

4.

f ( x) =

Chapter 3

2x - 5 3x 2 + 5

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

B( x) = 3x 2 + 5 B¢( x) = 6 x

T ( x) = 2 x - 5 T ¢( x) = 2

f ¢( x) = f ¢( x) =

f ¢( x) =

f ¢( x) =

5.

B( x)T ¢( x) - T ( x) B ¢( x)

[ B( x)]2 (3 x 2 + 5)(2) - (2 x - 5)(6 x) 2

é(3x 2 + 5) ù ë û

6 x 2 + 10 - 12 x 2 + 30 x 2

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

é(3 x 2 + 5) ù ë û

( )

f ( x) = 3e x (3ln x )

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

F ( x) = 3e x

S ( x) = 3ln x 3 S ¢( x) = x

F ¢( x) = 3e x

f ¢( x) = F ( x) S ¢( x) + S ( x) F ¢( x) 3 f ¢( x) = (3e x )( ) + (3ln x)(3e x ) x 9e x f ¢( x) = + 9e x ln x x

3-11 ..

Calculus 6.

Chapter 3

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

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

S ( x) = x 2 - 2 x S ¢( x) = 2 x - 2

F ( x) = 2 x3 - 5 x 2 F ¢( x) = 6 x 2 - 10 x

f ¢( x) = F ( x) S ¢( x) + S ( x) F ¢( x) f ¢( x) = (2 x3 - 5 x 2 )(2 x - 2) + ( x 2 - 2 x)(6 x 2 - 10 x) f ¢( x) = 4 x 4 - 4 x3 - 10 x3 + 10 x 2 + 6 x 4 - 10 x3 - 12 x3 + 20 x 2 f ¢( x) = 10 x 4 - 36 x3 + 30 x 2

7.

y=

x 2 - 3x + 4 4 x2 - 3

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

B( x) = 4 x 2 - 3 B ¢( x) = 8 x

T ( x) = x 2 - 3x + 4 T ¢( x) = 2 x - 3

f ¢( x) = f ¢( x) =

f ¢( x) =

f ¢( x) =

B( x)T ¢( x) - T ( x) B ¢( x)

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

é(4 x 2 - 3) ù ë û

8 x3 - 12 x 2 - 6 x + 9 - 8 x3 + 24 x 2 - 32 x 2

é(4 x 2 - 3) ù ë û 12 x 2 - 38 x + 9 2

é(4 x 2 - 3) ù ë û

3-12 ..

Calculus 8.

Chapter 3

f ( x) = 5 x log 7 x

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

F ( x) = 5 x

S ( x) = log 7 x

F ¢( x) = 5 x ln 5

S ¢( x) =

1 æ1 ö ç ÷ ln 7 è x ø

f ¢( x) = F ( x) S ¢( x) + S ( x) F ¢( x) æ 1 ö f ¢( x) = (5 x ) ç + (log 7 x)(5 x ln 5) è x ln 7 ÷ ø f ¢( x) =

9.

f ( x) =

5x + (5 x )(ln 5)(log 7 x) x ln 7

84 x x2 + 4

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

1

B( x) = x 2 + 4 B ¢( x) = 2 x

T ( x) = 8 x 4 T ¢( x) = 2 x f ¢( x) =

f ¢( x) =

- 34

B( x)T ¢( x) - T ( x) B ¢( x)

[ B( x)]2 ( x 2 + 4)(2 x

- 34

1

) - (8 x 4 )(2 x) 2

é( x 2 + 4)ù ë û 5

- 34

f ¢( x) =

2 x 4 + 8x

f ¢( x) =

- 14 x 4 + 8 x

5

- 16 x 4 2

é( x 2 + 4)ù ë û 5

- 34 2

é( x 2 + 4)ù ë û

=

2(- 7 x 2 + 4) 3

x 4 ( x 2 + 4) 2

3-13 ..

Calculus

10.

f ( x) =

Chapter 3

(2 x3 - 5)(3x 2 + 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) = 3x 2 + 1 S ¢( x) = 6 x

F ( x) = 2 x3 - 5 F ¢( x) = 6 x 2

T ¢( x) = F ( x) S ¢( x) + S ( x ) F ¢( x ) T ¢( x) = (2 x3 - 5)(6 x) + (3 x 2 + 1)(6 x 2 ) T ¢( x) = 12 x 4 - 30 x + 18 x 4 + 6 x 2 T ¢( x) = 30 x 4 + 6 x 2 - 30 x Now that we have the derivative for the numerator, follow the quotient rule.

B( x) = x 2 + 7 B ¢( x) = 2 x

T ( x) = (2 x3 - 5)(3x 2 + 1) T ¢( x) = 30 x 4 + 6 x 2 - 30 x f ¢( x) = f ¢( x) =

f ¢( x) =

f ¢( x) =

B( x)T ¢( x) - T ( x) B ¢( x)

[ B( x)]2 ( x 2 + 7)(30 x 4 + 6 x 2 - 30 x) - (2 x3 - 5)(3x 2 + 1)(2 x) 2

é( x 2 + 7) ù ë û

30 x6 + 210 x 4 + 6 x 4 + 42 x 2 - 30 x3 - 210 x - 12 x5 - 4 x 4 + 30 x3 + 10 x 2

é( x 2 + 7) ù ë û 30 x6 - 12 x5 + 212 x 4 + 42 x 2 - 200 x 2

é( x 2 + 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 and g ( x) is in the domain of f.

Theorem 1:

General Power Rule n

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

f ¢( x) = n [u ( x)] Theorem 2:

u ¢( x)

Chain Rule

If y = f (u ) and u = g ( x) define the composite function y = m( x) = f [ g ( x)], then

dy dy du = dx du dx

1.

provided

du dy and exist du dx

Find m( x) = f [ g ( x)] if f (u ) = u 3 and g ( x) = 3x 2 - 7 x + 17.

m( x ) = f [ g ( x ) ] m( x) = f é3x 2 - 7 x + 17 ù ë û m( x) = (3 x 2 - 7 x + 17)3

3-15 ..

Calculus 2.

Chapter 3

Find m( x) = f [ g ( x)] if f (u ) = ln u and g ( x) = 4 x 2 + 7 x - 35.

m( x ) = f [ g ( x ) ] m( x) = f é4 x 2 + 7 x - 35ù ë û m( x) = ln(4 x 2 + 7 x - 35) 3.

Given y = ( x 4 + 3x 2 - 3x + 5)3 , write this composite function in the form y = f (u ) and u = g ( x).

g ( x) = x 4 + 3 x 2 - 3 x + 5

y = u3

3

4.

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

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

y = eu

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

5.

f ( x) = (3x 4 - 5 x + 13)3 We must break the function into y and g(x) and then find each derivative.

y = u3 dy = 3u 2 du

u = g ( x) = 3x 4 - 5 x + 13 du = g ¢( x) = 12 x3 - 5 dx

Now, use the chain rule.

f ¢( x) =

dy du du dx

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

3-16 ..

Calculus 6.

Chapter 3

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

u = g ( x) = 9 x - 7

y = 4 ln u dy 4 = du u

du = g ¢( x) = 9 dx

Now, use the chain rule.

dy du du dx 4 f ¢( x) = (9) u 36 f ¢( x) = 9x - 7

f ¢( x) =

7.

f ( x) = 3e5 x- 2 We must break the function into y and g(x) and then find each derivative.

y = 3eu dy = 3eu du

u = g ( x) = 5 x - 2 du = g ¢( x) = 5 dx

Now, use the chain rule.

f ¢( x) =

dy du du dx

f ¢( x) = 3eu (5) f ¢( x) = 15eu f ¢( x) = 15e5 x - 2

3-17 ..

Calculus

8.

Chapter 3

y = 6( x 2 - 3x + 12)6 We must break the function into y and g(x) and then find each derivative.

u = g ( x) = x 2 - 3x + 12

y = 6u 6 dy = 36u 5 du

du = g ¢( x) = 2 x - 3 dx

Now, use the chain rule.

f ¢( x) =

dy du du dx

f ¢( x) = 36u 5 (2 x - 3) f ¢( x) = 36( x 2 - 3x + 12)5 (2 x - 3) f ¢( x) = 36(2 x - 3)( x 2 - 3x + 12)5

9.

1

3

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

1

y=u 3

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

dy 1 - 2 3 = u du 3

du = g ¢( x) = 4 x - 3 dx

Now, use the chain rule.

dy du du dx 1 -2 f ¢( x) = u 3 (4 x - 3) 3 -2 4x - 3 f ¢( x) = (2 x 2 - 3x) 3 3 4x - 3 f ¢( x) = 2 3(2 x 2 - 3 x) 3 4x - 3 f ¢( x) = 33 (2 x 2 - 3 x) 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 dy = 3u 2 du

u = g ( x) = ln(3x + 7) du = g ¢( x) = 3 dx 3 x +7

Since g ( x) is still a composite function, we must break it into pieces again.

v = h( x ) = 3 x + 7

g = ln v dg 1 = dv v

dv = h ¢( x) = 3 dx

Now, use the chain rule to find the first derivative of g.

dg dv dv dx 1 g ¢( x) = (3) v 3 g ¢( x) = 3x + 7

g ¢( x) =

Now, use the chain rule again.

dy du du dx æ 3 ö f ¢( x) = 3u 2 ç è 3x + 7 ÷ ø

f ¢( x) =

f ¢( x) =

9(ln(3x + 7)) 2 3x + 7

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 for the derivative of the variable y. dx 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 a y¢.

5. Solve for y¢by dividing.

1. Consider the equation

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

y = - 32 x 2 + 19 2

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

y ' = - 3x

y = - 32 x 2 + 19 2 b.) Use implicit differentiation to find the first derivative.

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

3-21 ..

Calculus

Chapter 3

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

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

3 y - 6 x3 + 8 = 0

y ¢= 6 x 2

3 y ¢- 18 x 2 + 0 = 0

y ¢= 6(3) 2 y ¢= 54

3 y ¢= 18 x 2 y ¢= 6 x 2

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

2 xy 2 + 3 y - 2 x 2 = 0; (1, 4)

2 xy 2 + 3 y - 2 x 2 = 0

y ¢=

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

3-22 ..

- 2 y2 + 4x 4 xy + 3

- 2(4)2 + 4(1) y ¢= 4(1)(4) + 3 28 y ¢= 19

Calculus

Chapter 3

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

5ln y = e x + x 2 - y 2 ; (0,3) x

2

2

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

e0 + 2(0) y ¢= 5 + 2(3) 3 1 y ¢= 23 3

y ¢=

æ5 ö y ¢ç + 2 y ÷ = e x + 2 x èy ø

3 23

ex + 2x y ¢= 5 + 2y y 5. Use implicit differentiation to find the first derivative and evaluate it at the given point. x 2 y3 + xy 2 - 6 y + 4ln x - 6 = 0; (- 2,3)

x 2 y 3 + xy 2 - 6 y + 4 ln x - 6 = 0 x 2 (3 y 2 y ¢) + y 3 (2 x) + x(2 yy ¢) + y 2 (1) - 6 y ¢+ 4x - 0 = 0 3x 2 y 2 y ¢+ 2 xy 3 + 2 xyy ¢+ y 2 - 6 y ¢+ 4x = 0 3x 2 y 2 y ¢+ 2 xyy ¢- 6 y ¢= - 2 xy 3 - y 2 - 4x y ¢(3x 2 y 2 + 2 xy - 6) = - 2 xy 3 - y 2 - 4x y ¢=

y ¢= y ¢=

- 2 xy 3 - y 2 - 4x 3x 2 y 2 + 2 xy - 6 - 2(- 2)(3)3 - (3)2 - -42

3(- 2)2 (3)2 + 2(- 2)(3) - 6 101 y ¢= 90

3-23 ..

- 2 xy 3 - y 2 - 4x 3x 2 y 2 + 2 xy - 6

Calculus

Chapter 3

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

2 y3 - 16 + 5 x 2 + 13 = 0; (3, 2) 2 y 3 - 16 + 5 x 2 + 13 = 0 1

(2 y 3 - 16) 2 + 5 x 2 + 13 = 0 1 2

1

(2 y 3 - 16) - 2 (6 y 2 y ') + 10 x + 0 = 0 1

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

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

1

y ¢=

- 10(3)(2(2)3 - 16) 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 + 6 y = x3 - 2 x + 7; (1, 0)

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

y ¢=

3x 2 - 2 4(4 + y )3 + 6

3(1)2 - 2

4(4 + 0)3 + 6 1 y ¢= 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

3x 2 + 9 xy - 2 y3 = 15,

dx = 4, x = 2, dt

dy given dt y = 1.

Find the first derivative using implicit differentiation:

x 2 + 8 xy - y 3 = 12 dx dy dx dy 6 x + 9 x + y (9 ) - 6 y 2 =0 dt dt dt dt Now substitute all known values into the first derivative and solve for

dx dy dx dy + 9 x + y (9 ) - 6 y 2 =0 dt dt 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 6x

3-25 ..

dy . dt

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 a = 4, = 0, b = 20, = ?, c » 20.396, = - 3. Use the Pythagorean Theorem to set up dt dt dt the relation between the variables and use implicit differentiation to solve.

a 2 + b2 = c2 da db dc 2a + 2b = 2c dt dt dt db 2(4)(0) + 2(20) = 2(20.396)( - 3) dt db 40 = - 122.376 dt db = - 3.06 dt 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 a = 12, = - 4, b = 16, = ?, c = 20, = 0. Use the Pythagorean Theorem to set up the dt dt dt relation between the variables and use implicit differentiation to solve.

a 2 + b2 = c2 da db dc 2a + 2b = 2c dt dt dt db 2(12)(- 4) + 2(16) = 2(20)(0) dt db - 96 + 32 =0 dt db 32 = 96 dt db =3 dt 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 =  r3 3 dV dr = 4 r 2 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 The percentage rate of change is 100 ´

f ¢( x) . f ( x)

f ¢( x) . f ( x)

Theorem 1: Elasticity of Demand If price and demand are related by x = f ( p), then the elasticity of demand is given by:

E ( p) = -

pf ¢( 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) = 42 x - 0.3x 2 , find the relative rate of change.

f '( x) 42 - 0.6 x = f ( x) 42 x - 0.3x 2

f ( x) = 42 x - 0.3x 2 f '( x) = 42 - 0.6 x

3-29 ..

Calculus

Chapter 3

2. Given f ( x) = 7 x 2 - 2ln x, find the relative rate of change when x = 4. (Round to three decimals if necessary.)

f ( x) = 7 x 2 - 2ln x f ¢( x) = 14 x - 2x f (4) = 7(4) 2 - 2 ln(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 - 7 x, find the relative rate of change when x = 60. (Round to three decimals if necessary.)

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 - 6 x 2 , find the percentage rate of change when x = 30. (Round to the nearest tenth if necessary.)

f ( x) = 4000 - 6 x 2 f ¢( x) = - 12 x f (30) = 4000 - 3(30)2 f (30) = 1300

f ¢(30) = - 12(30) f ¢(30) = - 360

3-30 ..

100 ´

f ¢( x) - 360 = 100 ´ f ( x) 1300 = 100 ´ - 0.277 = 27.7%

Calculus

Chapter 3

5. Given x = f ( p) = 12, 000 - 300 p, use the price–demand equation to find E ( p), the elasticity of demand.

f ( p) = 12,000 - 300 p f ¢( p) = - 300

pf ¢( p) f ( p) p(- 300) =12, 000 - 300 p 300 p = 12, 000 - 300 p p E ( p) = 40 - p

E ( p) = -

6. Given x = f ( p) = 850 - 3 p - 0.04 p 2 , 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. b. c.

p = 15 p = 35 p = 55

f ( p) = 850 - 3 p - 0.04 p 2 f ¢( p) = - 3 - 0.08 p

E ( p) = =E ( p) =

E (15) =

3(15) + 0.08(15)2 850 - 3(15) - 0.04(15)

E (35) =

2

3(35) + 0.08(35)2 850 - 3(35) - 0.04(35)

E (55) =

2

3(55) + 0.08(55)2 850 - 3(55) - 0.04(55)

2

850 - 3 p - 0.04 p 2 3 p + 0.08 p 2

850 - 3 p - 0.04 p 2

=

63 = 0.079; demand is inelastic.

796

=

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 - 4 p 2 , 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 - 4 p 2 f ¢( p) = - 8 p

E ( p) = == E ( p) =

pf ¢( p) f ( p) p(- 8 p) 2400 - 4 p 2 8 p2

2400 - 4 p 2 2 p2 600 - p 2

For the demand to be elastic, E ( p) > 1.

2 p2 600 - p 2

>1

2 p 2 > 600 - p 2 3 p 2 > 600 p 2 > 200 p > 10 2 For the demand to be inelastic, 0 < E ( p) < 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

8. Given x = f ( p) = 1600 - 2 p 2 , 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) = 1600 - 2 p 2

)

(

1

2

f ¢( p) = - 2 p 1600 - 2 p 2 f ¢( p) =

E ( p) = -

- 12

)

- 2p 1600 - 2 p 2 pf ¢( p) 1 = - pf ¢( p) × f ( p) f ( p)

æ ö - 2p 1 = - pç ÷ 2 çè 1600 - 2 p 2 ÷ ø 1600 - 2 p =

E ( p) =

2 p2 1600 - 2 p 2

p2 800 - p 2

For the demand to be elastic, E ( p) > 1.

p2 800 - p 2

>1

p 2 > 800 - p 2 2 p 2 > 800 p 2 > 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) = 0 or 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 - 27 x + 12 find: a.

f '( x)

f '( x) = 3x 2 - 27

4-1 ..

Calculus

Chapter 4

b. The critical values of the function

f '( x) = 3x 2 - 27 0 = 3 x 2 - 27 27 = 3 x 2 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: 2x + 6

f '( x)

f ( x) =

19 2x + 6

f ( x) = 19(2 x + 6) - 1 f '( x) = - 19(2 x + 6) - 2 (2) or -

38 (2 x +6)2

b. The critical values of the function

f '( x) = -

38 (2 x +6)2

f ¢( x) is always negative and cannot equal zero. f ¢( x) is undefined at x = - 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 3–5

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. The local extrema

3.

f ( x) = 2 x 2 - 16 x + 33 First find the derivative and the critical values as follows:

f '( x) = 4 x - 16 0 = 4 x - 16 16 = 4 x 4=x Choose a number on each side of the critical value to find the value of the first derivative: Test number: 3

Test number: 5

f '(3) = 4 x - 16 f '(3) = 4(3) - 16 f '(3) = 12 - 16 f '(3) = - 4

f '(5) = 4 x - 16 f '(5) = 4(5) - 16 f '(5) = 20 - 16 f '(5) = 4

a. The test number 5 has a first derivative that is a positive value, therefore, the function is increasing on the interval (4, ¥ ). b. The test number 3 has a first derivative that is a negative value, therefore, the 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 - 27 x + 12 First find the derivative and the critical values as follows:

f '( x) = 3x 2 - 27 0 = 3 x 2 - 27 27 = 3 x 2 9 = x2 ±3 = x The partition numbers (also critical values of f) are x = - 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 '(- 4) = 3(- 4) 2 - 27 f '(- 4) = 3(16) - 27 f '(- 4) = 48 - 27 f '(- 4) = 21

f '(0) = 3(0)2 - 27 f '(0) = 3(0) - 27 f '(0) = 0 - 27 f '(0) = - 27

f '(4) = 3(4)2 - 27 f '(4) = 3(16) - 27 f '(4) = 48 - 27 f '(4) = 21

a. The test numbers –4 and 4 have a first derivative that is a positive value, therefore, the function is increasing on the intervals (- ¥ , - 3) and (3, ¥ ). b. The test number 0 has a first derivative that is a negative value, therefore, the function is decreasing on the interval (- 3,3). c. Because the function changes from an increasing function to a decreasing function at the critical value of –3 the point (- 3, 66) is a local maximum. Because 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) = x 4 - 4 x3 + 3 First find the derivative and the critical values as follows:

f '( x) = 4 x3 - 12 x 2 0 = 4 x3 - 12 x 2 0 = 4 x 2 ( x - 3) 0,3 = x The partition numbers (also critical values of f) are x = 0 and x = 3. Choose a number within each interval to find the value of the first derivative: Test number: –1

Test number: 2

Test number: 4

f '(- 1) = 4(- 1)3 - 12(- 1)2 f '(- 1) = 4(- 1) - 12(1) f '(- 1) = - 4 - 12 f '(- 1) = - 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 at x = 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) = 2 x3 + 3x 2 - 12 x First find the derivative and the critical values as follows:

f '( x) = 6 x 2 + 6 x - 12 0 = 6 x 2 + 6 x - 12 0 = 6( x 2 + x - 2) 0 = 6( x + 2)( x - 1) - 2,1 = x The partition numbers (also critical values of f) are x = - 2 and x = 1. Choose a number within each interval to find the value of the first derivative: Test number: –3

Test number: 0

Test number: 2

f '(- 3) = 6(- 3) 2 + 6( - 3) - 12 f '(- 3) = 6(9) + 6(- 3) - 12 f '(- 3) = 54 - 18 - 12 f '(- 3) = 24

f '(0) = 6(0) 2 + 6(0) - 12 f '(0) = 6(0) + 6(0) - 12 f '(0) = 0 - 0 - 12 f '(0) = - 12

f '(2) = 6(2)2 + 6(2) - 12 f '(2) = 6(4) + 6(2) - 12 f '(2) = 24 + 12 - 12 f '(2) = 24

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, ¥ ). b. The test number 0 has a first derivative that is a negative value, therefore, the function is decreasing on the interval (- 2,1). c.

