[michael_sullivan,_kathleen_miranda]_instructor's_(bookzz.org).pdf

1Chapter P: Preparing for Calculus

P.1: Functions and Their Graphs

Concepts and Vocabulary

1. If f is a function defined by y = f(x), then x is called the independent variable and y is the

dependent variable.

2. True . The independent variable is sometimes referred to as the argument of the function.

3. False . If no domain is specified for a function f , then the domain of f is taken to be the largest setof real numbers for which the value f(x) is defined and is a real number.

4. False . The domain of the function f(x) =3(x2 1)x 1 is the set {x|x 6= 1}.

5. False . A function can have at most one y-intercept; otherwise, there would be more than one y-valuecorresponding to x = 0.

6. A set of points in the xy-plane is the graph of a function if and only if every vertical line intersectsthe graph in at most one point.

7. If the point (5,3) is on the graph of f , then f( 5 ) = 3 .

8. Let f(x) = ax2 + 4. For the point (1, 2) to be on the graph of f , we must have f(1) = 2. Butf(1) = a(1)2 + 4 = a+ 4, so we must have a+ 4 = 2, or a = 2 .

9. A function f is (a) increasing on an interval I if, for any choice of x1 and x2 in I, with x1 < x2,

then f(x1) < f(x2).

10. A function is (a) even if for every number x in its domain, the number x is also in the domain andf(x) = f(x). A function f is (b) odd if for every number x in its domain, the number x is alsoin the domain and f(x) = f(x).

11. False . Even functions have graphs that are symmetric with respect to the y-axis. Odd functions havegraphs that are symmetric with respect to the origin.

12. The average rate of change of f(x) = 2x3 3 from 0 to 2 is

f(2) f(0)2 0 =

2(2)3 3 (2(0)3 3)2

=13 (3)

2=

16

2= 8 .

Practice Problems

13. Let f(x) = 3x2 + 2x 4. Then

(a) f(0) = 3(0)2 + 2(0) 4 = 4 .(b) f(x) = 3(x)2 + 2(x) 4 = 3x2 2x 4 .(c) f(x) = (3x2 + 2x 4) = 3x2 2x+ 4 .(d) f(x+ 1) = 3(x+ 1)2 + 2(x+ 1) 4 = 3(x2 + 2x+ 1) + 2x+ 2 4 = 3x2 + 8x+ 1 .

2(e) f(x+h) = 3(x+h)2+2(x+h)4 = 3(x2+2xh+h2)+2x+2h4 = 3x2 + 6xh+ 3h2 + 2x+ 2h 4 .

14. Let f(x) =x

x2 + 1. Then

(a) f(0) =0

02 + 1= 0 .

(b) f(x) = x(x)2 + 1 =

x

x2 + 1.

(c) f(x) = xx2 + 1

.

(d) f(x+ 1) =x+ 1

(x+ 1)2 + 1=

x+ 1

x2 + 2x+ 2.

(e) f(x+ h) =x+ h

(x+ h)2 + 1=

x+ h

x2 + 2xh+ h2 + 1.

15. Let f(x) = |x|+ 4. Then

(a) f(0) = |0|+ 4 = 4 .(b) f(x) = | x|+ 4 = |x|+ 4 .

(c) f(x) = (|x|+ 4) = |x| 4 .

(d) f(x+ 1) = |x+ 1|+ 4 .

(e) f(x+ h) = |x+ h|+ 4 .

16. Let f(x) =

3 x. Then

(a) f(0) =

3 0 = 3 .

(b) f(x) =

3 (x) = 3 + x .

(c) f(x) = 3 x .

(d) f(x+ 1) =

3 (x+ 1) = 2 x .

(e) f(x+ h) =

3 (x+ h) = 3 x h .

17. Because f(x) = x31 is defined for any real number x, the domain of f is the set of all real numbers ,or in interval notation (,) .

18. Because x2 + 1 is never equal to zero for any real number x, f(x) =x

x2 + 1is defined for any real

number x. The domain of f is therefore the set of all real numbers , or in interval notation (,) .

19. Because the square root of a negative number is not a real number, the value of t2 9 must benonnegative. The solution of the inequality t2 9 0 is {t|t 3} {t|t 3}, so the domain of v isthe set of real numbers {t|t 3} {t|t 3} , or in interval notation (,3] [3,) .