4-6 ..

Calculus 7.

Chapter 4

f ( x) = x3 - 12 x + 4 First find the derivative and the critical values as follows:

f '( x) = 3 x 2 - 12 0 = 3 x 2 - 12 12 = 3 x 2 4 = x2 ±2 = x The partition numbers (also critical values of f) are x = - 2 and x = 2. Choose a number within each interval to find the value of the first derivative: Test number: –3

Test number: 0

Test number: 3

f '(- 3) = 3(- 3)2 - 12 f '(- 3) = 3(9) - 12 f '(- 3) = 27 - 12 f '(- 3) = 15

f '(0) = 3(0)2 - 12 f '(0) = 3(0) - 12 f '(0) = 0 - 12 f '(0) = - 12

f '(3) = 3(3) 2 - 12 f '(3) = 3(9) - 12 f '(3) = 27 - 12 f '(3) = 15

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, ¥ ). b. The test number 0 has a first derivative that is a negative value, therefore, the function is decreasing on the interval (- 2, 2). c.

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) = Definition:

d f '( x). dx

Concavity

The graph of a function f is concave upward on the interval (a, b) if f '( x) is increasing on (a b) and is concave downward on the interval (a, b) if f '( x) is decreasing on (a, b).

Summary:

Concavity

For the interval (a,b), if f ''( 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

If y = f ( x) is continuous on (a, b) and has an inflection point at x = c, then either 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 for f ''( x), and determine concavity.

Step 4 Sketch the graph of the function f.

4-9 ..

Calculus

Chapter 4

1. Given the function f ( x) = 3x 4 - 13x 2 + 13x + 1, find f ''( x).

f '( x) = 12 x3 - 26 x + 13 f ''( x) = 36 x 2 - 26

12

2. Given the function f ( x) = 2 x5 + 7 x 5 + 16, find f ''( x).

7

f '( x) = 10 x 4 + 84 x 5 5 2

f ''( x) = 40 x3 + 588 x 5 25

3. Given the function f ( x) = 2 x3 - 12 x 2 + 26 x - 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 x 2 - 24 x + 26 f ''( x) = 12 x - 24

f ''( x) = 12 x - 24 0 = 12 x - 24 24 = 12 x 2=x

f (2) = 2 x3 - 12 x 2 + 26 x - 2 f (2) = 2(2)3 - 12(2) 2 + 26(2) - 2 f (2) = 16 - 48 + 52 - 2 f (2) = 18

Now check the value of the second derivative on both sides of 2.

f ''(3) = 12(3) - 24 f ''(3) = 12

f ''(1) = 12(1) - 24 f ''(1) = - 12

Because the second derivative changes from negative to positive, the point (2, 18) is an inflection point.

4-10 ..

Calculus 4–6

Chapter 4

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) = x 4 - 54 x 2 First find the second derivative and the partition numbers as follows:

f '( x) = 4 x3 - 108 x f ''( x) = 12 x 2 - 108 0 = 12 x 2 - 108 108 = 12 x 2 9 = x2 ±3 = x The partition numbers are x = - 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 ''(- 4) = 12(- 4) 2 - 108 f ''(- 4) = 12(16) - 108 f ''(- 4) = 192 - 108 f ''(- 4) = 84

f ''(0) = 12(0)2 - 108 f ''(0) = 12(0) - 108 f ''(0) = 0 - 108 f ''(0) = - 108

f ''(4) = 12(4)2 - 108 f ''(4) = 12(16) - 108 f ''(4) = 192 - 108 f ''(4) = 84

a. The test numbers –4 and 4 have a second derivative that is a positive value, therefore the function is concave upward on the intervals (- ¥ , - 3) and (3, ¥ ). b. The test number 0 has a second derivative that is a negative value, therefore the function is concave downward on the interval (- 3,3). c. The partition numbers –3 and 3 are both inflection points because the function changes concavity at each number. The inflection points are (- 3, - 405) and (3, - 405).

4-11 ..

Calculus 5.

Chapter 4

f ( x) = 2 x3 - 6 x 2 + 12 x - 9 First find the second derivative and the partition numbers as follows:

f '( x) = 6 x 2 - 12 x + 12 f ''( x) = 12 x - 12 0 = 12 x - 12 12 = 12 x 1= x The partition number is x = 1. Choose a number on each side. Find the value of the second derivative at these test numbers Test number: 0

Test number: 2

f ''(0) = 12(0) - 12 f ''(0) = 0 - 12 f ''(0) = - 12

f ''(2) = 12(2) - 12 f ''(2) = 24 - 12 f ''(2) = 12

a. The test number 2 has a second derivative that is a positive value, therefore the function is concave upward on the interval (1, ¥ ). 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 at this number. The inflection point is (1, - 1).

4-12 ..

Calculus 6.

Chapter 4

f ( x) = 9e2 x - e3x First find the second derivative and the partition numbers as follows:

f '( x) = 18e2 x - 3e3 x f ''( x) = 36e2 x - 9e3 x 0 = 36e2 x - 9e3 x 9e3 x = 36e2 x ex = 4 x = ln 4 The partition number is x = ln 4. Choose a number on each side.Find the value of the second derivative at these test numbers Test number: 0

Test number: 2

f ''(0) = 36e2(0) - 9e3(0)

f ''(2) = 36e2(2) - 9e3(2)

f ''(0) = 36e0 - 9e0 f ''(0) = 36 - 9 f ''(0) = 27

f ''(2) = 36e4 - 9e6 f ''(2) = 1965.53 - 3630.86 f ''(2) = - 1665.33

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). b. The test number 2 has a second derivative that is a negative value, therefore the function is concave downward on the interval (ln 4, ¥ ). c. The partition number ln 4 is an inflection point because the function changes concavity at this number. The inflection point is (ln 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)( x 2 - 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 of f ( x) = 0.

f (0) = (0 + 2)((0) 2 - 0 - 4) f (0) = (2)(- 4) f (0) = - 8

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

or

x=-2

x2 - x - 4 = 0 x=

1 ± 17 2

Therefore, the x intercepts are (1+ 217 , 0), (1- 217 , 0) and (–2, 0), and the y intercept is (0, –8).

Step 2: Analyze the first derivative.

f '( x) = ( x + 2)(2 x - 1) + ( x 2 - x - 4)(1) f '( x) = 2 x 2 + 3x - 2 + x 2 - x - 4 0 = 3x 2 + 2 x - 6 - 1± 2 5 =x 3 The partition numbers are x = - 1+32 5 and x = - 1- 32 5 . Choose a number within each 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 '(- 2) = 3(4) - 4 - 6 f '(- 2) = 2

f '(0) = 3(0) 2 + 2(0) - 6 f '(0) = 3(0) - 0 - 6 f '(0) = - 6

f '(2) = 3(2) 2 + 2(2) - 6 f '(2) = 3(4) + 4 - 6 f '(2) = 10

Therefore, the function is increasing on the intervals (- ¥ , - 1- 32 5 ) and ( - 1+32 5 , ¥ ), decreasing on the interval ( - 1- 32 5 , - 1+32 5 ), and has a local maximum at x = - 1- 32 5 and a local minimum at x = - 1+32 5 .

4-14 ..

Calculus

Chapter 4

Step 3: Analyze the second derivative

f '( x) = 3 x 2 + 2 x - 6 f ''( x ) = 6 x + 2 0 = 6x + 2 - 2 = 6x - 13 = x The partition number is x = - 13 . Choose a number on each side of the partition number. Find the value of the second derivative at these numbers.

Test number: –1

Test number: 0

f ''(- 1) = 6 x + 2 f ''(- 1) = 6(- 1) + 2 f ''(- 1) = - 4

f ''(0) = 6 x + 2 f ''(0) = 6(0) + 2 f ''(0) = 2

Therefore, the function is concave upward on the interval (- 13 , ¥ ), concave downward on the interval ( - ¥ , - 13 ), and has an inflection point at x = - 13 .

Step 4: Sketch the function

4-15 ..

Calculus 8.

Chapter 4

f ( x) = 5 x 4 - 10 x3

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 of f ( x) = 0.

f (0) = 5(0)4 - 10(0)3 f (0) = 0 - 0 f (0) = 0

0 = 5 x 4 - 10 x3 0 = 5 x3 ( x - 2) x = 0, 2

Therefore, the x intercepts are (0, 0) and (2, 0), and the y intercept is (0, 0).

Step 2: Analyze the first derivative.

f '( x) = 20 x3 - 30 x 2 0 = 10 x 2 (2 x - 3) 3 0, = x 2 The partition numbers are x = 0 and x = 32 . Choose a number within each interval.

Find the value of the first derivative at these test numbers.

Test number: –1

Test number: 1

Test number: 2

f '(- 1) = 20(- 1)3 - 30(- 1) 2 f '(- 1) = 20(- 1) - 30(1) f '(- 1) = - 50

f '(1) = 20(1)3 - 30(1) 2 f '(1) = 20(1) - 30(1) f '(1) = - 10

f '(2) = 20(2)3 - 30(2) 2 f '(2) = 20(8) - 30(4) f '(2) = 40

Therefore, the function is increasing on the interval ( 32 , ¥ ) and decreasing on the intervals

(- ¥ , 0) and (0, 32 ). The function has a local minimum at x = 32 .

4-16 ..

Calculus

Chapter 4

Step 3: Analyze the second derivative

f '( x) = 20 x3 - 30 x 2 f ''( x) = 60 x 2 - 60 x 0 = 60 x 2 - 60 x 0 = 60 x( x - 1) 0,1 = x The partition numbers are x = 0 and x = 1. Choose a number on each side of the partition number. Find the value of the second derivative at these test numbers.

Test number: –1

Test number: 0.5

f ''(- 1) = 60(- 1)2 - 60(- 1) f ''(- 1) = 60(1) + 60 f ''(- 1) = 120

f ''(0.5) = 60(0.5) 2 - 60(0.5) f ''(2) = 60(2) 2 - 60(2) f ''(0.5) = 60(0.25) - 30 f ''(2) = 60(4) - 120 f ''(0.5) = - 15 f ''(2) = 120

Test number: 2

Therefore, the function is concave upward on the intervals (- ¥ , 0) and (1, ¥ ) and concave downward on the interval (0, 1), and has an inflection point at x = 0.

Step 4: Sketch the function

4-17 ..

Calculus 9.

Chapter 4

f ( x) = 8e x - e2 x

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 of f ( x) = 0.

0 = 8e x - e2 x

f (0) = 8e0 - e 2(0) f (0) = 8(1) - 1 f (0) = 7

0 = e x (8 - e x ) x = ln 8

Therefore, the x intercept is (ln 8, 0), and the y intercept is (0, 7).

Step 2: Analyze the first derivative.

f '( x) = 8e x - 2e2 x 0 = 8e x - 2e2 x 0 = 2e x (4 - e x ) ln 4 = x The partition number is x = ln 4. 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(7.39) - 2(54.60) f '(2) = - 50.08

Therefore, the function is increasing on the interval (- ¥ , ln 4) and decreasing on the interval (ln 4, ¥ ). The function has a local maximum at x = ln 4.

4-18 ..

Calculus

Chapter 4

Step 3: Analyze the second derivative

f '( x) = 8e x - 2e 2 x f ''( x) = 8e x - 4e 2 x 0 = 8e x - 4e2 x 0 = 4e x (2 - e x ) ln 2 = x

The partition number is x = ln 2. 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 ''(0) = 8e0 - 4e2(0) f ''(0) = 8(1) - 4(1) f ''(0) = 4

f ''(1) = 8e1 - 4e2(1) f ''(1) = 8(2.72) - 4(7.39) f ''(1) = - 7.81

Therefore, the function is concave upward on the interval (- ¥ , ln 2), concave downward on the interval (ln 2, ¥ ), and has an inflection point at x = ln 2.

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) = - 0.25 x 4 + 23x3 - 360 x 2 + 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 x 2 - 720 x N ''( x) = - 3 x 2 + 138 x - 720 0 = - 3( x 2 - 46 x + 240) 0 = - 3( x - 40)( x - 6) 6, 40 = x The partition numbers are x = 6 and x = 40. Choose a number on each side of the partition numbers within the given domain. Find the value of the second derivative at these test numbers.

Test number: 39

Test number: 41

f ''(39) = - 3(39) 2 + 138(39) - 720 f ''(39) = - 4563 + 5382 - 720 f ''(39) = 99

f ''(41) = - 3(41) 2 + 138(41) - 720 f ''(41) = - 5043 + 5658 - 720 f ''(41) = - 105

a. The test number 39 has a second derivative that is a positive value, therefore the 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

L’Hôpital’s Rule for 0/0 Indeterminate forms

Theorems:

(Version 1) For c a real number, if lim f ( x) = 0 and lim g ( x) = 0, then

1.

x® c

x® c

f ( x) f '( x) = lim x® c g ( x) x® c g '( x) lim

providing that the second limit exists or is ±¥ .

(Version 2) The first version of L’Hôpital’s rule remains valid if the symbol x ® c is replaced everywhere it occurs with one of the following symbols:

2.

x ® c-

x ® c+

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.

3.

Use L’Hôpital’s rule to find the limit in Problems 1 and 2.

1.

2.

3x 2 3(4)2 48 = lim lim 2 = = =4 x® 4 x + 4 x - 32 x® 4 2 x + 4 2(4) + 4 12 x3 - 64

lim

x® -¥

3x 4 + 17

x® ¥ 6 x

5

+ 25

= lim

12 x3

x® ¥ 30 x

4

12 =0 x® ¥ 30 x

= lim

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.

lim

3x 2 - 7

x® 2 2 x3 - 13

L’Hôpital’s rule does not apply because when the value is substituted into the function, the result is not 0/0.

lim

3x 2 - 7

x® 2 2 x

3

- 13

=

3(2)2 - 7 3

2(2) - 13

=

5 3

4.

Use L’Hôpital’s rule to find the following limit and then verify that limit by a second method. lim

x 2 - 10 x + 21

x® 7 x 2 - 2 x - 35

x 2 - 10 x + 21

2 x - 10 2(7) - 10 4 1 = = = 2(7) - 2 12 3 x® 7 2 x - 2

= lim

( x - 7)( x - 3) x® 7 x - 2 x - 35 x® 7 ( x - 7)( x + 5) x- 3 = lim x® 7 x + 5 7- 3 4 1 = = = 7 + 5 12 3 lim

2

= lim

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

x® - 1

x3 + 7 x - 6 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 + 7 x - 6 x2 - x

3x 2 + 7 3(- 1)2 + 7 10 10 = lim = = =2(- 1) - 1 - 3 3 x® - 1 2 x - 1

4-22 ..

Calculus

6.

Chapter 4

lim

x4 - 6 x - 3

x® 2 x3 + 2 x - 6

The value can be substituted into the function and not result in 0/0. Therefore, use simple substitution.

lim

x4 - 6 x - 3

x® 2 x3 + 2 x - 6

7.

lim

=

(2)4 - 6(2) - 3 (2)3 + 2(2) - 6

=

1 6

7 x3 + 4 x - 5

x® ¥ 9 x 4 - 6 x + 7

If the value ¥ is substituted into the function, the result will be ¥ / ¥ . Therefore, we use L’Hôpital’s rule.

lim

7 x3 + 4 x - 5

x® ¥ 9 x 4 - 6 x + 7

= lim

21x 2 + 4

x® ¥ 36 x3 - 6

= lim

42 x

x® ¥ 108 x 2

42 x® ¥ 216 x =0

= lim

8.

lim

x +3

x® - 3 x 2 - 2 x - 15

If the value –3 is substituted into the function, the result will be 0/0. Therefore, we use L’Hôpital’s rule.

lim

x +3

x® - 3 x 2 - 2 x - 15

1 1 1 = =8 x® - 3 2 x - 2 2(- 3) - 2

= lim

Alternative Solution:

lim

x +3

x® - 3 x 2 - 2 x - 15

x +3 x® - 3 ( x + 3)( x - 5) 1 = lim x® - 3 x - 5 1 1 = =- 3- 5 8

= lim

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.

f ( x) = Step 1:

x- 4 x+4 Analyze the function.

a.

The domain is all real numbers except where x + 4 = 0 Þ x = - 4. b.

The x intercepts are found by setting the function (or the numerator) equal to 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 or x = - 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 asymptote is y = 1.

4-25 ..

Calculus Step 2:

Chapter 4 Analyze the first derivative.

f '( x) = f '( x) =

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

8

f '(- 5) =

(- 5 + 4) 2 8 f '(- 5) = =8 (- 1) 2

Test number: –3

f '(- 3) =

8

(- 3 + 4) 2 8 f '(- 3) = 2 = 8 (1)

Therefore, the function is increasing on the intervals (- ¥ , - 4) and (- 4, ¥ ).

Step 3:

Analyze the second derivative

f '( x) = 8( x + 4) - 2 f ''( x) = - 16( x + 4) - 3 - 16 f ''( x) = ( 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) = 3 = - 16 (1)

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 where x - 3 = 0 Þ x = 3. b.

The x intercepts are found by setting the function (or the numerator) equal to 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 or x = 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 = -14 = - 4. c.

Step 2:

Analyze the first derivative.

f ( x) = f '( x) = f '( x) =

- 4x x- 3 ( x - 3)(- 4) - (- 4 x)(1) ( x - 3) 2 12 ( x - 3)2

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 first derivative at these test numbers.

4-27 ..

Calculus

Chapter 4

Test number: 2

f '(2) =

Test number: 4

12

f '(4) =

12

(4 - 3) 2 12 f '(4) = = 12 1

2

(2 - 3) 12 f '(2) = = 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) = f ''( x) = f ''( x) =

12 ( x - 3) 2 ( x - 3) 2 (0) - (12)(2( x - 3)(1) ( x - 3) 4 - 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) =

Test number: 4

- 24

(2 - 3) - 24 f ''(2) = -1 f ''(2) = 24

3

f ''(4) =

- 24

(4 - 3)3 - 24 f ''(4) = 1 f ''(4) = - 24

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 - 3 x Step 1:

2

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.

2

f (0) = e- 3(0) = 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) = - 6 xe- 3x

2

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

f '(- 1) = - 6(- 1)e- 3( - 1) f '(- 1) = 6e- 3 f '(- 1) = 0.299

Test number: 1 2

f '(1) = - 6(1)e- 3(1) f '(1) = - 6e- 3 f '(1) = - 0.299

4-29 ..

2

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

f ¢( x) = - 6 xe- 3 x

2 2

2

f ¢¢( x) = - 6 x(- 6 xe- 3 x ) + e - 3 x (- 6) 2

f ¢¢( x) = 6e- 3 x (6 x 2 - 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 ''(- 1) = 6e- 3( - 1) (6(- 1) 2 - 1)

f ''(0) = 6e- 3(0) (6(0) 2 - 1)

f ''(- 1) = 6e- 3 (5) f ''(- 1) = 1.49

f ''(0) = (6)e0 ( - 1) f ''(0) = - 6

Test number: 1 2

f ''(1) = 6e- 3(1) (6(1)2 - 1) f ''(1) = 6e- 3 (5) f ''(1) = 1.49 Therefore, the function is concave upward on the intervals (- ¥ , - 0.41) and (- 0.41, ¥ ). The function is concave downward on the interval (- 0.41, 0.41). The function has 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) ³ f ( 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) = 0 and f ''(c) > 0, then f (c) is the absolute minimum of f on I. If f '(c) = 0 and 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) = 3x 2 - 12 x + 17 [- 1,8] First find the critical value(s) for the function.

f ( x) = 3x 2 - 12 x + 17 f '( x) = 6 x - 12 0 = 6 x - 12 12 = 6 x 2=x Calculate the function value at the endpoints and the critical value.

f (- 1) = 3(- 1) 2 - 12(- 1) + 17 f (2) = 3(2) 2 - 12(2) + 17 f (- 1) = 3 + 12 + 17 f (2) = 12 - 24 + 17 f (- 1) = 32 f (2) = 5

f (8) = 3(8)2 - 12(8) + 17 f (8) = 192 - 96 + 17 f (8) = 113

Therefore, the absolute maximum is at the point (8, 113) and the absolute minimum is 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) = 5 x3 - 6 x 4

First find the critical value(s) for the function.

f ( x) = 5 x3 - 6 x 4

3x 2 = 0

f '( x) = 15 x 2 - 24 x3

x2 = 0 x=0

0 = 3x 2 (5 - 8 x)

5 - 8x = 0 - 8x = - 5 x = 85

Then use the second-derivative test at the critical values.

f ¢¢( x) = 30 x - 72 x 2

f ¢¢(0) = 30(0) - 72(0)2 f ¢¢(0) = 0

f ¢¢( 85 ) = 30( 85 ) - 72( 85 )2 f ¢¢( 85 ) = - 75 8 4-32

..

Calculus

Chapter 4

The second-derivative test fails at 0 and shows that there is a local maximum at 85 .

Calculate the function value at the critical values.

f (0) = 5(0)3 - 6(0)4 f (0) = 0 - 0 f (0) = 0

f ( 85 ) = 5( 85 )3 - 6( 85 ) 4 625 ) f ( 85 ) = 5( 125 ) - 6( 4096 512 625 - 3750 f ( 85 ) = 512 4096 625 f ( 85 ) = 2048

625 ) and the local minimum is at the Therefore, the local maximum is at the point ( 85 , 2048 625 is the absolute maximum; there is no absolute minimum. point (0, 0). Also, f ( 85 ) = 2048

f ( x) =

3.

x2 - 4 x2 + 1

First find the critical value(s) for the function.

f ( x) = f '( x) = f '( x) =

x2 - 4 x2 + 1 ( x 2 + 1)(2 x) - ( x 2 - 4)(2 x) ( x 2 + 1) 2 10 x 2

( x + 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

f '( x) = f ''( x) = f ''( x) = f ''(0) =

Chapter 4

10 x x4 + 2 x2 + 1 ( x 4 + 2 x 2 + 1)(10) - (10 x)(4 x3 + 4 x) ( x 4 + 2 x 2 + 1)2 - 30 x 4 - 20 x 2 + 10 ( x 4 + 2 x 2 + 1)2 - 30(0)4 - 20(0)2 + 10

((0)4 + 2(0)2 + 1) 2 10 f ''(0) = 2 = 10 > 0 (1)

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 + 2 x + . x First find the critical value(s) for the function.

f ( x) = 6 + 2 x + 7 x - 1 f '( x) = 2 - 7 x - 2 7 f '( x) = 2 - 2 x 7 0 = 2- 2 x

7 x2

=2

2 x2 = 7 x 2 = 72 x = 72 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 ''( x) = 14 x - 3

f ''(1.87) = 14(1.87)- 3 f ''(1.87) » 2.14

The second derivative of the critical value is positive, therefore, the absolute minimum is f ( x) = f ( 72 ) » 13.48.

4-34 ..

Calculus 5.

Chapter 4 Absolute minimum value on [0, ¥ ) for f ( x) = 10 x 2 - 8 x + 26.

First find the critical value(s) for the function.

f ( x) = 10 x 2 - 8 x + 26 f '( x) = 20 x - 8 0 = 20 x - 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 f ''( x) = 20 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

Chapter 4

6.

Absolute maximum value on (0, ¥ ) for f ( x) = 3x6 - 18 x 4 .

First find the critical value(s) for the function.

f ( x) = 3x 6 - 18 x 4 f '( x) = 18 x5 - 72 x3 0 = 18 x3 ( x 2 - 4) 18 x3 = 0 or x 2 - 4 = 0 x=0 x2 = 4 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) = 18 x5 - 72 x3 f ''( x) = 90 x 4 - 216 x 2 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 of f ( 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 A( y ) = (24 - y ) y A( y ) = 24 y - y 2

Find the first derivative:

A( y ) = 24 y - y 2 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 - 2 y 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 + 56 y

Step 2:

xy = 900 900 x= y x = 900 y

subject to xy = 900

C ( y ) = 28 x + 56 y C ( y ) = 28(900 y - 1 ) + 56 y -1

Find the first derivative:

C ( y ) = 25, 200 y - 1 + 56 y

C ( y ) = 25, 200 y - 1 + 56 y C '( y ) = - 25, 200 y - 2 + 56 0 = - 25, 200 y - 2 + 56 25, 200 y - 2 = 56 56 y 2 = 25, 200 y 2 = 450 y = 3 50 » 21.21 x = 900 = 900 » 424.43 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, 200 y - 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 value 3 50 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

P ( x) = 2400 + 20 x - x 2

Find the first derivative:

P( x) = 2400 + 20 x - x 2 P '( x) = 20 - 2 x 0 = 20 - 2 x 2 x = 20 x = 10

Step 3: Use the second-derivative test to determine whether the critical value produces a maximum value.

P '( x) = 20 - 2 x 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? subject to x £ 100

Step 1:

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

Step 2:

P( x) = (400 - 4 x)(100 + 2 x) - 12(400 - 4 x) P( x) = 35, 200 + 448 x - 8 x 2

Find the first derivative:

P( x) = 35, 200 + 448 x - 8 x 2 P '( x) = 448 - 16 x 0 = 448 - 16 x 16 x = 448 x = 28

Step 3: Use the second-derivative test to determine whether the critical value produces a maximum value.

P '( x) = 448 - 16 x 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) + k for some constant k.

Formulas and Properties of Indefinite Integrals For C and k both a constant

x n+1 n¹ -1 + C, n +1 x x òe dx = e + C 1 x¹ 0 ò dx = ln x + C , x òkf ( x)dx = k ò f ( x)dx ò[ f ( x) ± g ( x)]dx = ò f ( x)dx ± òg ( x)dx

1. òx n dx = 2.

3.

4.

5.

In Problems 1–3, find each indefinite integral and check by differentiating.

1. ò12x dx Check: y = 6 x 2 + C dy = 12 x + 0 dx dy = 12 x dx

y = ò12 x dx y=

12 x 2 +C 2

y = 6 x2 + C

5-1 ..

Calculus

Chapter 5 1

2. ò9x 2 dx 3

1

y = ò9 x 2 dx

Check: y = 6 x 2 + C 1

3

y=

9x 2 3

dy 3 6 x 2 = × +0 dx 2 1 1 dy = 9x 2 dx

+C

2 3

2 9x 2 y= × +C 3 1 3

y = 6x 2 + C

3. ò2e x dx Check: y = 2e x + C dy = 2e x + 0 dx dy = 2e x dx

y = ò2e x dx y = 2e x + C

4. Find all the antiderivatives for

dy = 3z - 1 + 7. dz

dy = 3z - 1 + 7 dz dy = (3z - 1 + 7)dz dy = ( 3z + 7)dz 3 òdy = ò( z + 7) dz

y = 3ln z + 7 z + C

5-2 ..

Calculus

Chapter 5

In Problems 5–8, find each indefinite integral.

5. òx3 ( x 2 + 2 x - 8) dx 3 2 5 4 3 òx ( x + 2 x - 8) dx = ò( x + 2 x - 8 x ) dx

æ3 6. òç 2 èx

æ3 òç 2 èx

=

x 6 2 x5 8 x 4 + +C 6 5 4

=

x 6 2 x5 + - 2 x4 + C 6 5

ö x5 ÷ dx ø 5 ö x5 ÷ dx = ò 3x - 2 - x 2 dx ø

(

)

7

3x - 1 x 2 = - 7 +C -1 2 =-

3 2 x7 +C x 7

æ5 x + 3 ö 7. òç 3 ÷ dx è x ø 3ö æ5 x + 3 ö æ5 òç 3 ÷ dx = òç 2 + 3 ÷ dx è x ø èx x ø

(

)

= ò 5 x - 2 + 3x - 3 dx 5 x - 1 3x - 2 = + +C -1 -2 5 3 =- +C x 2 x2

5-3 ..

Calculus

Chapter 5

æ5 x5 + 3x6e x ö 8. òç ÷ dx x6 è ø

æ5 x5 + 3x6e x ö æ5 xö dx òç ÷ dx = òçè x + 3e ÷ 6 ø x è ø = 5ln x + 3e x + C

In Problems 9–12, find the particular antiderivative of each derivative that satisfies the given conditions.

9. R '( x) = 8 x3 + 9 x 2 + 2; R(1) = 16

First find the indefinite integral of the function.

R( x) = ò(8 x3 + 9 x 2 + 2) dx R( x) =

8 x 4 9 x3 2 x1 + + +C 4 3 1

R( x) = 2 x 4 + 3x3 + 2 x + C Using the condition given, find the specific value of C.

R ( x) = 2 x 4 + 3 x3 + 2 x + C 16 = 2(1) 4 + 3(1)3 + 2(1) + C 16 = 2 + 3 + 2 + C 16 = 7 + C 9=C Substituting the specific value of C yields the particular equation R( x) = 2 x 4 + 3x3 + 2 x + 9.

5-4 ..

Calculus

10.

Chapter 5

dy = 5et + 4t - 8; dt

y (0) = 8

First find the indefinite integral of the function.

dy = (5et + 4t - 8)dt t òdy = ò(5e + 4t - 8) dt

y = 5et +

4t 2 8t1 +C 2 1

y = 5et + 2t 2 - 8t + C Using the condition given, find the specific value of C.

y = 5et + 2t 2 - 8t + C 8 = 5e0 + 2(0) 2 - 8(0) + C 8 = 5+0+0+C 8 = 5+C 3=C Substituting the specific value of C yields the particular equation y = 5et + 2t 2 - 8t + 3.

5-5 ..

Calculus

11.

Chapter 5

dD 5 x 2 - 9 = ; dx x2

D(9) = 50

First find the indefinite integral of the function.

æ5 x 2 - 9 ö dD = ç ÷dx è x2 ø dD = (5 - 9 x - 2 )dx -2 òdD = ò(5 - 9 x ) dx

5 x1 9 x - 1 +C 1 -1 9 D( x) = 5 x + + C x

D( x) =

Using the condition given, find the specific value of C.

9 D( x) = 5 x + + C x 9 50 = 5(9) + + C 9 50 = 45 + 1 + C 50 = 46 + C 4=C

9 Substituting the specific value of C yields the particular equation D( x) = 5 x + + 4. x

5-6 ..

Calculus

Chapter 5

12. h '( x) = 6 x - 1 + 7 x - 2 ; h(1) = 3 First find the indefinite integral of the function.

h '( x) = 6 x - 1 + 7 x - 2 h( x) = ò( 6x + 7 x - 2 ) dx 7 x- 1 +C -1 7 h( x) = 6 ln x + + C x

h( x) = 6 ln x -

Using the condition given, find the specific value of C.

7 h( x) = 6 ln x + + C x 7 3 = 6 ln 1 + + C 1 3 = 6(0) + 7 + C 3 = 7+C -4=C Substituting the specific value of C yields the particular equation 7 h( x) = 6ln x + - 4. x

5-7 ..

Calculus

Chapter 5

5-8 ..

Calculus

Chapter 5

Name ________________________________ Date ______________ Class ____________

Section 5-2 Integration by Substitution Goal: To find the indefinite integrals using general indefinite integral formulas

Formulas: General Indefinite Integral Formulas

[ f ( x)]n+1 1. ò[ f ( x)] f '( x)dx = + C, n +1 2. òe f ( x) f '( x)dx = e f ( x) + C 1 f '( x)dx = ln f ( x) + C 3. ò f ( x) n

u n+1 4. òu du = + C, n +1 5. òeu du = eu + C 1 6. ò du = ln u + C u n

Definition:

n¹ -1

n¹ -1

Differentials

If y = f ( x) defines a differentiable function, then 1. The differential dx of the independent variable x is an arbitrary real number.

2. The differential dy of the dependent variable y is defined as the product of dy = f '( x)dx f '( x) and dx: Procedure:

Integration by Substitution

1. Select a substitution that appears to simplify the integrand. In particular, try to select u so the du is a factor in the integrand.

2. Express the integrand entirely in terms of u and du, completely eliminating the original variable and its differential.

3. Evaluate the new integral if possible.

4. Express the antiderivative found in step 3 in terms of the original variable.

5-9 ..

Calculus

Chapter 5

In Problems 1–8, find each indefinite integral and check the result by differentiating.

1. ò(3x 2 + 7 x - 2)3 (6 x + 7) dx Let u = 3x 2 + 7 x - 2, therefore du = (6 x + 7)dx. Rewrite the original integral in terms of the variable u and solve.

y = ò(3x 2 + 7 x - 2)3 (6 x + 7) dx y = òu 3du u4 +C 4 1 y = (3x 2 + 7 x - 2)4 + C 4

y=

Check:

1 y = (3x 2 + 7 x - 2)4 + C 4 dy 1 d = (4)(3x 2 + 7 x - 2)3 (3x 2 + 7 x - 2) dx 4 dx dy = (3x 2 + 7 x - 2)3 (6 x + 7) dx 2. ò5 (3x 4 + 8 x - 2)(12 x3 + 8) dx Let u = 3x 4 + 8 x - 2, therefore du = (12 x3 + 8)dx. Rewrite the original integral in terms of the variable u and solve.

y = ò5 (3x 4 + 8 x - 2)(12 x3 + 8) dx 1

y = ò(3x 4 + 8 x - 2) 5 (12 x3 + 8) dx 1

y = òu 5 du 6

y=

u 5 6 5

+C

6 5 y = (3x 4 + 8 x - 2) 5 + C 6 5 y = 5 (3x 4 + 8 x - 2)6 + C 6

5-10 ..

Calculus

Chapter 5

Check: 5 y = 5 (3x 4 + 8 x - 2)6 + C 6 6 5 y = (3x 4 + 8 x - 2) 5 + C 6 1 d dy 5 æ6 ö 4 = ç ÷(3x + 8 x - 2) 5 (3x 4 + 8 x - 2) dx 6 è 5 ø dx 1 dy = (3x 4 + 8 x - 2) 3 (12 x3 + 8) dx dy 5 = (3x 4 + 8 x - 2)(12 x3 + 8) dx

3. ò

t3 + 2 2t 4 + 16t - 1

dt

Let u = 2t 4 + 16t - 1, therefore du = (8t 3 + 16)dt Þ du = 8(t 3 + 2)dt. Rewrite the original integral in terms of the variable u and solve.

1 8(t 3 + 2) y= ò 4 dt 8 2t + 16t - 1 1 1 y = ò du 8 u 1 y = ln u + C 8 ln 2t 4 + 16t - 1 y=

+C

8

Check:

ln 2t 4 + 16t - 1 y=

+C

8 1

dy d = (2t 4 + 16t - 1) dt 8(2t 4 + 16t - 1) dx dy 1 = (8t 3 + 16) 4 dt 8(2t + 16t - 1) dy t3 + 2 = 4 dt 2t + 16t - 1 5-11 ..

Calculus

Chapter 5

4. òe- 0.9x dx Let u = - 0.9 x, therefore du = - 0.9dx. Rewrite the original integral in terms of the variable u and solve.

y = òe- 0.9 x dx 1 - 0.9 x y=(- 0.9dx) òe 0.9 1 u y=òe du 0.1 1 u y=e +C 0.9 10 - 0.9 x y=e +C 9 Check:

y = - 10 e - 0.9 x + C 9 dy 10 - 0.9 x d =e (- 0.9 x) dx 9 dx dy = - 10 e - 0.9 x (- 0.9) 9 dx dy = e - 0.9 x dx

5. òx( x + 7)7 dx Let u = x + 7, therefore du = dx. If we rewrite the original integral in terms of the variable u, the substitution would not be complete. We also need u - 7 = x. Now rewrite the original integral in terms of u and solve.

y = òx( x + 7)7 dx y = ò(u - 7)u 7 du y = ò(u8 - 7u 7 ) du u 9 7u8 y= +C 9 8 ( x + 7)9 7( x + 7)8 y= +C 9 8 5-12 ..

Calculus

Chapter 5

Check: ( x + 7)9 7( x + 7)8 y= +C 9 8 dy 1 d 7 d = (9)( x + 7)8 ( x + 7) - (8)( x + 7)7 ( x + 7) dt 9 dx 8 dx dy = ( x + 7)8 - 7( x + 7)7 dt dy = ( x + 7)7 ( x + 7 - 7) dt dy = x( x + 7)7 dt 6. ò

2(ln(3x 2 ))4 dx x 6x

Let u = ln(3x 2 ), therefore du =

3x

dx Þ du = 2

terms of the variable u and solve.

2(ln(3x 2 )) 4 y=ò dx x y = òu 4 du y=

u5 +C 5 5

ln 3 x ) ( y= +C 2

5

Check:

(

ln 3 x 2

y=

5

) +C

5 dy 1 d = (5)(ln 3 x 2 ) 4 (ln 3 x 2 ) dt 5 dx dy 6x = (ln 3 x 2 ) 4 2 dt 3x 2 4 dy 2(ln 3 x ) = dt x

5-13 ..

2 dx. Rewrite the original integral in x

Calculus

Chapter 5

1 - 15 7. ò 6 e x dx x Let u = - x - 5 , therefore du = 5 x - 6dx Þ du = 56 dx. Rewrite the original integral in x

terms of the variable u and solve.

1 - 15 y = ò 6 e x dx x 1 - 15 5 y = òe x × 6 dx 5 x 1 y = òeu du 5 1 u y = e +C 5 1 - 15 y = e x +C 5 Check:

1 - 15 y = e x +C 5 dy 1 - 1 x5 d = e (- x - 5 ) dt 5 dx 1 dy 1 - x5 = e (5 x - 6 ) dt 5 - 1 5

dy e x = 6 dt x

5-14 ..

Calculus

8.

Chapter 5

dy = 12 x 2 (2 x3 + 7)5 Þ dy = 12 x 2 (2 x3 + 7)5 dx dx

Let u = 2 x3 + 7, therefore du = 6 x 2 dx. Rewrite the original integral in terms of the variable u and solve.

y = ò12 x 2 (2 x3 + 7)5 dx y = ò2(2 x3 + 7)5 6 x 2 dx y = 2 òu 5 du 2u 6 y= +C 6 1 y = (2 x3 + 7)6 + C 3 Check:

1 y = (2 x3 + 7)6 + C 3 dy 1 d = (6)(2 x3 + 7)5 (2 x3 + 7) dt 3 dx dy = 2(2 x3 + 7)5 (6 x 2 ) dt dy = 12 x 2 (2 x3 + 7)5 dt

9. The indefinite integral can be found in more than one way. Given the integral, 2 2 ò2 x( x + 3) dx, first use the substitution method to find the indefinite integral and then find it without using substitution.