20. Because the expression x 1 appears under the square root and in the denominator, the value of x 1must be positive; that is, x 1 > 0. The solution of this inequality is {x|x > 1}, so the domain of g isthe set of real numbers {x|x > 1} , or in interval notation (1,) .

321. Because division by zero is not defined, x3 4x = x(x2 4) = x(x 2)(x+ 2) cannot be zero; that is,x 6= 0, x 6= 2, and x 6= 2. Thus, the domain of h is the set of real numbers {x|x 6= 2, x 6= 0, x 6= 2} .

22. Because the square root of a negative number is not a real number, the value of t + 1 must benonnegative. The solution of the inequality t + 1 0 is {t|t 1}. Moreover, because division byzero is not defined, t 5 cannot be 0; that is, t cannot be equal to 5. The intersection of the sets{t|t 1} and {t|t 6= 5} is the set {t| 1 t < 5} {t|t > 5}. The domain of s is therefore the set ofreal numbers {t| 1 t < 5} {t|t > 5} , or in interval notation [1, 5) (5,) .

23. Let f(x) = 3x+ 1. Then

f(x+ h) f(x)h

=3(x+ h) + 1 (3x+ 1)

h=3x 3h+ 1 + 3x 1

h=3hh

= 3 .

24. Let f(x) =1

x+ 3. Then

f(x+ h) f(x)h

=1

x+h+3 1x+3h

(x+ h+ 3)(x+ 3)(x+ h+ 3)(x+ 3)

=x+ 3 (x+ h+ 3)h(x+ h+ 3)(x+ 3)

=h

h(x+ h+ 3)(x+ 3)= 1

(x+ h+ 3)(x+ 3).

25. Let f(x) =x+ 7. Then

f(x+ h) f(x)h

=

x+ h+ 7x+ 7

hx+ h+ 7 +

x+ 7

x+ h+ 7 +x+ 7

=x+ h+ 7 (x+ 7)

h(x+ h+ 7 +

x+ 7)

=h

h(x+ h+ 7 +

x+ 7)

=1

x+ h+ 7 +x+ 7

.

26. Let f(x) =2x+ 7

. Then

f(x+ h) f(x)h

=

2x+h+7

2x+7

hx+ h+ 7

x+ 7

x+ h+ 7x+ 7

=2x+ 7 2x+ h+ 7hx+ h+ 7

x+ 7

x+ 7 +

x+ h+ 7

x+ 7 +x+ h+ 7

=2(x+ 7) 2(x+ h+ 7)

hx+ h+ 7

x+ 7(

x+ 7 +

x+ h+ 7)

=2h

hx+ h+ 7

x+ 7(

x+ 7 +

x+ h+ 7)

= 2x+ h+ 7

x+ 7(

x+ 7 +

x+ h+ 7)

.

27. Let f(x) = x2 + 2x. Then

f(x+ h) f(x)h

=(x+ h)2 + 2(x+ h) (x2 + 2x)

h=x2 + 2xh+ h2 + 2x+ 2h x2 2x

h

=2xh+ h2 + 2h

h=h(2x+ h+ 2)

h= 2x+ h+ 2 .

428. Let f(x) = (2x+ 3)2. Then

f(x+ h) f(x)h

=(2(x+ h) + 3)2 (2x+ 3)2

h=

4(x+ h)2 + 12(x+ h) + 9 (4x2 + 12x+ 9)h

=4x2 + 8xh+ 4h2 + 12x+ 12h+ 9 4x2 12x 9

h

=8xh+ 4h2 + 12h

h=h(8x+ 4h+ 12)

h= 8x+ 4h+ 12 .

29. This graph fails the vertical line test (for example, the vertical line x = 2 intersects the graph in two

points), so it does not represent a function .

30. This graph passes the vertical line test, and so it does represent a function .

(a) From the graph, the domain of the function is the set of all real numbers or the interval

(,) , and the range of the function is the set of all positive real numbers or the inter-val (0,) .

(b) The graph has no x-intercepts, and the y-intercept is 1. Thus, the only intercept is (0, 1) .

(c) The graph is not symmetric with respect to the x-axis, the y-axis, or the origin.

31. This graph passes the vertical line test, and so it does represent a function .

(a) From the graph, the domain of the function is the set {x| pi x pi} or the closed interval[pi, pi] , and the range of the function is the set {y| 1 y 1} or the closed interval [1, 1] .

(b) The graph has x-intercepts ofpi2

and a y-intercept of 1, so the intercepts are(pi

2, 0)

and (0, 1) .

(c) The graph is symmetric with respect to the y-axis but not symmetric with respect to the x-axisor the origin.

32. This graph fails the vertical line test (for example, the vertical line x = 2 intersects the graph in two

points), so it does not represent a function .

33. (a) f(1) = 1 + 3 = 2; f(0) = 0 + 3 = 3; f(1) = 5; f(8) = 8 + 2 = 6(b) The graph of f consists of three pieces corresponding to each equation in the definition. On the

interval [2, 1), the graph is the line y = x + 3, and on the interval (1,), the graph is the liney = x+ 2. The graph also contains the point (1, 5).

-2 -1 1 2 3

-1

1

2

3

4

5

5(c) The individual components of f have domains of {x| 2 x < 1}, {x|x = 1}, and {x|x > 1}.The domain of f is the union of these three sets; that is, {x|x 2} in set notation or [2,)in interval notation. Moreover, the individual components of f have ranges of {y|1 y < 4},{y|y = 5}, and {y|y < 1}. The range of f is the union of these three sets; that is, {y|y = 5 or y < 4}in set notation or (, 4) {5} in interval notation. The x-intercept is 2, and the y-interceptis 3, so the intercepts are (2, 0) and (0, 3) .

34. (a) f(1) = 2(1) + 5 = 3; f(0) = 3; f(1) = 5(1) = 5; f(8) = 5(8) = 40

(b) The graph of f consists of three pieces corresponding to each equation in the definition. On theinterval [3, 0), the graph is the line y = 2x+ 5, and on the interval (0,), the graph is the liney = 5x. The graph also contains the point (0,3).

-3 -2 -1 1 2 3

-15

-10

-5

5

(c) The individual components of f have domains of {x| 3 x < 0}, {x|x = 0}, and {x|x > 0}.The domain of f is the union of these three sets; that is, {x|x 3} in set notation or [3,)in interval notation. Moreover, the individual components of f have ranges of {y| 1 y < 5},{y|y = 3}, and {y|y < 0}. The range of f is the union of these three sets; that is, {y|y < 5} inset notation or (, 5) in interval notation. The x-intercept is 5

2, and the y-intercept is 3,

so the intercepts are

(5

2, 0

)and (0,3) .

35. (a) f(1) = 1 + (1) = 0; f(0) = 02 = 0; f(1) = 12 = 1; f(8) = 82 = 64

(b) The graph of f consists of two pieces corresponding to each equation in the definition. On theinterval (, 0), the graph is the line y = 1 + x, and on the interval [0,), the graph is theparabola y = x2.

6-5 -4 -3 -2 -1 1 2 3

-4

-2

2

4

6

8

(c) The individual components of f have domains of {x|x < 0} and {x|x 0}. The domain of f isthe union of these two sets; that is, all real numbers or the interval (,) . Moreover, theindividual components of f have ranges of {y|y < 1} and {y|y 0}. The range of f is the unionof these two sets; that is, all real numbers or the interval (,) . The x-intercepts are 1and 0, and the y-intercept is 0, so the intercepts are (1, 0) and (0, 0) .

36. (a) f(1) = 11 = 1; f(0) =3

0 = 0; f(1) = 3

1 = 1; f(8) = 3

8 = 2

(b) The graph of f consists of two pieces corresponding to each equation in the definition. On theinterval (, 0), the graph is y = 1x , and on the interval [0,), the graph is y = 3

x.

-4 -2 2 4 6 8 10

-8

-6

-4

-2

2

4

(c) The individual components of f have domains of {x|x < 0} and {x|x 0}. The domain of f isthe union of these two sets; that is, all real numbers or the interval (,) . Moreover, theindividual components of f have ranges of {y|y < 0} and {y|y 0}. The range of f is the unionof these two sets; that is, all real numbers or the interval (,) . The x-intercept is 0, andthe y-intercept is 0, so the only intercept is (0, 0) .

37. Because the graph of f includes the point (0, 3), f(0) = 3 ; because the graph also includes the point

(6,3), f(6) = 3 .

38. Because the graph of f lies above the x-axis at x = 3, f(3) is positive .

739. Because the graph of f lies below the x-axis at x = 4, f(4) is negative .

40. Because the graph of f includes the points (3, 0), (6, 0), and (10, 0), f(x) = 0 forx = 3, x = 6, and x = 10 .

41. Because the graph of f lies above the x-axis for 3 < x < 6 and for 10 < x 11, f(x) > 0 for3 < x < 6 and for 10 < x 11 . In interval notation, this can be written as (3, 6) (10, 11] .