Using substitution, let u = x 2 + 3, therefore du = 2 x dx. Rewrite the original integral in terms of the variable u and solve.

y = ò2 x( x 2 + 3) 2 dx y = òu 2 du u3 y= +C 3 1 y = ( x 2 + 3)3 + C 3 5-15 ..

Calculus

Chapter 5

Expanding the final function yields the following:

1 y = ( x 2 + 3)3 + C 3 1 y = ( x 2 + 3)( x 2 + 3)( x 2 + 3) + C 3 1 6 y = ( x + 9 x 4 + 27 x 2 + 27) + C 3 y=

x6 + 3x 4 + 9 x 2 + C 3

Note: The “new” value of C above is the addition of the constant 9 and the original constant C from the integration.

Without using substitution to find the given indefinite integral, you first would multiply the integrand as follows

2 x( x 2 + 3) 2 = 2 x( x 2 + 3)( x 2 + 3) = 2 x( x 4 + 6 x 2 + 9) = 2 x5 + 12 x3 + 18 x Now find the indefinite integral using the above function as the integrand.

y = ò(2 x5 + 12 x3 + 18 x) dx y=

2 x 6 12 x 4 18 x 2 + + +C 6 4 2

y=

x6 + 3x 4 + 9 x 2 + C 3

5-16 ..

Calculus

Chapter 5

Name ________________________________ Date ______________ Class ____________

Section 5-3 Differential Equations; Growth and Decay Goal: To solve differential equations that involve growth and decay.

Theorem 1:

Exponential Growth Law

dQ = rQ and Q(0) = Q0 , then Q = Q0e rt , dt where Q0 = amount of Q at t = 0 r = relative growth rate (expressed as a decimal) t = time Q = quantity at time t If

Table 1:

Exponential Growth

Description Unlimited growth

Model

Solution

dy = ky dt

y = cekt

k, t > 0 y(0) = c Exponential decay

y = ce- kt

dy = - ky dt

k, t > 0 y(0) = c Limited growth

dy = k (M - y) dt

y = M (1 - e- kt )

k, t > 0 y(0) = 0 Logistic Growth

dy = ky ( M - y ) dt

k, t > 0

y(0) = 1M +c

5-17 ..

y=

M 1 + ce - kMt

Calculus

Chapter 5

In Problems 1–4, find the general or particular solution, as indicated, for each differential equation.

1.

dy = 8x- 5 dx

The differential equation can be found by using the integration properties from Section 6.1.

y = ò8 x - 5 dx y=

8x- 4 +C -4

y = - 2 x- 4 + C

2.

5 dy = - 10 x 4e- 2 x ; dx

y(0) = 9

The differential equation can be found by using the integration properties from Section 6.2. Let u = - 2 x5 , therefore du = - 10 x4dx. Rewrite the original integral in terms of the variable u and solve.

5

y = ò- 10 x 4e- 2 x dx y = òeu du y = eu + C 5

y = e- 2 x + C Using the condition given, find the specific value of C. 5

y = e- 2 x + C 5

9 = e- 2(0) + C 9 = e0 + C 9 = 1+ C 8=C 5

Substituting the specific value of C yields the particular equation y = e- 2 x + 8.

5-18 ..

Calculus

3.

Chapter 5

dy = - 8y dx

The differential equation is in the form of exponential decay; therefore, using the exponential decay model in Table 1 yield’s the following solution:

y = ce- 8x

4.

dx = 3x; dt

x(0) = 6

The differential equation is in the form of unlimited growth; therefore, using the unlimited growth model in Table 1 yield’s the following solution:

x = ce3t Using the condition given, find the specific value of C.

x = ce3t 6 = ce3(0) 6 = ce0 6=c Substituting the specific value of C yields the particular solution x = 6e3t .

5. Find the amount A in an account after t years if

dA = 0.02 A dt

and

A(0) = 8000

The exponential growth law (unlimited growth) applies to the situation. Since we are given the initial amount, the amount in an account after t years would be:

A = 8000e0.02t

5-19 ..

Calculus

Chapter 5

6. A single injection o a drug is administered to a patient. The amount Q in the body decreases at a rate proportional to the amount present. For a particular drug, the rate dQ is 6% per hour. Thus, = - 0.06Q and Q(0) = Q0 where t is time in hours. dt a. If the initial injection is 5 milliliters [ Q(0) = 5 ], find Q = Q(t ) satisfying both conditions. b. How many milliliters (to two decimal places) are in the body after 8 hours? c. How many hours (to two decimal places) will it take for half the drug to be left in the body?

a. By the exponential growth law (exponential decay) for the given conditions, we have the equation Q(t ) = 5e- 0.06t . b. After 8 hours, t = 8, therefore

Q(8) = 5e- 0.06(8) Q(8) = 5e- 0.48 Q(8) = 5(0.61878) Q(8) » 3.09 Therefore, after 8 hours, approximately 3.09 milliliters will remain in the body. c.

Half the initial injection is 2.5 milliliters. We need to find t such that Q(t ) = 2.5.

Q(t ) = 5e - 0.06t 2.5 = 5e- 0.06t 0.5 = e - 0.06t ln 0.5 = ln e - 0.06t ln 0.5 = - 0.06t 11.55 » t Therefore, it will take approximately 11.55 hours for the amount of the drug to be half of the initial amount.

5-20 ..

Calculus

Chapter 5

7. A company is trying to expose a new product to as many people as possible through radio ads. Suppose that the rate of exposure to new people is proportional to the number of those who have not heard of the product out of L possible listeners. No one is aware of the product at the start of the campaign, and after 15 days, 50% of L dN are aware of the product. Mathematically, = k ( L - N ), N (0) = 0, and dt N (15) = 0.5L. a. Solve the differential equation. b. What percent of L will have been exposed after 7 days of the campaign? c. How many days will it take to expose 75% of L? a. By the exponential growth law (limited growth) for the given conditions, we have the equation

N (t ) = L(1 - e - kt ) 0.5L = L(1 - e - 15k ) 0.5 = 1 - e- 15k - 0.5 = - e- 15k 0.5 = e- 15k ln 0.5 = ln e- 15k ln 0.5 = - 15k 0.046 = k Therefore, the solution is N (t ) = L(1 - e- 0.046t ). b. After 7 days, t = 7, therefore

N (7) = L(1 - e- 0.046(7) ) N (7) = L(1 - 0.724698) N (7) = L(0.275302) N (7) » 0.275L Therefore, after 7 days, approximately 27.5% of L will have been exposed to the product.

5-21 ..

Calculus c.

Chapter 5 Find t such that N (t ) = 0.75L.

N (t ) = L(1 - e - 0.046t ) 0.75 L = L(1 - e - 0.046t ) 0.75 = 1 - e - 0.046t - 0.25 = - e- 0.046t 0.25 = e- 0.046t ln 0.25 = ln e - 0.046t ln 0.25 = - 0.046t 30.1 » t Therefore, it will take approximately 30 days to expose 75% of L to the new product.

5-22 ..

Calculus

Chapter 5

Name ________________________________ Date ______________ Class ____________

Section 5-4 The Definite Integral Goal: To calculate the values of definite integrals using the properties.

Theorem:

Limits of Left and Right Sums

If f ( x) > 0 and is either increasing or decreasing on [a, b], then its left and right sums approach the same real number as n ® ¥ .

Theorem:

Limit of Riemann Sums

If f is a continuous function on [a, b], then the Riemann sums for f on [a, b] approach a real number limit I as n ® ¥ .

Definition:

Definite Integral

Let f be a continuous function on [a, b]. The limit I of Riemann sums for f on [a, b] is called the definite integral of f from a to b and is denoted as òab f ( x)dx.

Properties:

Definite Integrals

1. òaa f ( x)dx = 0 2. òab f ( x)dx = - òba f ( x)dx 3. òab kf ( x)dx = k òab f ( x)dx, k a constant 4. òab [ f ( x) ± g ( x)]dx = òab f ( x)dx ± òab g ( x)dx 5. òab f ( x)dx = òac f ( x)dx + òcb f ( x)dx

5-23 ..

Calculus

Chapter 5

In Problems 1 and 2, calculate the indicated Riemann sum Sn for the function f ( x) = 13 - x3.

1. Partition [–1, 9] into five subintervals of equal length, and for each subinterval [ xk - 1, xk ], let ck = ( xk - 1 + xk ) / 2.

x =

c1 =

9 - (- 1) 10 = =2 5 5

- 1 +1 =0 2

c2 =

c3 =

f (2) = 13 - (2)3 f (2) = 5

f (0) = 13 - (0)3 f (0) = 13

5+7 =6 2

c4 =

1+ 3 =2 2

c5 =

3+5 =4 2

f (4) = 13 - (4)3 f (4) = - 51

7+9 =8 2

f (8) = 13 - (8)3 f (8) = - 499

f (6) = 13 - (6)3 f (6) = - 203

S5 = f (c1 ) ×Dx + f (c2 ) ×Dx + f (c3 ) ×Dx + f (c4 ) ×Dx + f (c5 ) ×Dx S5 = (13)(2) + (5)(2) + (- 51)(2) + (- 203)(2) + (- 499)(2) S5 = 26 + 10 - 102 - 406 - 998 S5 = - 1470 2. Partition [–4, 8] into four subintervals of equal length, and for each subinterval [ xk - 1, xk ], let ck = (2 xk - 1 + xk ) / 3.

x =

c1 =

8 - (- 4) 12 = =3 4 4

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

c2 =

2(- 1) + 2 =0 3

f (0) = 17 - 2(0) 2 f (0) = 17

f (- 3) = 17 - 2(- 3)2 f (- 3) = - 1

5-24 ..

Calculus

Chapter 5

c3 =

2(2) + 5 =3 3

c4 =

2(5) + 8 =6 3

f (6) = 17 - 2(6) 2 f (6) = - 55

f (3) = 17 - 2(3)2 f (3) = - 1

S4 = f (c1 ) ×x + f (c2 ) ×x + f (c3 ) ×x + f (c4 ) ×x S4 = (- 1)(3) + (17)(3) + ( - 1)(3) + ( - 55)(3) S4 = - 3 + 51 - 3 - 165 S4 = - 120

In Problems 3–7, calculate the definite integral, given that 4

4 2 ò0 x dx =

ò0 x dx = 8

64 3

7 2 ò4 x dx =

3. ò04 3x dx 4

4

ò0 3x dx = 3ò0 x dx = 3(8) 4

ò0 3x dx = 24

4. ò04 (2 x + x 2 ) dx 4 4 4 2 2 ò0 (2 x + x ) dx = 2ò0 x dx + ò0 x dx 64 = 2(8) + 3 64 = 16 + 3 112 4 2 ò0 (2 x + x ) dx = 3

5-25 ..

279 3

Calculus

Chapter 5

5. ò07 2x 2 dx 7 2 7 2 ò0 2 x dx = 2 ò0 x dx

(

= 2 ò04 x 2 dx + ò47 x 2 dx

)

æ64 279 ö = 2ç + ÷ è3 3 ø æ343 ö = 2ç è 3 ÷ ø 686 7 2 ò0 2 x dx = 3

6. ò40 7x dx 0

4

ò4 7 x dx = - 7 ò0 x dx = - 7(8) 0

ò5 7 x dx = - 56

7. ò77 ( x2 + 2 x - 15) dx Since the upper and lower limits of integration are the same, the value of the integral would be ò77 ( x2 + 2 x - 15) dx = 0.

5-26 ..

Calculus

Chapter 5

Name ________________________________ Date ______________ Class ____________

Section 5-5 The Fundamental Theorem of Calculus Goal: To use the fundamental theorem of calculus to solve problems.

Theorem 1:

Fundamental Theorem of Calculus

If f is a continuous function on [a, b], and F is any antiderivative of f, then b

òa f ( x)dx = F (b) - F (a) Definition:

Average Value of a Continuous Function f over [a, b]

1 b ò f ( x)dx b- a a

In Problems 1–7, evaluate the integrals.

1. ò15 (8x3 + 3) dx 5

æ 4

3x ö + ÷ 1ø è 4 1

8x 5 3 ò1 (8 x + 3) dx = ç

(

= 2 x 4 + 3x

5

)1

= (2(5) 4 + 3(5)) - (2(1) 4 + 3(1)) = 1265 - 5 5

3

ò1 (8 x + 3) dx = 1260

5-27 ..

Calculus

Chapter 5

2. ò03 (2e x ) dx 3

x

ò0 (2e ) dx = 2e

x3 0 3

= 2e - 2e0 = 2e3 - 2 3

x

ò0 (2e ) dx » 38.17

3. ò22 (2 x3 + 5x 2 - x + 27) dx Since the upper limit and lower limit of integration are the same value, the definite integral value will be ò22 (2 x3 + 5x 2 - x + 27) dx = 0.

22 4. ò13

22

ò13

5 dx x +3

5 -1 22 dx = ò13 5( x + 3) 2 dx x +3 22

1

=

5( x + 3) 2 1

2

13 1 22

= 10( x + 3) 2

13 1

1

= 10(22 + 3) 2 - 10(13 + 3) 2 = 10(5) - 10(4) 22

ò13

5 dx = 10 x +3

5-28 ..

Calculus

Chapter 5

5. ò01 2 x3 (3x 4 + 2)6 dx We will solve the problem using the substitution method of integration. Let u = 3x + 2, therefore du = 12 x3dx Þ du = 6(2 x3dx). Rewrite the original integral in terms of the variable u and solve.

4

1 3 4 6 ò0 2 x (3x + 2) dx =

1 x=1 6 ò u du 6 x =0

u7 = 42

x =1

x =0

(3x 4 + 2)7 = 42 4

=

1

0 7

(3(1) + 2) (3(0) 4 + 2)7 42 42

1 3 4 6 ò0 2 x (3x + 2) dx » 1857.07

6.

2 2 3x - 2 x + 5 ò1 3 2

dx

x - x + 5x - 3

We will solve the problem using the substitution method of integration. Let u = x3 - x 2 + 5 x - 3, therefore du = (3x 2 - 2 x + 5) dx. Rewrite the original integral in terms of the variable u and solve.

2 2 3x - 2 x + 5 ò1 3 2

x - x + 5x - 3

dx = òxx==12

1 du u

= ln u

x =2 x =1

= ln x3 - x 2 + 5 x - 3

2 1

3

2

= ln (2) - (2) + 5(2) - 3 - ln (1)3 - (1) 2 + 5(1) - 3 = ln 11 - ln 2 2 2 3x - 2 x + 5 ò1 3 2

dx » 1.70

x - x + 5x - 3

5-29 ..

Calculus

Chapter 5

7. ò12 3x 6 x 2 + 1 dx We will solve the problem using the substitution method of integration. Let u = 6 x 2 - 1, therefore du = 12 xdx Þ du = 4(3x)dx. Rewrite the original integral in terms of the variable u and solve.

2 2 ò1 3x 6 x + 1 dx =

1 x =2 12 ò u du 4 x=1 x =2

1 2 3 = × u 2 4 3 x =1

2

3 1 = (6 x 2 + 1) 2 6 1

3 3 1 1 = (6(2)2 + 1) 2 - (6(1) 2 + 1) 2 6 6 125 343 = 6 6

2 2 ò1 3x 6 x + 1 dx » 17.75

In Problems 8 and 9, find the average value of the function over the given interval.

8.

f ( x) = 8 x3 - 16 x + 4; [1, 4] 1 b 1 4 3 f ( x) dx = (8 x - 16 x + 4) dx ò b- a a 4 - 1 ò1 4 1 = 2 x4 - 8x2 + 4 x 1 3 1 = ((2(4) 4 - 8(4) 2 + 4(4)) - (2(1) 4 - 8(1) 2 + 4(1)) ) 3 1 = (400 - (- 2)) 3 = 134

(

)

5-30 ..

Calculus

Chapter 5

9. g ( x) = 2e- 0.3x

[0,10]

We will solve the problem using the substitution method of integration. Let u = - 0.3x, therefore du = - 0.3dx. Rewrite the original integral in terms of the variable u and solve

1 b 1 10 - 0.3 x ) dx òa f ( x) dx = ò (2e b- a 10 - 0 0 2 x =10 u (e ) du ò (10)(- 0.3) x =0 =-

2 æ u x =10 ö e ø 3 çè x =0 ÷

2 æ - 0.3 x 10 ö e ÷ 0 ø 3 çè 2 = - e- 0.3(10) - e- 0.3(0) 3 2 = - e- 3 - 1 3 » 0.63 -

( (

)

)

10. The total cost (in dollars) of manufacturing x units of a product is C ( x) = 30, 000 + 250 x. a. Find the average cost per unit if 400 units are produced. [Hint: Recall that C ( x) is the average cost per unit.]

C ( x) x 30, 000 + 250 x C ( x) = x 30, 000 C ( x) = + 250 x

C ( x) =

30, 000 + 250 400 C (400) = 75 + 250

C (400) =

C (400) = 325

Therefore, the average cost for 400 units is $325 per unit.

5-31 ..

Calculus

Chapter 5

b. Find the average value of the cost over the interval [0, 400].

1 b 1 8 òa f ( x) dx = ò2 (30, 000 + 250 x) dx b- a 400 - 0 400 ö 1 æ = 30, 000 x + 125 x 2 ç ÷ 0 ø 400 è 1 ((30, 000)(400) + 125(400)2 ) - ((30, 000)(0) + 125(0)2 ) 400 1 = (32, 000, 000 - 0) 400 = 80, 000

(

=

)

Therefore, the average value of the cost will be $80,000.

11. A company manufactures a product and the research department produced the marginal cost function

C '( x) = 300 -

x 4

0 £ x £ 800

where C '( x) is in dollars and x is the number of units produced per month. Compute the increase in cost going from a production level of 400 units per month to 800 units per month. Set up a definite integral and evaluate it.

b

800

òa C '( x)dx = ò400 (300 -

x ) dx 4 800

æ x2 ö = ç300 x 8 ÷ è ø 400 æ (800)2 ö æ (400)2 ö = ç300(800) - ç300(400) 8 ÷ 8 ÷ è ø è ø = 160, 000 - 100, 000 b

òa C '( x)dx = 60, 000 Therefore, the increase in cost going from 400 units per month to 800 units per month is $60,000.

5-32 ..

Calculus

Chapter 6

Name ________________________________ Date ______________ Class ____________

Section 6-1 Area Between Curves Goal: To find the area between two curves

Theorem:

Area Between Two Curves

If f and g are continuous and f ( x) ³ g ( x) over the interval [a, b], then the area bounded by y = f ( x) and y = g ( x) for a £ x £ b is given exactly by

A = òab [ f ( x) - g ( x)] dx Definition:

Gini Index of Income Concentration

If y = f ( x) is the equation of a Lorenz curve, then

Gini index = 2ò01[ x - f ( x)] dx

In Problems 1–7, find the area bounded by the graphs of the indicated equations over the given interval (when stated). Compute answers to three decimals, if necessary.

1. y = 4 x3 + 5; y = 0; 1 £ x £ 5

A = ò15 [4 x3 + 5 - 0] dx A = ò15 [4 x3 + 5] dx A = x4 + 5x

(

5 1

) (

A = (5)4 + 5(5) - (1)4 + 5(1)

)

A = (625 + 25) - (1 + 5) A = 644

6–1 ..

Calculus

Chapter 6

2. y = - 2 x3 + 9; y = 0; - 3 £ x £ 2

A = ò-23[- 2 x3 + 9 - 0] dx A = ò-23[- 2 x3 + 9] dx x4 A=+ 9x 2

2

-3

æ (2)4 ö æ (- 3) 4 ö A = ç+ 9(2) ÷- ç+ 9(- 3) ÷ 2 è 2 ø è ø A = (- 8 + 18) - (-

81 - 27) 2

A = 10 - (- 67.5) A = 77.5

3. y = 5e x - 3; y = 0; 0 £ x £ 2

A = ò02 [5e x - 3 - 0] dx A = ò02 [5e x - 3] dx A = 5e x - 3 x

(

2

2 0

) (

A = 5e - 3(2) - 5e0 - 3(0)

)

A = (36.9452805 - 6) - (5) A » 25.945

6–2 ..

Calculus

Chapter 6

4. y = 3x + 7; y = - 4; - 7 £ x £ 0

A = ò-07 [3x + 7 - (- 4)] dx A = ò-07 [3x + 11] dx 3x 2 A= + 11x 2

0

-7

æ3(0) ö æ3(- 7) 2 ö A=ç + 11(0) ÷- ç + 11(- 7) ÷ è 2 ø è 2 ø 2

147 - 77) 2 A = 0 - (- 3.5) A = 3.5

A = (0 - 0) - (

5. y = x 2 - 3; y = 6 The first step in the problem is to find where the two functions intersect. Set both equations equal to each other to find the intersection points.

x2 - 3 = 6 x2 = 9 x = ±3 A graph of y = x 2 - 3 and y = 6 shows that 6 ³ x 2 - 3 on the interval [–3, 3]. Now find the area using the definite integral from –3 to 3.

A = ò-33[6 - ( x 2 - 3)] dx A = ò-33[- x 2 + 9] dx x3 A=+ 9x 3

3

-3

æ (3) ö æ (- 3)3 ö A = ç+ 9(3) ÷- ç+ 9(- 3) ÷ 3 è 3 ø è ø 3

A = (- 9 + 27) - (9 - 27) A = 18 - (- 18) A = 36 6–3 ..

Calculus

Chapter 6

6. y = x 2 - 2; y = x - 5; - 3 £ x £ 2

A = ò-23[( x 2 - 2) - ( x - 5)] dx A = ò-23[ x 2 - x + 3] dx x3 x 2 A= + 3x 3 2

2

-3

æ(2)3 (2)2 ö æ(- 3)3 (- 3) 2 ö A=ç + 3(2) ÷- ç + 3(- 3) ÷ 2 2 è 3 ø è 3 ø 9 æ8 ö æ ö A = ç - 2 + 6 ÷- ç- 9 - - 9 ÷ è3 ø è 2 ø A = 6.6666667 - (- 22.5) A » 29.167

7. y = 2 x + 3; y = 2 x - 4; - 5 £ x £ 3

A = ò-35 [(2 x + 3) - (2 x - 4)] dx A = ò-35 [7] dx 3

A = 7x - 5 A = 7(3) - 7(- 5) A = 21 + 35 A = 56

6–4 ..

Calculus

Chapter 6

In Problems 8 and 9, use a graphing calculator to graph the equations and find relevant intersection points. Then find the area bounded by the curves. Compute answers to three decimal places.

8. y = x3 ; y = 9 x

Based on the graph above, the points of intersection are –3, 0, and 3, therefore the area between the curves is

A = ò-03[( x3 ) - (9 x)] dx + ò03[(9 x) - ( x3 )] dx A = ò-03[ x3 - 9 x] dx + ò03[9 x - x3 ] dx 0 ö æ 2 3ö æ 4 x 9 x2 9x x4 ÷+ ç ÷ A=ç 2 2 4 ÷ çè 4 ÷ ç -3ø è 0ø

æ(0) 4 9(0) 2 ö æ(- 3) 4 9( - 3) 2 ö æ(0) 4 9(0) 2 ö æ(3) 4 9(3) 2 ö A=ç - ç +ç - ç 2 ÷ 2 ÷ 2 ÷ 2 ÷ è 4 ø è 4 ø è 4 ø è 4 ø

æ81 81 ö æ81 81 ö A = (0 - 0) - ç + (0 - 0) - ç ÷ è4 2ø è4 2÷ ø A = 0 - (- 20.25) + 0 - (- 20.25) A = 40.5

6–5 ..

Calculus

Chapter 6

9. y = x 2 + 2 x - 3; y = - x 2 - 3x + 4;

Based on the graph above, the points of intersection are –3.5 and 1, therefore the area between the curves is

A = ò-13.5 [(- x 2 - 3x + 4) - ( x 2 + 2 x - 3)] dx A = ò-13.5 [- 2 x 2 - 5 x + 7] dx - 2 x3 5 x 2 A= + 7x 3 2

1

- 3.5

æ- 2(1) 5(1) ö æ- 2(- 3.5)3 5( - 3.5) 2 ö A=ç + 7(1) ÷- ç + 7(- 3.5) ÷ 2 3 2 è 3 ø è ø 3

2

æ- 2 ö æ85.75 61.25 ö A = ç - 2.5 + 7 ÷- ç - 24.5 ÷ è3 ø è 3 ø 2 A = 3.8333333 - (- 26.5416667) A = 30.375

6–6 ..

Calculus

Chapter 6

Name ________________________________ Date ______________ Class ____________

Section 6-2 Applications in Business and Economics Goal: To solve application problems that involve business and economics

Definitions: 1. Total Income for a Continuous Income Stream If f (t ) is the rate of flow of a continuous income stream, then the total income produced during the period from t = a to t = b is

Total income = òab f (t ) dt 2. Future Value of a Continuous Income Stream If f (t ) is the rate of flow of a continuous stream, 0 £ t £ T , and if the income is continuously invested at a rate r, compounded continuously, then the future value FV at the end of T years is given by

FV = ò0T f (t )er (T - t ) dt = erT ò0T f (t )e- rt dt 3. Consumers’ Surplus If ( x , p) is a point on the graph of the price–demand equation p = D( x) for a particular product, then the consumers’ surplus CS at a price level of p is

CS = ò0x [ D( x) - p] dx 4. Producers’ Surplus If ( x , p) is a point on the graph of the price–supply equation p = S ( x), then the producers’ surplus PS at a price level of p is PS = ò0x [ p - S ( x)] dx

6–7 ..

Calculus

Chapter 6

In Problems 1–3, evaluate each definite integral to two decimals.

1. ò04 e- 5t dt

1 - 5t 4 - 5t ò0 e dt = - e 5

4 0

æ 1 ö æ 1 ö = ç- e- 5(4) ÷- ç- e- 5(0) ÷ è 5 ø è 5 ø = 0.20 2. ò07 e0.03(3- 2t ) dt 7 0.03(3- 2)

ò0 e

dt = e0.03(3) ò07 e- 0.06t dt 7ö æ 1 = e0.09 çe- 0.06t ÷ è 0.06 0ø

1 - 0.06(0) ö æ 1 - 0.06(7) = e0.09 çe + e ÷ è 0.06 ø 0.06 » e0.09 (- 10.95078 + 16.66667 ) » e0.09 (5.71589) » 6.25 3. ò040 600e0.04t e0.07(40- t ) dt 40

ò0 600e

0.04t 0.07(40- t )

e

dt = ò040 600e0.04t e 2.8- 0.07t dt = ò040 600e2.8- 0.03t dt = 600e2.8 ò040 e- 0.03t dt

æ

40 1 - 0.03t ö e ÷ è 0.03 0 ø

40 0.04t 0.07(40- t ) e dt = 600e2.8 çò0 600e

1 - 0.03(0) ö æ 1 - 0.03(40) = 600e2.8 çe + e ÷ è 0.03 ø 0.03 » 600e2.8 (- 10.03981 + 33.33333) » 600e2.8 (23.29352) » 229,832.29 6–8 ..

Calculus

Chapter 6

4. Find the total income produced by a continuous income stream in the first 6 years if the rate of flow if f (t ) = 400e0.05t .

Total income = òab f (t ) dt = ò06 400e0.05t dt 6

400 0.05t = e 0.05 0 400 0.05(6) 400 0.05(0) e e 0.05 0.05 » $2799

=

5. Find the future value, at 1.75% interest, compounded continuously for 7 years, of the continuous income stream with rate of flow f (t ) = 3000e0.05t .

FV = erT ò0T f (t )e - rt dt FV = e(0.0175)(7) ò07 3000e0.05t e - 0.0175t dt FV = 3000e0.1225 ò07 e0.0325t dt FV =

3000e0.1225 æ 0.0325t 7 ö e ÷ 0ø 0.0325 çè

3000e0.1225 0.0325(7) 0.0325(0) (e -e ) 0.0325 FV = $26, 654

FV =

6. Find the consumers’ surplus at a price level of p = $160 for the price–demand equation p = D( x) = 460 - 0.06 x.

First find x by substituting the value p = $160 into the price–demand equation.

p = 460 - 0.06 x 160 = 460 - 0.06 x - 300 = - 0.06 x 5000 = x

6–9 ..

Calculus

Chapter 6

Now use the consumers’ surplus definition.

CS = ò0x [ D( x) - p ] dx CS = ò05000 [460 - 0.06 x - 160] dx CS = ò05000 [300 - 0.06 x] dx

CS = 300 x - 0.03 x 2

5000 0

CS = (300(5000) - 0.03(5000) 2 ) - (300(0) - 0.03(0) 2 ) CS = 1,500, 000 - 750, 000 - 0 - 0 CS = $750, 000

7. Find the producers’ surplus at a price level of p = $98 for the price–supply equation

p = S ( x) = 10 + 0.1x + 0.0003x 2 .

First find x by substituting the value p = $98 into the supply–demand equation.

p = 10 + 0.1x + 0.0003 x 98 = 10 + 0.1x + 0.0003 x

2 2

0 = - 88 + 0.1x + 0.0003 x

2

Using the quadratic equation will yield that x = 400.

Now use the producers’ surplus definition.

PS = ò0x [ p - S ( x)] dx PS = ò0400 [98 - (10 + 0.1x + 0.0003 x 2 )] dx PS = ò0400 [88 - 0.1x - 0.0003 x 2 ] dx

PS = 88 x - 0.05 x 2 - 0.0001x3

400 0

2

PS = (88(400) - 0.05(400) - 0.0001(400)3 ) - (88(0) - 0.05(0)2 - 0.0001(0)3 ) PS = (35, 200 - 8000 - 6400) - (0 - 0 - 0) PS = $20,800

6–10 ..

Calculus

Chapter 6

8. Find the consumers’ surplus and the producers’ surplus at the equilibrium price level for the given price–demand and price–supply equations. Include a graph that shows the equilibrium point ( x, p). Round all values to the nearest integer if necrssary.

p = D( x) = 55 - 0.3x

p = S ( x) = 10 + 0.15 x

The equilibrium point is also the point of intersection. Through a process that narrows the window required to see the point of intersection, the graph will look as follows

Therefore, the equilibrium point is (100, 25).

The consumers’ surplus will be

CS = ò0x [ D( x) - p ] dx CS = ò0100 [55 - 0.3x - 25] dx CS = ò0100 [30 - 0.3x] dx

CS = 30 x - 0.15 x 2

100 0

CS = (30(100) - 0.15(100)2 ) - (30(0) - 0.15(0)2 ) CS = (3000 - 1500) - (0 - 0) CS = $1500

6–11 ..

Calculus

Chapter 6

The producers’ surplus will be

PS = ò0x [ p - S ( x)] dx PS = ò0100 [25 - (10 + 0.15 x)] dx PS = ò0100 [15 - 0.15 x] dx

PS = 15 x - 0.075 x 2

100 0

PS = (15(100) - 0.075(100) 2 ) - (15(0) - 0.075(0) 2 ) PS = (1500 - 750) - (0 - 0) PS = $750

6–12 ..

Calculus

Chapter 6

Name ________________________________ Date ______________ Class ____________

Section 6-3 Integration by Parts Goal: To solve integration problems using the integration by parts formula

Formula:

Integration by Parts òu dv = uv - òv du

Summary:

Integration by Parts: Selection of u and dv

For òu dv = uv - òv du, 1. The product u dv must equal the original integrand.

2. It must be possible to integrate dv (preferably by using standard formulas or simple substitution) 3. The new integral òv du should not be more complicated than the original òu dv.

4. For integrals involving x p eax , try u = x p and

dv = eax dx 5. For integrals involving x p (ln x)q , try u = (ln x)q

and

dv = x p dx

6–13 ..

Calculus

Chapter 6

In Problems 1 and 2, integrate by parts. Assume that x > 0 whenever the natural logarithm function is involved.

1.

7x òxe dx

Let u = x and dv = e7x dx , then du = dx and v = 17 e7 x . Substituting these terms into the integration by parts formula yields

òu dv = uv - òv du

2.

òxe

7x

dx = x( 17 e7 x ) - ò17 e7 x dx

òxe

7x

dx = 7x e7 x - 71 71 e7 x + C

òxe

7x

dx =

(

)

x 7x 1 7x e e +C 7 49

3 òx ln x dx

Let u = ln x and dv = x3dx, then du = 1x dx and v = 14 x 4 . Substituting these terms into the integration by parts formula yields

òu dv = uv - òv du 3 1 4 1 4 1 òx ln x dx = ln x( 4 x ) - ò 4 x ( x ) dx 3 3 1 4 1 òx ln x dx = 4 x ln x - 4 òx dx

( )

3 1 4 1 1 4 òx ln x dx = 4 x ln x - 4 4 x + C 3 òx ln x dx =

1 4 1 4 x ln x x +C 4 16

6–14 ..

Calculus

Chapter 6

Problems 3–8 are mixed – some may require use of the integration-by-parts formula along with techniques learned earlier. Assume that g ( x) > 0 whenever ln g ( x) is involved.

5

ò1 ln 3x dx

3.

Use integration by parts by letting u = ln 3 x and dv = dx, then du = 1x dx and v = x.

Substitute these terms into the integration by parts formula yields 5

5

5

1

ò1 ln 3x dx = (ln 3x)( x) 1 - ò1 x( x ) dx 5

= (ln 3x)( x) 1 - ò151 dx 5

= x ln 3x - x 1

= (5ln 3(5) - (5)) - (1ln 3(1) - (1)) = 8.540251 - 0.098612 5

ò1 ln 3x dx » 8.442

x ò9 xe dx

4.

Use integration by parts by letting u = 9 x and dv = e x dx , then du = 9dx and v = e x .

Substituting these terms into the integration by parts formula yields x x x ò9 xe dx = 9 x(e ) - 9òe dx x x x ò9 xe dx = 9 xe - 9e + C

3

5 x ò3x e dx

5.

3

Use integration by parts by letting u = x3 and dv = 3x 2e x dx , then du = 3x 2 dx and 3

v = e x . Substituting these terms into the integration by parts formula yields 3

3

3

5 x 3 x x 2 ò3x e dx = x (e ) - òe (3x ) dx 3

3

3

5 x 3 x x ò3x e dx = x e - e + C

6–15 ..

Calculus

Chapter 6

4e 4 x + 3

dx ò 4x e + 3x Use substitution by letting u = e 4 x + 3 x and du = 4e4 x + 3. Substitute these terms into the original equation yields 6.

4e4 x + 3 1 dx = ò du ò 4x u e + 3x = ln u + C = ln e4 x + 3x + C

7.

2

ò0 xe

- 2 x2

dx

Use substitution by letting u = - 2 x 2 and du = - 4 x. Substitute these terms into the original equation yields 2

ò0 xe

- 2 x2

dx = -

1 x =2 u ò e du 4 x =0 x =2

=-

1 u e 4 x =0

=-

1 - 2 x2 e 4 0

2

1 - 2(2)2 1 - 2(0)2 e + e 4 4 » - 0.00008387 + 0.25 » 0.2499

=-

8.

8 ò( x + 7) ( x - 3) dx

Use integration by parts by letting u = x - 3 and dv = ( x + 7)8 dx , then du = dx and

v = 19 ( x + 7)9 . Substituting these terms into the integration by parts formula yields 8 9 9 1 1 ò( x + 7) ( x - 3) dx = ( x - 3)( 9 ( x + 7) ) - ò9 ( x + 7) dx

= 19 ( x - 3)( x + 7)9 - 19 ò( x + 7)9 dx

(

)

1 ( x + 7)10 + C = 19 ( x - 3)( x + 7)9 - 19 10 8 9 10 1 1 ò( x + 7) ( x - 3) dx = 9 ( x - 3)( x + 7) - 90 ( x + 7) + C

6–16 ..

Calculus

Chapter 6

Name ________________________________ Date ______________ Class ____________

Section 6-4 Other Integration Methods Goal: To use the Trapezoidal Rule and Simpson’s Rule to approximate definite integrals.

Use integration tables to solve indefinite integrals

Trapezoidal Rule: Let f be a function defined on an interval [a, b]. Partition [a, b] into n subintervals of equal length Dx = (b - a) / n with endpoints a = x0 < x1 < x2 < L < xn = b Then Tn = [ f ( x0 ) + 2 f ( x1) + 2 f ( x2 ) + L + f ( xn )]Dx / 2 b

is an approximation of ò f ( x)dx. a

Simpson’s Rule: Let f be a function defined on an interval [a, b]. Partition [a, b] into 2n subintervals of equal length Dx = (b - a) / n with endpoints a = x0 < x1 < x2 < L < xn = b Then S2n = [ f ( x0 ) + 4 f ( x1) + 2 f ( x2 ) + 4 f ( x1) + L + f ( xn )]Dx / 3 b

is an approximation of ò f ( x)dx. a

Process:

Using the integration tables

1. Determine which category has the form of the given integrand (integrals involving trigonometric functions, integrals involving ln u, etc.)

2. Examine the integrals within the category to see if one represents exactly the integrand in the given problem. a. If an exact match is found, use the formula to find the indefinite integral. b. If an exact match is not found, use substitution to change the integral into one that appears in the table. Use the formula to find the indefinite integral.

Replace the original function after the formula is used.

6–17 ..

Calculus

Chapter 6

Use Table II (Integration Table) to find each integral in Problems 1–5.

x2

1. ò

2

dx

16 x - 25 Examining the form of the integral, we should look at formulas 40–45, which involve

u - a2 . Based on this, we let u 2 = 16 x2 and a 2 = 25 so that 16 x2 - 25 = u 2 - a 2 .

2

1 u 2 , and du = 4dx. Using formula 44 and the substitutions, the Then we have u = 4 x, x 2 = 16

integral will be written 1

x2

2

u 1 dx = ò 16 du ò 4 u 2 - 52 16 x 2 - 25

2. ò13

=

1 u2 du ò 64 u 2 - 52

=

1 æ1 2 2 2 2 2 ö +C çè u u - 5 + 5 ln u + u - 5 ÷ ø 64 2

=

1 æ1 ö 2 2 +C çè 4 x 16 x - 25 + 25ln 4 x + 16 x - 25 ÷ ø 64 2

1 dx x(7 + 3 x)

Examining the form of the integral, we should look at formulas 3–12, which involve a + bu. Based on this, we let u = x. Therefore, we use formula 9 with a = 7 and b = 3. Use this information to rewrite and solve the integral as follows: 3

ò1

1 1 dx = òxx==13 du x(7 + 3x) u (a + bu ) x =3

1 u = ln a a + bu x =1 3

1 x = ln 7 7 + 3x 1 =

1 3 1 1 ln - ln 7 7 + 3(3) 7 7 + 3(1)

» 0.0898 6–18 ..

Calculus

3. ò

Chapter 6

16 x 2 + 36 16 x 2

dx

Examining the form of the integral, we should look at formulas 32–39, which involve

u 2 + a2 . Based on this, we let u 2 = 16 x2 and 16 x2 + 36 = u 2 + a 2 . Then we have u = 4 x and du = 4dx. Using formula 35 and the substitutions, the integral can be written as

ò

4. ò01

16 x 2 + 36 16 x 2

dx =

1 u 2 + 62 du ò 4 u2

1æ = ç4 çè

ö u 2 + 62 2 2 + ln u + u + 6 ÷ ÷ u ø

1æ = ç4 çè

ö 16 x 2 + 36 2 + ln 4 x + 16 x + 36 ÷ ÷ 4x ø

3 + 5x dx 4 + 7x

Examining the form of the integral, we should look at formulas 15–20, which involve a + bu and c + du. Based on this, we let u = x. Therefore, we use formula 20 with a = 3, b = 5, c = 4, and d = 7. Use this information to rewrite and solve the integral as follows: 1 3 + 5x

ò0

4 + 7x

dx = òxx==01

a + bu du c + du x =1

bu ad - bc = + ln c + du d d2 x =0 5 x (3)(7) - (5)(4) = + ln 4 + 7 x 7 (7) 2

1

0

1

=

5x 1 + ln 4 + 7 x 7 49 0

5(1) 1 5(0) 1 + ln 4 + 7(1) + ln 4 + 7(0) 7 49 7 49 » 0.7915

=

6–19 ..