42. The points on the graph of f have x-coordinates between 6 and 11 inclusive. The domain of f istherefore the set of real numbers {x| 6 x 11} or the closed interval [6, 11] .

43. The points on the graph of f have y-coordinates between 3 and 3 inclusive. The range of f is thereforethe set of real numbers {y| 3 y 3} or the closed interval [3, 3] .

44. Because the graph of f includes the points (3, 0), (6, 0), and (10, 0), the x-intercepts are 3, 6, and 10 .

45. Because the graph of f includes the point (0, 3), the y-intercept is 3 .

46. The graph of the horizontal line y =1

2will intersect the graph of f three times .

47. The graph of the vertical line x = 5 will intersect the graph of f once .

48. Because the graph of f includes the point (x, 3) for 0 x 4, f(x) = 3 for 0 x 4 .

49. Because the graph of f includes the points (5,2) and (8,2), f(x) = 2 for x = 5 and forx = 8 .

50. The function f is increasing on the intervals (6, 0) and (8, 11) .

51. The function f is decreasing on the interval (4, 8) .

52. The function f is constant on the interval (0, 4) .

53. The function f is nonincreasing (that is, the function is constant or decreasing) on the interval (0, 8) .

54. The function f is nondecreasing (that is, the function is constant or increasing) on the intervals

(6, 4) and (8, 11) .

55. Because division by zero is not defined, x 6 cannot be equal to 0. The domain of g is therefore theset {x|x 6= 6} .

56. Note that g(3) =3 + 2

3 6 = 5

3. Because g(3) 6= 14, the point (3, 14) is not on the graph of g.

57. If x = 4, then g(x) = g(4) =4 + 2

4 6 = 3 ; therefore, the point (4,3) is on the graph of g.

58. If g(x) = 2, thenx+ 2

x 6 = 2. Multiplying both sides of this equation by x 6 yields

x+ 2 = 2(x 6) = 2x 12,

so that x = 14 . Accordingly, the point (14, 2) is on the graph of g.

859. The x-intercepts occur when g(x) = 0. The value of g can only be zero when its numerator is zero;therefore, g(x) = 0 only when x + 2 = 0. Thus, the graph of g has only one x-intercept, and this

intercept is 2 .

60. Because g(0) =0 + 2

0 6 = 1

3, the y-intercept is 1

3.

61. The domain of h is {x|x 6= 1}, so for every number x in the domain of h, the number x is also inthe domain. Moreover,

h(x) = x(x)2 1 =

x

x2 1 = h(x),

so that the function h is odd . Because h is an odd function, its graph is symmetric with respect to

the origin .

62. The domain of f is all real numbers, so for every number x in the domain of f , the number x is alsoin the domain. Moreover,

f(x) = 3

3(x)2 + 1 = 3

3x2 + 1 = f(x),

so that the function f is even . Because f is an even function, its graph is symmetric with respect to

the y-axis .

63. The domain of G is {x|x 0}. The number x = 1 is in the domain of G, but x = 1 is not; therefore,the function G is neither even nor odd . Because G is neither an even nor an odd function, its graph

is not symmetric with respect to either the y-axis or the origin .

64. The domain of F is {x|x 6= 0}, so for every number x in the domain of F , the number x is also inthe domain. Moreover,

F (x) = 2(x)| x| = 2x

|x| = F (x),

so that the function F is odd . Because F is an odd function, its graph is symmetric with respect to

the origin .

65. Let f(x) = 2x2 + 4.(a) The average rate of change of f from 1 to 2 is

f(2) f(1)2 1 =

4 22 1 =

61

= 6 .

(b) The average rate of change of f from 1 to 3 is

f(3) f(1)3 1 =

14 23 1 =

162

= 8 .

(c) The average rate of change of f from 1 to 4 is

f(4) f(1)4 1 =

28 24 1 =

303

= 10 .

(d) The average rate of change of f from 1 to x for x 6= 1 is

f(x) f(1)x 1 =

2x2 + 4 2x 1 =

2(1 x2)x 1 =

2(1 x)(1 + x)x 1 = 2(1 + x) .