Calculus

5. ò

Chapter 6

1 49 - 16 x 2

dx

Examining the form of the integral, we should look at formulas 13 and 14 which involve a 2 - u 2 . Based on this, we let u 2 = 16 x2 and 49 - 16 x2 = a2 - u 2 . Then we have a = 7, u = 4 x, and du = 4dx. Using formula 14 and the substitutions, the integral can be written as

ò

1 49 - 16 x

dx = 2

1 1 du ò 2 4 7 - u2

=

1 æ1 u+a ö ln ç ø 4 è 2a u - a ÷

=

1æ 1 4x + 7 ö ln ç 4 è 2(7) 4 x - 7 ÷ ø

=

1 4x + 7 ln 56 4 x - 7

8

6. Use the trapezoidal rule with n = 4 to approximate ò 1 + x3 dx.

0

Partition [0, 8] into 4 equal subintervals of width (8 – 0) / 4 = 2. The endpoints are x0 = 0, x1 = 2, x2 = 4, x3 = 6, and x4 = 8. We calculate the value of the function at each endpoint:

x 0 2 4 6 8

f(x) 1 3 8.0623 14.7309 22.6495

By the trapezoidal rule:

Tn = [ f (0) + 2 f (2) + 2 f (4) + 2 f (6) + f (8)]Dx / 2 = [1 + 3 + 8.0623 + 14.7309 + 22.6495](1) = 49.44

6–20 ..

Calculus

7.

Chapter 6

7 1 Use Simpson’s rule with n = 3 to approximate ò dx.

2 1 1+ x

Partition [1, 7] into 6 equal subintervals of width (7 – 1) / 6 = 1. The endpoints are x0 = 1, x1 = 2, x2 = 3, x3 = 4, x4 = 5, x5 = 6, and x6 = 7. We calculate the value of the function at each endpoint:

x 1 2 3 4 5 6 7

f(x) .5 .2 .1 .0588 .0385 .0270 .02

By the simpson’s rule:

S6 = [ f (1) + 4 f (2) + 2 f (3) + 4 f (4) + 2 f (5) + 4 f (6) + f (7)]Dx / 3 = [.5 + .2 + .1 + .0588 + .0385 + .0270 + .02](1/ 3) = .31

6–21 ..

Calculus

Chapter 6

6–22 ..

Calculus

Chapter 7

Name ________________________________ Date ______________ Class ____________

Section 7-1 Functions of Several Variables Goal: To work with functions that involve more than one variable

Definition: An equation of the form z = f ( x, y) describes a function of two independent variables if, for each permissible ordered pair (x, y), there is one and only one value of z determined by f ( x, y ).

Examples of such functions are: Area of a rectangle

A( x, y ) = xy

V ( r , h) =  r h 2

Volume of a right circular cylinder nt

æ rö A( P, r , t , n) = P ç1 + ÷ è nø

1.

Compound Interest

If f ( x, y ) = 4 x3 + 2 y 2 + 5x + 15, find f (2,3).

f ( x, y ) = 4 x 2 + 2 y 2 + 5 x + 15 f (2,3) = 4(2) 2 + 2(3) 2 + 5(2) + 15 f (2,3) = 4(4) + 2(9) + 10 + 15 f (2,3) = 59 2.

8x + 5 y , find g (0, 2).

7 x + 10 y 8x + 5 y g ( x, y ) = 7 x + 10 y 8(0) + 5(2) g (0, 2) = 7(0) + 10(2) 0 + 10 g (0, 2) = 0 + 20 g (0, 2) = 0.5

If g ( x, y ) =

7–1 ..

Calculus 3.

Chapter 7 If f ( x, y, z ) = 5 x3 + 2 y 2 - 4 z 3 + 5, find f (1, 2,3).

f ( x, y , z ) = 5 x 3 + 2 y 2 - 4 z 3 + 5 f (1, 2,3) = 5(1)3 + 2(2) 2 - 4(3)3 + 5 f (1, 2,3) = 5(1) + 2(4) - 4(27) + 5 f (1, 2,3) = 5 + 8 - 108 + 5 f (1, 2,3) = - 90

4.

Find A(2000,0.03,10) for A( P, r , t ) = P + Prt.

A( P, r , t ) = P + Prt A(2000, 0.03,10) = 2000 + 2000(0.03)(10) A(2000, 0.03,10) = 2000 + 600 A(2000, 0.03,10) = 2600

5.

Find v(10, 6,5) for v(l , w, h) = lwh.

v(l , w, h) = lwh v(10, 6,5) = (10)(6)(5) v(10, 6,5) = 300

In Problems 6 and 7, find the indicated function f of a single variable.

6.

f ( y ) = G(3, y ) for G( x, y ) = x3 + 2 xy 2 - 4 y + 3. f ( y ) = G ( x, y ) = x3 + 2 xy 2 - 4 y + 3 f ( y ) = G (2, y ) = (3)3 + 2(3) y 2 - 4 y + 3 f ( y ) = G (2, y ) = 27 + 6 y 2 - 4 y + 3 f ( y ) = G (2, y ) = 6 y 2 - 4 y + 30

7–2 ..

Calculus 7.

Chapter 7

f ( x) = N ( x,3x) for N ( x, y ) = x 2 + y 2 + 7 xy - 2 x - 7.

f ( x) = N ( x, y ) = x 2 + y 2 + 7 xy - 2 x - 7 f ( x) = N ( x,3x) = x 2 + (3 x) 2 + 7 x(3 x) - 2 x - 7 f ( x) = N ( x,3x) = x 2 + 9 x 2 + 21x 2 - 2 x - 7 f ( x) = N ( x,3x) = 31x 2 - 2 x - 7

8.

The Cobb-Douglas production function for a company is given by

f ( x, y ) = 40 x0.2 y 0.8 where x is the utilization of labor and y is the utilization of capital. If the company uses 1500 units of labor and 1200 units of capital, how many units will be produced by the company?

f ( x, y ) = 40 x 0.2 y 0.8 f (1500,1200) = 40(1500)0.2 (1200)0.8 f (1500,1200) » 40(4.3174)(290.6330) f (1500,1200) » 50,191

7–3 ..

Calculus

Chapter 7

7–4 ..

Calculus

Chapter 7

Name ________________________________ Date ______________ Class ____________

Section 7-2 Partial Derivatives Goal: To find partial derivatives of functions of several variables

Definition:

Partial Derivative

With Respect to x:

¶z f ( x, y + h ) - f ( x , y ) = f x ( x, y ) = lim ¶x h h® 0

With Respect to y:

¶z f ( x, y + h ) - f ( x, y ) = f y ( x, y ) = lim ¶y k k® 0

Definition:

Second–Order Partial Derivatives

If z = f ( x, y), then

f xx = f xx ( x, y ) =

¶2 z ¶x

2

=

¶ æ¶ z ö ç ÷ ¶x è ¶x ø

f xy = f xy ( x, y ) =

¶2 z ¶ æ¶ z ö = ç ÷ ¶ y¶ x ¶ y è ¶ x ø

f yx = f yx ( x, y ) =

¶2 z ¶ æ¶ z ö = ç ÷ ¶ x¶ y ¶ x è ¶ y ø

f yy = f yy ( x, y ) =

¶2 z ¶y

2

=

¶ æ¶ z ö ¶ y çè ¶ y ÷ ø

7–5 ..

Calculus

Chapter 7

In Problems 1–3, find the indicated first-order partial derivative for each function z = f ( x, y).

1.

f y ( x, y ) if f ( x, y ) = 9 x3 y + 5 xy 4 - 4 y + 8 x - 12.

We are looking for the partial derivative with respect to y. Therefore, we consider only the y as a variable and all other variables as constants.

f ( x, y ) = 9 x3 y + 5 xy 4 - 4 y + 8 x - 12 f y ( x, y ) = 9 x3 + 20 xy 3 - 4

2.

f x ( x, y ) if f ( x, y ) = 7 xy - 2 x + 8 y - 11xy 3 + 17.

We are looking for the partial derivative with respect to x. Therefore, we consider only the x as a variable and all other variables as constants.

f ( x, y) = 7 xy - 2 x + 8 y - 11xy3 + 17 f x ( x, y) = 7 y - 2 - 11y 3

3.

¶z if z = 2 y 2 - 2 xy + 7 x - 5. ¶x

We are looking for the partial derivative with respect to x. Therefore, we consider only the x as a variable and all other variables as constants.

z = 2 y 2 - 2 xy + 7 x - 5 ¶z = - 2y + 7 ¶x

7–6 ..

Calculus 4.

Chapter 7 Find the value of f x (0,3) if f ( x, y ) = 3x3 y 2 + 5 xy - 8 x + 2 y + 15.

We will first find the partial derivative with respect to x. Therefore, we consider only the x as a variable and all other variables as constants.

f ( x, y) = 3x3 y 2 + 5 xy - 8 x + 2 y + 15 f x ( x, y) = 9 x 2 y 2 + 5 y - 8 Now find the value using the partial derivative.

f x ( x, y ) = 9 x 2 y 2 + 5 y - 8 f x (0,3) = 9(0) 2 (3) 2 + 5(3) - 8 f x (0,3) = 15 - 8 f x (0,3) = 7

5.

Find the value of f y (3,1) if f ( x, y ) = 5 x 2 - 7 x 2 y + 2 xy - 8 y + 6.

We will first find the partial derivative with respect to y. Therefore, we consider only the y as a variable and all other variables as constants.

f ( x, y ) = 5 x 2 - 7 x 2 y + 2 xy - 8 y + 6 f y ( x, y ) = - 7 x 2 + 2 x - 8 Now find the value using the partial derivative.

f y ( x, y ) = - 7 x 2 + 2 x - 8 f y (3,1) = - 7(3) 2 + 2(3) - 8 f y ( x, y ) = - 63 + 6 - 8 f y ( x, y ) = - 65

7–7 ..

Calculus

Chapter 7

In Problems 6–8, find the indicated second-order partial derivative for each function f ( x, y ).

6.

f xx ( x, y ) if f ( x, y ) = 7 x 2 + 2 y 2 - 8 xy + 17

We will first find the partial derivative with respect to x. Therefore, we consider only the x as a variable and all other variables as constants.

f ( x, y ) = 7 x 2 + 2 y 2 - 8 xy + 17 f x ( x, y ) = 14 x - 8 y Now using the function just found, find the partial derivative with respect to x.

f x ( x, y ) = 14 x - 8 y f xx ( x, y ) = 14 7.

2

2

f xy ( x, y ) if f ( x, y) = e3x +2 y +6 xy

We will first find the partial derivative with respect to x. Therefore, we consider only the x as a variable and all other variables as constants.

2

2

f ( x, y ) = e3 x +2 y +6 xy 2

2

f x ( x, y ) = (6 x + 6 y )(e3 x +2 y +6 xy ) Now using the function just found, find the partial derivative with respect to y.

2

2

f x ( x, y ) = (6 x + 6 y )(e3 x +2 y +6 xy ) 2

2

2

2

f xy ( x, y ) = (6 x + 6 y )(4 y + 6 x)(e3 x +2 y +6 xy ) + (e3 x +2 y +6 xy )(6) 2

2

f xy ( x, y ) = [(6 x + 6 y )(4 y + 6 x) + 6](e3 x +2 y +6 xy ) 2

2

f xy ( x, y ) = (24 xy + 36 x 2 + 24 y 2 + 36 xy + 6)(e3 x +2 y +6 xy ) 2

2

f xy ( x, y ) = (36 x 2 + 24 y 2 + 60 xy + 6)(e3 x +2 y +6 xy )

7–8 ..

Calculus 8.

Chapter 7

f yx ( x, y ) if f ( x, y) = (7 x + 2)7

We will first find the partial derivative with respect to y. Therefore, we consider only the y as a variable and all other variables as constants.

f ( x, y ) = (7 x + 2 y )7 f y ( x, y ) = 7(7 x + 2 y )6 (2) f y ( x, y ) = 14(7 x + 2 y )6 Now using the function just found, find the partial derivative with respect to x.

f y ( x, y ) = 14(7 x + 2 y )6 f yx ( x, y ) = 84(7 x + 2 y )5 (7) f yx ( x, y ) = 588(7 x + 2 y )5

In Problems 9–12, find the indicated function or value if N ( x, y ) = 12 x3 + 5 xy - 7 y 2 + 4 x - 9 y + 17 and S ( x, y ) = x 2 ln y + 3 y 2e2 x .

9.

N yy ( x, y) N ( x, y ) = 12 x3 + 5 xy - 7 y 2 + 4 x - 9 y + 17 N y ( x, y ) = 5 x - 14 y - 9 N yy ( x, y ) = - 14

10.

S y (4, - 4) S ( x, y ) = x 2 ln y + 3 y 2e 2 x x2 S y ( x, y ) = + 6 ye 2 x y (4)2 + 6(- 4)e2(4) -4 S y (4, - 4) » - 4 - 71,542.9917

S y (4, - 4) =

S y (4, - 4) » - 71,546.9917

7–9 ..

Calculus 11.

Chapter 7 S x ( x, y )

S ( x, y) = x 2 ln y + 3 y 2e2 x S x ( x, y ) = 2 x ln y + 6 y 2e2 x 12.

N xx (7,3) N ( x, y ) = 12 x3 + 5 xy - 7 y 2 + 4 x - 9 y + 17 N x ( x, y ) = 36 x 2 + 5 y + 4 N xx ( x, y ) = 72 x N xx (7,3) = 72(7) N xx (7,3) = 504

13.

A firm produces two types of toy trains each week, x of type A and y of type B. The weekly revenue and cost functions (in dollars) are

R( x, y ) = 50 x + 70 y + 0.07 xy - 0.04 x 2 - 0.04 y 2 C ( x, y) = 7 x + 9 y + 18,000 Find the profit function and then find Px (800,1000) and Py (800,1000), and interpret the results.

P ( x, y ) = R ( x, y ) - C ( x, y ) P( x, y ) = 50 x + 70 y + 0.07 xy - 0.04 x 2 - 0.04 y 2 - (7 x + 9 y + 18, 000) P( x, y ) = 43x + 61y + 0.07 xy - 0.04 x 2 - 0.04 y 2 - 18, 000 P( x, y ) = 43 x + 61 y + 0.07 xy - 0.04 x 2 - 0.04 y 2 - 18, 000 Px ( x, y ) = 43 + 0.07 y - 0.08 x Px (800,1000) = 43 + 0.07(1000) - 0.08(800) Px (800,1000) = 43 + 70 - 64 Px (800,1000) = 49

P( x, y ) = 43x + 61y + 0.07 xy - 0.04 x 2 - 0.04 y 2 - 18, 000 Py ( x, y ) = 61 + 0.07 x - 0.08 y Py (800,1000) = 61 + 0.07(800) - 0.08(1000) Py (800,1000) = 61 + 56 - 80 Py (800,1000) = 37 7–10 ..

Calculus

Chapter 7

Name ________________________________ Date ______________ Class ____________

Section 7-3 Maxima and Minima Goal: To find local maxima and minima of functions of several variables

Theorem 1:

Local Extrema and Partial Derivatives

Let f (a, b) be a local extremum (a local maximum or a local minimum) for the function f. If both f x and f y exist at (a, b), then f x (a, b) = 0 and f y (a, b) = 0.

Theorem 2:

Second-Derivative Test for Local Extremum

If 1.

2.

z = f ( x, y) f x (a, b) = 0 and f y (a, b) = 0 [(a, b) is a critical point]

3.

All second–order partial derivatives of f exist in some circular region containing (a, b) as center.

A = f xx (a, b), B = f xy (a, b), C = f yy (a, b)

4.

Then Case 1: If AC - B 2 > 0 and A < 0, then f (a, b) is a local maximum.

Case 2: If AC - B 2 > 0 and A > 0, then f (a, b) is a local minimum.

Case 3: If AC - B 2 < 0, then f has a saddle point at (a, b).

Case 4: If AC - B 2 = 0, the test fails.

1.

If f ( x, y) = 17 x - 12 y + 27, find f x ( x, y ) and f y ( x, y ), and explain, using

Theorem 1, why f ( x, y) has no local extremum.

f ( x, y ) = 17 x - 12 y + 27 f x ( x, y ) = 17

f ( x, y ) = 17 x - 12 y + 27 f y ( x, y ) = - 12

Since the partial derivative with respect to x and y cannot equal zero, there is no local extremum.

7–11 ..

Calculus

Chapter 7

In Problems 2–5, use theorem 2 to find the local extremum.

2. f ( x, y ) = 5 x 2 - 3 y 2 + 2 x - 3 y + 17

f ( x, y ) = 5 x 2 - 3 y 2 + 2 x - 3 y + 17 f x ( x, y ) = 10 x + 2

f ( x, y ) = 5 x 2 - 3 y 2 + 2 x - 3 y + 17 f y ( x, y ) = - 6 y - 3

0 = 10 x + 2 - 2 = 10 x

0 = - 6y - 3 3 = - 6y

- 15 = x

- 12 = y

Based on the above partial derivatives, there is a critical value at the point (- 15 , - 12 ).

Now find the values for A, B, and C.

A = f xx (a, b)

B = f xy (a, b)

C = f yy (a, b)

A = 10

B=0

C =-6

AC - B 2 = (10)(- 6) - (0)2 AC - B 2 = - 60 Since the above value is negative, the function f has a saddle point at (- 15 , - 12 ).

3.

f ( x, y ) = 2 x 2 + 4 y 2 + 7 x - 3 y - 12 f ( x, y ) = 2 x 2 + 4 y 2 + 7 x - 3 y - 12 f x ( x, y ) = 4 x + 7

f ( x, y ) = 2 x 2 + 4 y 2 + 7 x - 3 y - 12 f y ( x, y ) = 8 y - 3

0 = 4x + 7 - 7 = 4x

0 = 8y - 3 3 = 8y

- 74 = x

3= y 8

Based on the above partial derivatives, there is a critical value at the point (- 52 , 16 ).

Now find the values for A, B, and C.

A = f xx (a, b)

B = f xy (a, b)

C = f yy (a, b)

A=4

B=0

C =8

AC - B 2 = (4)(8) - (0)2 AC - B 2 = 32 Since the above value is positive and the value of A is positive, the function has a minimum at the point (- 74 , 83 ).

7–12 ..

Calculus 4.

Chapter 7

f ( x, y ) = 4 xy + 5 y 2 - 4 x - 7 y + 19 f ( x, y ) = 4 xy + 5 y 2 - 4 x - 7 y + 19 f x ( x, y ) = 4 y - 4 0 = 4y - 4 4 = 4y 1= y

f ( x, y ) = 4 xy + 5 y 2 - 4 x - 7 y + 19 f y ( x, y ) = 4 x + 10 y - 7 0 = 4 x + 10(1) - 7 - 3 = 4x - 34 = x

Based on the above partial derivatives, there is a critical value at the point (- 34 , 1).

Now find the values for A, B, and C.

A = f xx (a, b)

B = f xy (a, b)

C = f yy (a, b)

A=0

B=4

C = 10

AC - B 2 = (0)(10) - (4)2 AC - B 2 = - 16 Since the above value is negative, the function f has a saddle point at (- 34 , 1).

7–13 ..

Calculus 5.

Chapter 7

f ( x, y) = 12 x 4 + y 2 - 4 xy + 3 f ( x, y ) = 12 x 4 + y 2 - 4 xy + 3

f ( x, y ) = 12 x 4 + y 2 - 4 xy + 3

f x ( x, y ) = 2 x 3 - 4 y

f y ( x, y ) = 2 y - 4 x

Setting both equations equal to zero and solving will yield the critical points of (0, 0), (2, 4) and (–2, –4). Now find the equations for A, B, and C. A = f xx (a, b)

B = f xy (a, b)

C = f yy (a, b)

A = 6 x2

B=-4

C=2

For the critical point (0, 0), the value of A will be 0. Therefore,

AC - B 2 = (0)(2) - (- 4) 2 AC - B 2 = - 16 Since the above value is negative, the function has a saddle point at the point (0, 0).