966. Let s(t) = 20 0.8t2.

(a) The average rate of change of s from 1 to 4 is

s(4) s(1)4 1 =

7.2 19.24 1 =

123

= 4 .

(b) The average rate of change of s from 1 to 3 is

s(3) s(1)3 1 =

12.8 19.23 1 =

6.42

= 3.2 .

(c) The average rate of change of s from 1 to 2 is

s(2) f(1)2 1 =

16.8 19.22 1 =

2.41

= 2.4 .

(d) The average rate of change of s from 1 to t for t 6= 1 is

s(t) s(1)t 1 =

20 0.8t2 19.2t 1 =

0.8(1 t2)t 1 =

0.8(1 t)(1 + t)t 1 = 0.8(1 + t) .

67. For 1 x < 0, the graph is a line through the points (1, 1) and (0, 0). The slope of this lineis m = 1, and the y-intercept is 0; therefore, the equation for this component of the function isy = x. For 0 x 2, the graph is a line through the points (0, 0) and (2, 1). The slope of this line ism = 12 , and the y-intercept is 0; therefore, the equation for this component of the function is y =

12x.

Combining these two equations, the definition for this piecewise function is

f(x) =

{ x, if 1 x < 012x, if 0 x 2.

The domain of f is the set {x| 1 x 2} or the closed interval [1, 2] , and the range is the set{y|0 y 1} or the closed interval [0, 1] .

68. For x 0, the graph is a line with slope 1 through the origin, and for 0 < x 2, the graph is ahorizontal line at height y = 1. The definition for this piecewise function is therefore

f(x) =

{x, if x 01, if 0 < x 2.

The domain of f is {x|x 2} in set notation or (, 2] in interval notation, and the range is{y|y = 1 or y 0} in set notation or (, 0] {1} in interval notation.

69. For x < 1, the graph is a horizontal line at height y = 1; at x = 1, the value of the function is 0; andfor 1 < x 2, the graph is a line with slope 1 and x-intercept 2. The definition for this piecewisefunction is therefore

f(x) =

1, if x < 10, if x = 12 x, if 1 < x 2.

The domain of f is {x|x 2} in set notation or (, 2] in interval notation, and the range is{y|y = 1 or 0 y < 1} in set notation or {1} [0, 1) in interval notation.

10

70. For x < 1, the graph is a horizontal line at height y = 2; at x = 1, the value of the function is0; and for 1 < x 1, the graph is a line with slope 2 and y-intercept 1. The definition for thispiecewise function is therefore

f(x) =

2, if x < 10, if x = 12x 1, if 1 < x 1.The domain of f is {x|x 1} in set notation or (, 1] in interval notation, and the range is{y|y = 2 or 3 y < 1} in set notation of [3, 1) {2} in interval notation.

71. The definition for this piecewise function is

f(x) =

x3, if 2 < x < 1

0, if x = 1x2, if 1 < x 3.

The domain of f is the set {x| 2 < x 3} or the interval (2, 3] , and the range is the set{y| 1 < y 9} or the interval (1, 9] .

72. The definition for this piecewise function is

f(x) =

{x2 1, if 3 x 0x3, if 0 < x 2.

The domain of f is the set {x| 3 x 2} or the closed interval [3, 2] , and the range is the set{y| 1 y 8} or the closed interval [1, 8] .

73. (a) The cost of manufacturing 100 road bikes is

C(100) = 0.004(100)3 0.6(100)2 + 250(100) + 100, 500 = 123, 500;the cost of manufacturing 101 road bikes is

C(101) = 0.004(101)3 0.6(101)2 + 250(101) + 100, 500 123, 750.60.The average rate of change of C from 100 to 101 road bikes is therefore

C(101) C(100)101 100

123, 750.60 123, 5001

= $250.60 per bike .

(b) The cost of manufacturing 500 road bikes is

C(500) = 0.004(500)3 0.6(500)2 + 250(500) + 100, 500 = 575, 500;the cost of manufacturing 501 road bikes is

C(501) = 0.004(501)3 0.6(501)2 + 250(501) + 100, 500 578, 155.40.The average rate of change of C from 500 to 501 road bikes is therefore

C(501) C(500)501 500

578, 155.40 575, 5001

= $2655.40 per bike .

(c) Answers will vary. One possible interpretation is that the unit cost per road bike increases as thenumber of road bikes manufactured increases.

11

74. (a) The cost of producing 500 tons of steel is

C(500) =1

10(500)2 + 5(500) + 1500 = 29000;

the cost of producing 501 tons of steel is

C(501) =1

10(501)2 + 5(501) + 1500 = 29105.1;

The average rate of change of C from 500 to 501 tons of steel is therefore

C(501) C(500)501 500 =

29105.1 290001

= $105.10 per ton .

(b) The cost of producing 1000 tons of steel is

C(1000) =1

10(1000)2 + 5(1000) + 1500 = 106, 500;

the cost of producing 1001 tons of steel is

C(1001) =1

10(1001)2 + 5(1001) + 1500 = 106, 705.1;

The average rate of change of C from 1000 to 1001 tons of steel is therefore

C(1001) C(1000)1001 1000 =

106.705.1 106, 5001

= $205.10 per ton .

(c) Answers will vary. One possible interpretation is that the unit cost per ton of steel increases asthe number of tons produced increases.

12

P.2: Library of Functions; Mathematical Modeling


1. The function f(x) = x2 is (a) increasing on the interval (0,).

2. True . The floor function f(x) = bxc is an example of a step function.

3. True . The cube function is odd and is increasing on the interval (,).

4. False . The cube root function is odd, but is increasing on the interval (,).

5. False . The domain and range of the reciprocal function are all real nonzero numbers.

6. A number r for which f(r) = 0 is called a zero of the function f .

7. If r is a zero of even multiplicity of a function f , the graph of f (b) touches the x-axis at r.

8. True . The x-intercepts of the graph of a polynomial function are called zeros of the function.

9. True . The function f(x) =[x+ 3x2 pi]2/3 is an algebraic function.

10. False . The domain of every rational function is the set of all real numbers except those for which thevalue of the denominator polynomial is zero.

Practice Problems

11. This is the graph of a cube function (D) .

12. This is the graph of an identity function (B) .

13. This is the graph of a reciprocal function (F) .

14. This is the graph of a constant function (A) .

15. This is the graph of a square function (C) .

16. This is the graph of a square root function (E) .

17. This is the graph of an absolute value function (G) .

18. This is the graph of a cube root function (H) .

19. Let f(x) = b2xc. Then:(a) f(1.2) = b2(1.2)c = b2.4c = 2 .(b) f(1.6) = b2(1.6)c = b3.2c = 3 .(c) f(1.8) = b2(1.8)c = b3.6c = 4 .

20. Let f(x) =x

2

. Then:

(a) f(1.2) =

1.2

2

= d0.6e = 1 .

13

(b) f(1.6) =

1.6

2

= d0.8e = 1 .

(c) f(1.8) =1.8

2

= d0.9e = 0 .

21. Let f(x) = 3(x 7)(x+ 4)3.(a) The zeros of f are: x = 7 with multiplicity one and x = 4 with multiplicity three .(b) The x-intercepts of the graph of f are the zeros 7 and 4. The y-intercept is

f(0) = 3(0 7)(0 + 4)3 = 1344 .(c) Because both x-intercepts are zeros with odd multiplicity, the graph of f crosses the x-axis at

each x-intercept.

22. Let f(x) = 4x(x2 + 1)(x 2)3.(a) The zeros of f are: x = 0 with multiplicity one and x = 2 with multiplicity three . Note that

the factor x2 + 1 is never equal to zero for any real number x.

(b) The x-intercepts of the graph of f are the zeros 0 and 2. The y-intercept is

f(0) = 4(0)(02 + 1)(0 2)3 = 0 .(c) Because both x-intercepts are zeros with odd multiplicity, the graph of f crosses the x-axis at

each x-intercept.

23. The graph of f has x-intercepts of 0, 1, and 2, so the function must have factors of x, x 1, and x 2,which is true for all of the listed functions. Because the graph crosses the x-axis at each x-intercept,the zeros must each have odd multiplicity. This rules out functions (b) and (d). Finally, f becomesunbounded in the positive direction as x becomes unbounded in the positive direction, and f becomesunbounded in the negative direction as x becomes unbounded in the negative direction. This rules out

function (a). Thus, functions (c), (e), and (f) might have the given graph.