For the critical point (2, 4), the value of A will be 24. Therefore,

AC - B 2 = (24)(2) - (- 4)2 AC - B 2 = 32 Since the above value is positive and the value of A is positive, the function has a minimum at the point (2, 4).

For the critical point (–2, –4), the value of A will be 24. Therefore,

AC - B 2 = (24)(2) - (- 4)2 AC - B 2 = 32 Since the above value is positive and the value of A is positive, the function has a minimum at the point (–2, –4).

7–14 ..

Calculus

Chapter 7

6.

A company produces two types of toy cars per day: x units of type A and y units of type B. If the revenue and cost equations for the day are (in thousands of dollars)

R( x, y) = 20 x + 25 y

C ( x, y ) = 2 x 2 - 5 xy + 4 y 2 - 5 x + 9 y + 7 determine how many of each type of toy car should be produced per day to maximize profit? What is the maximum profit?

P ( x, y ) = R ( x , y ) - C ( x , y ) P( x, y ) = 20 x + 25 y - (2 x 2 - 5 xy + 4 y 2 - 5 x + 9 y + 7) P( x, y ) = - 2 x 2 + 5 xy - 4 y 2 + 25 x + 16 y - 7 Px ( x, y ) = - 4 x + 5 y + 25 Py ( x, y ) = 5 x - 8 y + 16 Setting both equations equal to zero and solving will yield the critical point of (40, 27). Now find the values for A, B, and C.

A = Pxx (a, b)

B = Pxy (a, b)

C = Pyy (a, b)

A=-4

B=5

C =-8

AC - B 2 = (- 4)(- 8) - (5)2 AC - B 2 = 7 Since the above value is positive and the value of A is negative, the function has a maximum at the point (40, 27). Therefore, you should produce 40 type A toy cars and 27 type B toy cars to maximize your profit. The maximum profit will be P( x, y ) = - 2 x 2 + 5 xy - 4 y 2 + 25 x + 16 y - 7 P(40, 27) = - 2(40) 2 + 5(40)(27) - 4(27) 2 + 25(40) + 16(27) - 7 P(40, 27) = $706

7–15 ..

Calculus

Chapter 7

7.

A rectangular box with no top and two parallel partitions must hold a volume of 125 cubic inches. Find the dimensions that will require the least amount of material.

Based on the information in the problem, we will let x = length, y = width, and z = height. Therefore, M = xy + 2 xz + 4 yz and the volume is V = xyz = 125. Solve the volume equation for z and reduce the original equation down to two variables yields

M ( x, y ) = xy + 250 + 500 y x Use the new equation for M and follow the procedure outlined in theorem 2.

M ( x, y ) = xy + 250 + 500 y x

M ( x, y ) = xy + 250 + 500 y x

M x ( x, y ) = y - 500 2

M y ( x, y ) = x - 250 2 y

x

Setting both equations equal to zero and solving will yield the critical point of (10, 5).

Now find the values for A, B, and C.

A = M xx (a, b)

B = Pxy (a, b)

C = Pyy (a, b)

A = 1000 3

B =1

C = 500 3

x

y

A =1

C=4

AC - B 2 = (1)(4) - (1)2 AC - B 2 = 3 Since the above value is positive and the value of A is positive, the function has a minimum at the point (10, 5). Therefore, the length of the box should be 10 inches, the width of the box should be 5 inches, and solving for the variable z will yield a height of 2.5 inches.

7–16 ..

Calculus

Chapter 7

Name ________________________________ Date ______________ Class ____________

Section 7-4 Maxima and Minima Using Lagrange Multipliers Goal: To find maxima and minima of functions of several variables using Lagrange multipliers Theorem 1: Method of Lagrange Multipliers for Functions of Two Variables Any local maxima or minima of the function z = f ( x, y) subject to the constraint g ( x, y) = 0 will be among those points ( x0 , y0 ) for which ( x0 , y0 , 0 ) is a solution of the system Fx ( x, y,  ) = 0 Fy ( x, y,  ) = 0

F ( x, y,  ) = 0 where F ( x, y,  ) = f ( x, y) +  g ( x, y), provided that all partial derivatives exist.

Procedure: Method of Lagrange Multipliers: Key Steps Step 1: Write the problem in the form Maximize (or minimize) z = f ( x, y) subject to g ( x, y) = 0 Step 2: Form the function F: F ( x, y,  ) = f ( x, y) +  g ( x, y) Step 3: Find the critical points of F; that is, solve the system

Fx ( x, y,  ) = 0 Fy ( x, y,  ) = 0 F ( x, y,  ) = 0 Step 4: If ( x0 , y0 , 0 ) is the only critical point of F, we assume that ( x0 , y0 ) will always produce the solution to the problems we consider. If F has more than one critical point, we evaluate z = f ( x, y) at ( x0 , y0 ) for each critical point ( x0 , y0 , 0 ) of F. For the problems we consider , we assume that the largest of these values is the maximum value of f ( x, y ), subject to the constraint g ( x, y) = 0 and the smallest is the minimum value of f ( x, y ), subject to the constraint g ( x, y) = 0.

7–17 ..

Calculus

Theorem 2:

Chapter 7

Method of Lagrange Multipliers for Functions of Three Variables

Any local maxima or minima of the function w = f ( x, y, z ) subject to the constraint g ( x, y, z ) = 0 will be among the set of points ( x0 , y0 , z0 ) for which ( x0 , y0 , z0 , 0 ) is a solution of the system Fx ( x, y, z ,  ) = 0

Fy ( x, y, z,  ) = 0 Fz ( x, y, z ,  ) = 0

F ( x, y, z,  ) = 0 where F ( x, y, z,  ) = f ( x, y, z) +  g ( x, y, z), provided that all partial derivatives exist.

1. Use Theorem 1 to explain why no minima exist.

Minimize:

f ( x, y) = 8 x - 5 y

subject to:

5x + 8 y = 9

The problem will need to be rewritten in standard form first.

Minimize: subject to:

f ( x, y) = 8 x - 5 y 5x + 8 y - 9 = 0

Now find the function F using the definition.

F ( x, y , l ) = f ( x , y ) + l g ( x , y ) F ( x, y, l ) = (8 x - 5 y ) + l (5 x + 8 y - 9) F ( x, y , l ) = 8 x - 5 y + 5 x l + 8 y l - 9 l Find the three partial derivatives.

Fx ( x, y, l ) = 8 + 5l Fy ( x, y, l ) = - 5 + 8l Fl ( x, y, l ) = 5 x + 8 y - 9 Therefore, Fx = 0 and Fy = 0 have no simultaneous solution.

7–18 ..

Calculus

Chapter 7

In Problems 2–5, use the method of Lagrange multipliers.

2. Maximize:

f ( x, y) = 9 xy

subject to:

2x + 2 y = 6

Step 1:

Rewrite the problem.

Maximize: subject to:

Step 2:

f ( x, y) = 9 xy 2x + 2 y - 6 = 0

Form the function F.

F ( x, y , l ) = f ( x , y ) + l g ( x , y ) F ( x, y, l ) = (9 xy ) + l (2 x + 2 y - 6) F ( x, y, l ) = 9 xy + 2 xl + 2 y l - 6 l Step 3:

Find the critical points.

Fx ( x, y, l ) = 9 y + 2 l = 0 Fy ( x, y, l ) = 9 x + 2 l = 0 Fl ( x, y, l ) = 2 x + 2 y - 6 = 0 Therefore, y = - 29l and x = - 29l .

2x + 2 y - 6 = 0

y=-(

- 49l - 49l - 6 = 0

y=

- 89l = 6

- 2(6.75) ) 9

3 2

x=-( x=

- 2(6.75) ) 9

3 2

- 8l = 54 l = - 6.75 Step 4:

æ3 3 27 ö Since the solution ç , , ÷ is the only critical point of F, the è2 2 4 ø

æ3 3 ö point ç , ÷will result in the maximum value of the function. è2 2 ø

7–19 ..

Calculus

Chapter 7

3. Minimize:

f ( x, y ) = 2 x 2 + 2 y 2

subject to:

4 x + 5 y = 20

Step 1:

Rewrite the problem.

f ( x, y ) = 2 x 2 + 2 y 2 4 x + 5 y - 20 = 0

Minimize: subject to: Step 2:

Form the function F.

F ( x , y ,  ) = f ( x , y ) +  g ( x, y ) F ( x, y,  ) = (2 x 2 + 2 y 2 ) +  (4 x + 5 y - 20) F ( x, y,  ) = 2 x 2 + 2 y 2 + 4 x + 5 y - 20 Step 3:

Find the critical points.

Fx ( x, y,  ) = 4 x + 4 = 0 Fy ( x, y,  ) = 4 y + 5 = 0 F ( x, y,  ) = 4 x + 5 y - 20 = 0 Therefore, x = -  and y = - 54 .

4 x + 5 y - 20 = 0 4(-  ) + 5(- 54 ) - 20 = 0 - 4 - 6.25 = 20 - 10.25 = 20 80 =41 Step 4:

x=-

y = - 54

æ 80 ö æ 5 öæ 80 ö x = - çy = ÷ çè- ÷ç ÷ è 41 ø 4 øè 41 ø x=

80 41

y=

100 41

æ80 100 80 ö Since the solution ç , ,÷is the only critical point of F, the è 41 41 41 ø

æ80 100 ö point ç , will result in the minimum value of the function. è 41 41 ÷ ø

7–20 ..

Calculus 4. Minimize: subject to: Step 1:

Chapter 7

f ( x, y , z ) = x 2 + 2 y 2 + z 2 2 x - 2 y + 3z = - 21 Rewrite the problem.

Minimize: f ( x, y , z ) = x 2 + 2 y 2 + z 2 subject to: 2 x - 2 y + 3z + 21 = 0 Step 2: Form the function F.

F ( x, y , z ,  ) = f ( x, y , z ) +  g ( x, y , z ) F ( x, y, z,  ) = ( x 2 + 2 y 2 + z 2 ) +  (2 x - 2 y + 3z + 21) F ( x, y, z,  ) = x 2 + 2 y 2 + z 2 + 2 x - 2 y + 3z + 21 Step 3:

Find the critical points.

Fx ( x, y, z,  ) = 2 x + 2 = 0 Fy ( x, y, z,  ) = 4 y - 2 = 0 Fz ( x, y, z,  ) = 2 z + 3 = 0 F ( x, y, z,  ) = 2 x - 2 y + 3z + 21 = 0 Therefore, x = -  , y = 0.5 , and z = - 1.5 .

2 x - 2 y + 3z + 21 = 0 2(-  ) - 2(0.5 ) + 3(- 1.5 ) + 21 = 0 - 2 -  - 4.5 = - 21 - 7.5 = - 21 14 = 5 x=- y = 0.5(2.8) y = 1.4 æ14 ö x=- ç ÷ è5ø 7 y= 5 14 x=5 Step 4:

z = - 1.5(2.8) z = - 4.2 21 z=5

æ 14 7 21 14 ö Since the solution ç, ,, ÷is the only critical point of F, è 5 5 5 5ø

æ 14 7 21 ö the point ç, ,÷ will result in the minimum value of the function. è 5 5 5ø 7–21 ..

Calculus

Chapter 7

5. Maximize:

f ( x, y, z ) = 5 xyz

subject to:

3x + 2 y + 3z = 90

Step 1:

Rewrite the problem.

Maximize: subject to:

Step 2:

f ( x, y, z ) = 5 xyz 3x + 2 y + 3z - 90 = 0

Form the function F.

F ( x , y , z ,  ) = f ( x , y , z ) +  g ( x, y , z ) F ( x, y, z ,  ) = (5 xyz ) +  (3x + 2 y + 3z - 90) F ( x, y, z ,  ) = 5 xyz + 3x + 2 y + 3z - 90 Step 3:

Find the critical points.

Fx ( x, y, z,  ) = 5 yz + 3 = 0 Fy ( x, y, z,  ) = 5 xz + 2 = 0 Fz ( x, y, z,  ) = 5 xy + 3 = 0 F ( x, y, z,  ) = 3x + 2 y + 3z - 90 = 0 Simplifying the first two equations will yield the result y = 1.5 x and simplifying the first and third equations will yield the result z = x.

3x + 2 y + 3z - 90 = 0 3x + 2(1.5 x) + 3( x) - 90 = 0 3x + 3x + 3x = 90 9 x = 90 x = 10

Step 4:

y = 1.5 x y = 1.5(10) y = 15

z=x z = 10

5 yz + 3 = 0 5(15)(10) + 3 = 0  = - 250

Since the solution (10,15,10, - 250) is the only critical point of F, the

point (10,15,10) will result in the maximum value of the function.

7–22 ..

Calculus

Chapter 7

Name ________________________________ Date ______________ Class ____________

Section 7-5 Method of Least Squares Goal: To find the equation of a line using the method of least squares

Theorem:

Least Squares Approximation

For a set of n points ( x1, y1),( x2 , y2 ), ( xn , yn ), the coefficients of the least squares line y = ax + b are the solutions of the system of normal equations n æn 2ö æn ö x a + x b = xk yk å å å k÷ k÷ çè çè k =1 ø k =1 ø k =1

n æn ö x a + nb = yk å å k÷ çè k =1 ø k =1

and are given by the formulas

a=

æn ö æn öæ n ö n ç å xk yk ÷- ç å xk ÷ç å yk ÷ èk =1 ø è k =1 øè k =1 ø 2

æn ö æn ö n ç å xk2 ÷- ç å xk ÷ èk =1 ø èk =1 ø

n æn ö å yk - a ç å xk ÷ è k =1 ø b = k =1 n

7–23 ..

Calculus

Chapter 7

In Problems 1 and 2, find the least squares line. Graph the data and the least squares line.

1.

Totals

x

y

xy

1 2 3 4 10

4 5 4 8 21

4 10 12 32 58

x2 1 4 9 16 30

The normal equations would be 30a + 10b = 58

10a + 4b = 21 The solutions for a and b are

a=

æn ö æn öæ n ö n ç å xk yk ÷- ç å xk ÷ç å yk ÷ èk =1 ø è k =1 øè k =1 ø 2

æn ö æn ö n ç å xk2 ÷- ç å xk ÷ èk =1 ø è k =1 ø

=

4(58) - (10)(21) 4(30) - (10)2

= 1.1

n æn ö å yk - a ç å xk ÷ è k =1 ø 21 - (1.1)(10) b = k =1 = = 2.5 n 4

Therefore, the equation of the least squares line is y = 1.1x + 2.5.

7–24 ..

Calculus

Chapter 7

2.

Totals

x

y

xy

1 2 3 4 10

3 0 –2 –1 0

3 0 –6 –4 –7

x2 1 4 9 16 30

The normal equations would be 30a + 10b = - 7

10a + 4b = 0 The solutions for a and b are

a=

æn ö æn öæ n ö n ç å xk yk ÷- ç å xk ÷ç å yk ÷ è k =1 ø è k =1 øè k =1 ø 2

æn ö æn ö n ç å xk2 ÷- ç å xk ÷ èk =1 ø è k =1 ø

=

4(- 7) - (10)(0) 4(30) - (10) 2

= - 1.4

n æn ö å yk - a ç å xk ÷ èk =1 ø 0 - (- 1.4)(10) b = k =1 = = 3.5 n 4

Therefore, the equation of the least squares line is y = - 1.4 x + 3.5.

7–25 ..

Calculus

Chapter 7

In Problems 3 and 4, find the least squares line and use it to estimate y for the indicated value of x. Round answers to two decimal places, if necessary.

3.

Totals

x

y

xy

3 5 7 9 11 35

2 8 6 7 9 32

6 40 42 63 99 250

x2 9 25 49 81 121 285

Estimate y when x = 7.

The normal equations would be 285a + 35b = 250

35a + 5b = 32 The solutions for a and b are

a=

æn ö æn öæ n ö n ç å xk yk ÷- ç å xk ÷ç å yk ÷ è k =1 ø è k =1 øè k =1 ø 2

æn ö æn ö n ç å xk2 ÷- ç å xk ÷ è k =1 ø è k =1 ø

=

5(250) - (35)(32) 5(285) - (35) 2

= 0.65

n æn ö å yk - a ç å xk ÷ è k =1 ø 32 - (0.65)(35) b = k =1 = = 1.85 n 5

Therefore, the equation of the least squares line is y = 0.65 x + 1.85 and the value of y will be estimated to y = 0.65(7) + 1.85 = 6.4.

7–26 ..

Calculus

Chapter 7

4.

Totals

x

y

xy

–6 –3 0 3 6 0

12 9 8 5 1 35

–72 –27 0 15 6 –78

x2 36 9 0 9 36 90

Estimate y when x = 9.

The normal equations would be 90a + 0b = - 78

0a + 5b = 35 The solutions for a and b are

a=

æn ö æn öæ n ö n ç å xk yk ÷- ç å xk ÷ç å yk ÷ èk =1 ø è k =1 øè k =1 ø 2

æn ö æn ö n ç å xk2 ÷- ç å xk ÷ èk =1 ø è k =1 ø

=

5(- 78) - (0)(35) 5(90) - (0) 2

= - 0.87

n æn ö å yk - a ç å xk ÷ èk =1 ø 35 - (- 0.87)(0) b = k =1 = =7 n 5

Therefore, the equation of the least squares line is y = - 0.87 x + 7 and the value of y will be estimated to y = - 0.87(9) + 7 = - 0.83.

7–27 ..

Calculus

Chapter 7

5. The U.S. Centers for Disease Control and Prevention (CDC) use a measure called body mass index (BMI) to determine whether a person is obese. A BMI between 25.0 and 29.9 is considered overweight, and a BMI of 30.0 or more is considered obese. The following table shows the percent of United States males 20 years and older who were obese in the years indicated, judging on the basis of BMI. (Source: CDC, May 19, 2010)

Year 2000 2002 2004 2006 2008

Percent Obese 27.5 27.8 31.1 33.3 32.3

Using 0 for 2000, find the least squares line and use it to predict the percent of males 20 years and older that will be obese in the year 2014.

Totals

x

y

xy

0 2 4 6 8 20

27.5 27.8 31.1 33.3 32.3 152

0 55.6 124.4 199.8 258.4 638.2

x2 0 4 16 36 64 120

The normal equations would be 120a + 20b = 638.2

20a + 5b = 152 The solutions for a and b are

a=

æn ö æn öæ n ö n ç å xk yk ÷- ç å xk ÷ç å yk ÷ èk =1 ø è k =1 øè k =1 ø 2

æn ö æn ö n ç å xk2 ÷- ç å xk ÷ è k =1 ø è k =1 ø

=

5(638.2) - (20)(152) 5(120) - (20) 2

= 0.755

n æn ö å yk - a ç å xk ÷ èk =1 ø 152 - (0.755)(20) b = k =1 = = 27.38 n 5

Therefore, the equation of the least squares line is y = 0.755 x + 27.38 and the value of y when x = 14 will be estimated to y = 0.755(14) + 27.38 = 37.95. Therefore, approximately 38 percent of the male population will be considered obese in the year 2014.

7–28 ..

Calculus

Chapter 7

6. The table gives the winning distance in the discus throw in the Olympic Games from 1980 to 2008.

Year 1980 1984 1988 1992 1996 2000 2004 2008

Distance (ft) 66.64 66.60 68.82 65.12 69.40 69.30 69.89 68.82

Use a graphing calculator to find the least squares line for the data, letting x = 0 for 1980. Estimate the winning distance in the discus throw in the Olympic Games of 2020.

Rewrite the table using x for the year and y for the distance.

x 0 4 8 12 16 20 24 28

Y 66.64 66.60 68.82 65.12 69.40 69.30 69.89 68.82

On your graphing calculator, go into the “STAT” mode and enter the x values to list 1 and the y values to list 2. Next go to the “CALC” mode within “STAT”. Option number 4 is “LinReg(ax+b). Hit the enter key twice for the following result:

7–29 ..

Calculus

Chapter 7

Therefore, the equation of the least squares line is y = 0.111x + 66.514 and the value of y when x = 40 will be estimated to y = 0.111(40) x + 66.514 = 70.95 Therefore, a distance of approximately 70.95 feet will be the winning distance in the year 2020.

7–30 ..

Calculus

Chapter 7

Name ________________________________ Date ______________ Class ____________

Section 7-6 Double Integrals over Rectangular Regions Goal: To evaluate indefinite and definite double integrals

Definition:

Double Integral

The double integral of a function f ( x, y) over a rectangle R = {( x, y) | a £ x £ b, c £ y £ d } is b d dx òò f ( x, y ) dA = òa éëòc f ( x, y) dy ù û R

= òcd éòab f ( x, y ) dx ù dy ë û Definition:

Average Value over Rectangular Regions

The average value of the function f ( x, y) over the rectangle R = {( x, y) | a £ x £ b, c £ y £ d } is 1 òò f ( x, y ) dA (b - a )(d - c) R

In Problems 1 and 2, find each antiderivative. Then use the antiderivative to evaluate the definite integral.

1. a.

4 3 ò25x y dx

a. ò25 x 4 y3 dx = 5 x5 y3 + E ( y )

b.

2 4 3 ò0 25x y dx

b.

2 4 3 5 3 ò0 25 x y dx = 5 x y

2

0 5 3

= 5(2) y - 5(0)5 y3 = 125 y3

7–31 ..

Calculus

Chapter 7

2. a.

ò

ln x xy

dy 2

b.

ln x

ln x - 2 a. ò 2 dy = ò y dy x xy

b.

2 ln x dy 2

ò1

xy

ln x 2 ln x ò1 2 dy = -

xy 1

xy

ln x y - 1 + C ( x) x -1 ln x =+ C ( x) xy

=

2

=-

ln x æ ln x ö - x(2) çè x(1) ÷ ø

=-

ln x ln x ln x + = 2x x 2x

In Problems 3 and 4, evaluate each iterated integral.

3. ò16 ò-42 (8x + 6 y + 7) dx dy x =4 ù 6 4 6é 2 dy ò1 ò- 2 (8 x + 6 y + 7) dx dy = ò1 ê4 x + 6 xy + 7 x x =- 2 ú ë û

= ò16 é(4(4) 2 + 6(4) y + 7(4)) - (4( - 2) 2 + 6( - 2) y + 7( - 2)) ù dy ë û = ò16 [(64 + 24 y + 28) - (16 - 12 y - 14) ] dy = ò16 [90 + 36 y ] dy = 90 y + 18 y 2

6 1

= (90(6) + 18(6)2 ) - (90(1) + 18(1)2 ) 6 4

ò1 ò- 2 (8 x + 6 y + 7) dx dy = - 216

7–32 ..

Calculus

Chapter 7

4. ò-32 ò04 20x3 y 4 dy dx 3 4 3 é 3 5 y =4 ù 3 4 dx ò- 2 ò0 20 x y dy dx = ò- 2 ê4 x y y =0 ú ë û

= ò-32 é(4 x3 (4)5 ) - (4 x3 (0)5 ) ù dx ë û = ò-32 é(4(1024) x3 ) - (0)ù dx ë û 3 é 3ù = ò- 2 4096 x dx ë û = 1024 x 4

3 -2 4

= (1024(3) ) - (1024(- 2)4 ) 3 4 3 4 ò- 2 ò0 20 x y dy dx = 66,560

5. Use both orders of iteration to evaluate the double integral.

3 3

R = {( x, y) | - 3 £ x £ 0, 0 £ y £ 4}

òò( x y ) dA; R

4 0

3 3

1 4 é 4 3 x =0 ù dy ò (x y ) x =- 3 ú 4 0 êë û 1 = ò04 é((0)4 y 3 ) - ((- 3)4 y 3 ) ù dy û 4 ë

ò0 ò- 3 ( x y ) dx dy =

= ò04 é- 81 y 3 ù dy ë 4 û 81 y 4 = - 16

4 0 4

81 (4) ) - ( - 81 (0) 4 ) = (- 16 16

= - 1296 0

4

3 3

1 0 é 3 4 y =4 ù dx ò (x y ) y =0 ú 4 - 3 êë û 1 = ò-03 é( x3 (4) 4 ) - ( x3 (0) 4 ) ù dx û 4 ë

ò- 3 ò0 ( x y ) dy dx =

= ò-03 é64 x3 ù dx ë û = 16 x 4

0 -3 4

= (16(0) ) - (16(- 3) 4 ) = - 1296 7–33 ..

Calculus

Chapter 7

6. Find the average value of the function over the given rectangle.

f ( x, y ) = 3x 2 - 4 y3 ; R = {( x, y) | 0 £ x £ 4, - 3 £ y £ 1} Using the average value formula yields the result

1 1 1 é 3 1 4 2 3 3 x =4 ù dy ò- 3 ò0 (3 x - 4 y ) dx dy = ò- 3 êx - 4 xy x =0 ú (4 - 0)(1 - (- 3)) 16 ë û 1 = ò-13 é64 - 16 y 3 ù dy û 16 ë 1 1 = (64 y - 4 y 4 ) -3 16 1 = ((64 - 4) - (- 192 - 324)) 16 = 36

7. Find the volume of the solid under the graph of the function over the given rectangle.

f ( x, y) = 4 x + 8 y - 7; R = {( x, y ) | - 2 £ x £ 2, - 1 £ y £ 3}

x =2 ù 3 2 3 é 2 dy ò- 1 ò- 2 (4 x + 8 y - 7) dx dy = ò- 1 ê2 x + 8 xy - 7 x x =- 2 ú ë û

= ò-31[(8 + 16 y - 14) - (8 - 16 y + 14) ] dy = ò-31[32 y - 28] dy = 16 y 2 - 28 y

3 -1

= (16(3) 2 - 28(3)) - (16(- 1) 2 - 28(- 1)) = 16 cubic units

7–34 ..

Calculus

Chapter 7

Name ________________________________ Date ______________ Class ____________

Section 7-7 Double Integrals over More General Regions Goal: To evaluate double integrals over non–rectangular regions

Definition:

Regular Regions

A region R in the xy plane is a regular x region if there exist functions f(x) and g(x) and numbers a and b such that

R = {( x, y) | g ( x) £ y £ f ( x), a £ x £ b} A region R is a regular y region if there exist h(y) and k(y) and numbers c and d such that

R = {( x, y ) | h( y) £ x £ k ( y ), c £ y £ d } Definition:

Double Integration over Regular Regions

Regular x Region If R = {( x, y) | g ( x) £ y £ f ( x), a £ x £ b}, then b f ( x) dx òòF ( x, y ) dA = òa éëòg ( x ) F ( x, y ) dy ù û R

Regular y Region If R = {( x, y) | h( y) £ x £ k ( y), c £ y £ d }, then d k ( y) dy òòF ( x, y ) dA = òc éëòh( y ) F ( x, y ) dx ù û R

7–35 ..

Calculus

Chapter 7

1. Evaluate the following integral.

6 x

2

ò1 ò0 (3x + y ) dy dx y3 6 x 6 2 ò1 ò0 (3x + y ) dy dx = ò1 (3xy + 3 )

= ò16 (3x( x) +

y= x

dx y =0

( x )3 (0)3 ) (3 x (0) + ) dx 3 3

3

= ò16 (3x 2 + x3 ) dx 6

4

x ) = ( x3 + 12

1

(6)4

(1)4

= ((6)3 + 12 ) - ((1)3 + 12 ) 1 + 1) = ( 2592 + 1296 ) - (12 12 12 12 6 x 2 3886 ò1 ò0 (3x + y ) dy dx = 12 » 323.83

2. Use the description of the region R to evaluate the integral.

R = {( x, y) | 0 £ x £ y, 0 £ y £ 6}

òò(2 x + y - 3) dA; R 6 y

1 6

ò0 ò0 (2 x + y - 3) dx dy = 4 ò0 (2 x + y - 3)

2 x= y x =0 2

dy

= 14 ò06 (2 y + y - 3) - (2(0) + y - 3) 2 dy = 14 ò06 (3 y - 3) 2 - ( y - 3) 2 dy = 14 ò06 (8 y 2 - 12 y ) dy 8 y3 = 14 ( 3 - 6 y 2 ) 8(6)

6 0

3

8(0)3

= 14 ( 3 - 6(6) 2 ) - 14 ( 3 - 6(0) 2 ) 6 y

ò0 ò0 (2 x + y - 3) dx dy = 90

7–36 ..

Calculus

Chapter 7

3. Graph the region R bounded by the graphs of the indicated equations. Describe R in set notation with double inequalities, and evaluate the indicated integral.

y = 1 + 6 x - x 2 , y = 1; òò15 x dA R

The region bounded by the equations is (the window shown is quadrant 1 only)

Therefore, the region R is defined by R = {( x, y ) |1 £ y £ 1 + 6 x - x 2 , 0 £ x £ 6}.

6 1+6 x - x 2

ò0 ò1

y =1+6 x - x 2

(15 x) dy dx = ò06 (15 xy ) y =1

dx

= ò06 (15 x(1 + 6 x - x 2 )) - (15 x(1)) dx = ò06 (90 x 2 - 15 x3 ) dx 4

= (30 x3 - 154x ) = (30(6)3 6 1+6 x - x 2

ò0 ò1

6 0

4 15(6)4 3 15(0) ) (30(0) ) 4 4

(15 x) dy dx = 1620

7–37 ..

Calculus

Chapter 7

4. Graph the region of integration, reverse the order of integration, and then evaluate the integral with the order reversed.

4 x

3

ò0 ò0 ( y - x) dy dx The region of integration will look as follows:

Therefore, reversing the order of integration will yield

1 4 4 x =4 dy ò0 ( y - x) x= y 4 1 = - ò04 ( y - 4) 4 - ( y - y ) 4 dy 4 1 = - ò04 ( y - 4) 4 dy 4 4 1 =( y - 4)5 0 20 1 256 4 4 3 ((0)5 - (- 4)5 ) = ò0 òy ( y - x) dx dy = 20 5 4 4

3

ò0 òy ( y - x) dx dy = -

7–38 ..

Calculus

Chapter 7

5. Reverse the order of integration and then evaluate the integral with the order reversed.

Do not attempt to evaluate the integral in the original form.

25 4 3 ò0 ò y ( x + 4) dx dy

5 x2 ò0 ò0 (

x

3

+ 4) dy dx = ò05 y

3

y = x2

x +4

dx y =0

= ò05 x 2 x3 + 4 - 0 x3 + 4 dx = ò05 x 2 x3 + 4 dx 3 5 1 = ( x3 + 4) 2 3 0 3 3 1 1 = ((5)3 + 4) 2 - ((0)3 + 4) 2 3 3 2

5 x 3 ò0 ò0 ( x + 4) dy dx » 488.39

7–39 ..

Calculus

Chapter 7

6. Use a graphing calculator to graph the region R bounded by the graphs of the indicated equations. Use approximation techniques to find intersection points correct to two decimal places. Describe R in set notation with double inequalities, and evaluate the indicated integral correct to two decimal places.

y = x + 3, y = x 2 , x = 0;

òò12 xy dA R

The region bounded by the above equations is

The point of intersection is found by using the intersect feature on the calculator. The point of intersection is (2.30, 5.30). Therefore, the region of integration R is defined by R = {( x, y ) | x 2 £ y £ x + 3, 0 £ x £ 2.30}.

2.30 x +3 2.30 2 y = x +3 òx 2 (12 xy ) dy dx = ò0 (6 xy ) y = x2

ò0

dx

= ò02.30 (6 x( x + 3) 2 ) - (6 x( x 2 ) 2 ) dx = ò02.30 (- 6 x5 + 6 x3 + 36 x 2 + 54 x) dx

= - x6 + 32 x 4 + 12 x3 + 27 x 2

2.30 0

= (- (2.30)6 + 32 (2.30) 4 + 12(2.30)3 + 27(2.30) 2 ) (- (0)6 + 32 (0)4 + 12(0)3 + 27(0)2 ) 2.30 x +3 òx2 (12 xy ) dy dx » 182.77

ò0

7–40 ..

Calculus

Chapter 8

Name ________________________________ Date ______________ Class ____________

Section 8-1 Trigonometric Functions Review Goal: To review the basic concepts involved with trigonometric functions

Conversion Factors:

1 =

 rad

1 rad =

180

Definition:

180



Sine and Cosine Functions with Real-Number Domains

sin x = sin( x radians) and Definition:

cos x = cos( x radians)

Four Other Trigonometric Functions

tan x =

sin x , cos x ¹ 0 cos x

sec x =

1 , cos x ¹ 0 cos x

cot x =

cos x , sin x ¹ 0 sin x

csc x =

1 , sin x ¹ 0 sin x

In Problems 1 and 2, convert the given degree measure to radian measure.

1.

210o

210o =

2.

210o p 210p 7p = = rad 1 180 180 6

- 225o - 225o p - 225p 5p - 225 = = =rad 1 180 180 4 o

8–1 ..

Calculus

Chapter 8

In Problems 3 and 4, convert the given radian measure to degree measure.

3.

p 10

p p 180o 180o p = = = 18o 10 10 p 10p

4.

-

3p 4

-

3p - 3p 180o - 540o p = = = - 135o 4 4 p 4p

In Problems 5–10, find the exact value of each expression.

5.

sin 30

1 Using Figure 8 on page 497 gives sin 30 = .

2 6.

cos 45 Using Figure 8 on page 497 gives cos 45 =

7.

2 .

2

cos  Using Figure 6 on page 496 gives cos  = - 1.

8.

æ 2 ö sin çè 3 ÷ ø

3 æ 2 ö =.

Using Figures 5 and 8 on pages 496 and 497, respectively gives sin ç÷ è 3 ø 2

8–2 ..

Calculus 9.

Chapter 8

cos(- 60 ) 1 Using Figures 5 and 8 on pages 496 and 497, respectively gives cos(- 60 ) = .

2

10.

sin135 Using Figures 5 and 8 on pages 496 and 497, respectively gives sin135 =

2 .

2

In Problems 11–14, use a calculator in radian or degree mode, as appropriate, to find the value of each expression to four decimal places.

11.

tan(- 73o )

The angle measurement is in degrees, we must make sure that the calculator is in degree mode. Therefore, tan(- 73o ) = - 3.2709.

12.

sin(2.13)

The angle measurement is in radians, we must make sure that the calculator is in radian mode. Therefore, sin(2.13) = 0.8477.

13.

æ7p ö sec ç ÷ è 8 ø

The angle measurement is in radians, we must make sure that the calculator is in 1 1 æ7p ö radian mode. Therefore, sec ç ÷ = = = - 1.0824.

7 p è 8 ø cos( ) - 0.923880 8

14.

cos(169o )

The angle measurement is in degrees, we must make sure that the calculator is in degree mode. Therefore, cos(169o ) = - 0.9816.

8–3 ..

Calculus

Chapter 8

8–4 ..

Calculus

Chapter 8

Name ________________________________ Date ______________ Class ____________

Section 8-2 Derivatives of Trigonometric Functions Goal: To find the first derivatives of the trigonometric functions

Derivatives: Basic Form

d sin x = cos x dx

d cos x = - sin x dx

d tan x = sec2 x dx

d cot x = - csc2 x dx

d sec x = sec x tan x dx

d csc x = - csc x cot x dx

Derivatives: Generalized Form For u = u ( x),

d du sin u = cos u dx dx

d du cos u = - sin u dx dx

d du tan u = sec2 u dx dx

d du cot u = - csc2 u dx dx

d du sec u = sec u tan u dx dx

d du csc u = - csc u cot u dx dx

8–5 ..

Calculus

Chapter 8

In Problems 1–6, find the indicated derivative.

1.

d (cos(5 x + 3)) dx The angle measurement is given in function form, therefore use the generalized form.

d du cos u = - sin u dx dx d d cos(5 x + 3) = - sin(5 x + 3) (5 x + 3) dx dx = - sin(5 x + 3)(5) = - 5sin(5 x + 3)

2.

d (12 tan x) dx The angle measurement is given in basic form, therefore use the basic form.

d tan x = sec2 x dx d 12 tan x = 12sec2 x dx

3.

d [7 x(sin(7 x + 2))] dx

The function is a product and the trigonometric function has an angle measurement in function form. Therefore use the product rule and the generalized trigonometric derivative.

d [7 x(sin(7 x + 2))] = 7 x(cos(7 x + 2)(7) + sin(7 x + 2)(7) dx = 49 x cos(7 x + 2) + 7 sin(7 x + 2)

8–6 ..

Calculus

4.

Chapter 8

d ætan t ö ç ÷ dt è sec t ø

The function is a quotient and the trigonometric function has an angle measurement in basic form. Therefore use the quotient rule and the basic trigonometric derivatives.

d ætan t ö sec t (sec 2 t ) - tan t (sec t tan t ) ç ÷= dt è sec t ø (sec t ) 2 = =

5.

sec3 t - tan 2 t sec t sec2 t sec2 t - tan 2 t sec t

d (tan(3p u ) du The angle measurement is given in function form, therefore use the generalized form.

d (tan(3p u ) = sec2 (3p u )(3p ) du = 3p sec2 (3p u )

6.

d (sec t )6 dt

The function is a power and the trigonometric function has an angle measurement in basic form. Therefore use the power rule and the basic trigonometric derivatives.

d (sec t )6 = 6(sec t )5 (sec t tan t ) dt = 6 tan t sec6 t

8–7 ..

Calculus

Chapter 8

8–8 ..

Calculus

Chapter 8

Name ________________________________ Date ______________ Class ____________

Section 8-3 Integration of Trigonometric Functions Goal: To evaluate indefinite and definite integrals that involve trigonometric functions

Integrals:

Indefinite Integrals of Sine and Cosine

òsin u du = - cos u + C

òcos u du = sin u + C

In Problems 1–5, find each indefinite integral.

1.

òcos(5 x) dx

Let u = 5 x, therefore du = 5dx. Use the method of substitution and the integral of the cosine to solve the problem.

1 òcos(5 x) dx = òcos u du 5 1 = (sin u + C ) 5 sin(5 x) = +C 5

8–9 ..

Calculus 2.

Chapter 8

5 ò(cos x) sin x dx

Let u = cos x, therefore du = - sin x dx. Use the method of substitution to solve the problem.

5 5 ò(cos x) sin x dx = - òu du

3.

=-

u6 +C 6

=-

(cos x)6 +C 6

2 ò(6 x + 5)sin(3x + 5 x - 7) dx

Let u = 3x 2 + 5 x - 7, therefore du = (6 x + 5)dx. Use the method of substitution and the integral of the sine to solve the problem.

2 ò(6 x + 5) sin(3x + 5 x - 7) dx = òsin u du = - cos u + C

= - cos(3x 2 + 5 x - 7) + C

4.

ò

sin 5 x dx cos 5 x

Let u = cos(5 x), therefore du = - 5sin(5x) dx. Use the method of substitution to solve the problem.

ò

sin 5 x 1 1 dx = - ò du cos 5 x 5 u 1 = - ln u + C 5 1 = - ln cos 5 x + C 5

8–10 ..

Calculus

5.

Chapter 8

ò

csc x dx sec x

Rewrite the problem in terms of the sine and cosine functions.

ò

csc x cos x dx = ò dx sec x sin x

Let u = sin x, therefore du = cos x dx. Use the method of substitution to solve the problem.

ò

cos x 1 dx = ò du sin x u = ln u + C = ln sin x + C

In Problems 6–8, evaluate each of the definite integrals. 

6.



2 cos x dx = sin x 2

ò0

0

= sin  - sin 0 2

= 1- 0 

ò02 cos x dx = 1

3

7.

3

ò2 sin x dx = - cos x 2 4

4

= - cos( 32 ) - (- cos( 4 )) = 0 - (- 0.7071) 3

ò2 sin x dx = 0.7071 4

8–11 ..

Calculus

8.

Chapter 8

7

7

ò3 sin x dx = - cos x 3

= - cos 7 - (- cos 3) = 0.753902 - (- 0.989992) 7

ò3 sin x dx » 1.7439 Be sure that your calculator is in radian mode to solve this problem.

8–12 ..