24. The graph of f has x-intercepts of 0, 1, and 2, so the function must have factors of x, x 1, and x 2,which is true for all of the listed functions. Because the graph crosses the x-axis at x = 0 and at x = 2but just touches the x-axis at x = 1, the zeros 0 and 2 must have odd multiplicity while the zero 1must have even multiplicity. This rules out functions (a), (b), and (d). Finally, f becomes unboundedin the positive direction as x becomes unbounded in either direction. Each of the remaining functions

satisfies this condition. Thus, functions (c), (e), and (f) might have the given graph.

25. The domain of R is the set {x|x 6= 3} . Because 0 is in the domain of R and R(0) = 0, the y-interceptis 0. The zeros of R are solutions of the equation 5x2 = 0, or x = 0. Because 0 is in the domain of R,

the x-intercept is 0. Thus, the only intercept is (0, 0) .

26. The domain of H is the set {x|x 6= 2, x 6= 4} . Because 0 is in the domain of H and H(0) = 0, they-intercept is 0. The zeros of H are solutions of the equation 4x2 = 0, or x = 0. Because 0 is in thedomain of H, the x-intercept is 0. Thus, the only intercept is (0, 0) .

27. Because the denominator x2 + 4 is never equal to zero for any real number x, the domain of R is the

set of all real numbers or the interval (,) . Because 0 is in the domain of R and R(0) = 0,the y-intercept is 0. The zeros of R are solutions of the equation 3x2 x = x(3x 1) = 0, or x = 0and x = 13 . Because both 0 and

13 are in the domain of R, the x-intercepts are 0 and

13 . Thus, the

intercepts are (0, 0) and

(1

3, 0

).

14

28. The domain of R is the set {x|x 6= 3} . Because 0 is in the domain of R and R(0) = 12 , the y-interceptis 12 . The zeros of R are solutions of the equation 3(x

2 x 6) = 3(x 3)(x + 2) = 0, or x = 3 andx = 2. However, x = 3 is not in the domain of R, so the only x-intercept is 2. Thus, the interceptsare (2, 0) and

(0,

1

2

)29. (a) Let A(x) denote the area of the rectangles as a function of x. From the diagram, each rectangle

has length x and height y. Because the point (x, y) lies on the graph of y = 16 x2, it followsthat the height of each rectangle is 16 x2. Therefore,

A(x) = xy = x(16 x2) = 16x x3 .

(b) Because the point (x, y) must lie in the first quadrant, x 0. As y must also be greater than orequal to 0, x can only be as large as the positive zero of 16 x2. Thus, the domain of A is theset {x|0 x 4} or the closed interval [0, 4] .

30. (a) Let A(x) denote the area of the rectangles as a function of x. From the diagram, each rectanglehas length 2x and height y. Because the point (x, y) is in the first quadrant and lies on the graphof a circle of radius 2 centered at the origin, whose equation is x2 + y2 = 4, it follows that theheight of each rectangle is y =

4 x2. Therefore,

A(x) = 2xy = 2x

4 x2 .

(b) Let p(x) denote the perimeter of the rectangles as a function of x. From the diagram, eachrectangle has length 2x and height y. Because the point (x, y) is in the first quadrant and lies onthe graph of a circle of radius 2 centered at the origin, whose equation is x2 + y2 = 4, it followsthat the height of each rectangle is y =

4 x2. Therefore,

p(x) = 2(2x) + 2y = 4x+ 2

4 x2 .

31. (a) A scatter plot of the data is shown below. The pattern in the points suggests a quadratic

relationship between the height of the ball and the horizontal distance traveled.

20 40 60 80 100 120 140 160 180 200Horizontal distance (feet)

20

30

40

50

60

70

80

Heigh

t (fe

et)

(b) Using the QuadReg function on a TI-84 Plus calculator yields

h(x) = 0.0037x2 + 1.03x+ 5.7

15

as the quadratic function that best fits these data. The ball will hit the ground when h(x) = 0.Using the quadratic formula, this will happen when

x =1.03(1.03)2 4(0.0037)(5.7)

2(0.0037) =1.03 1.070.0074

283.8,5.4

The value x = 5.4 may be disregarded. The ball will therefore hit the ground after traveling ahorizontal distance of approximately 283.8 feet .

(c) When the ball has traveled a horizontal distance of 10 feet, the height will be approximately

h(10) = 0.0037(10)2 + 1.03(10) + 5.7 = 15.63 feet .

32. (a) A scatter plot of the data is shown below. The pattern in the points suggests a cubic relationshipbetween the percentage of the population without a high school diploma and age.

30 40 50 60 70 80Age (years)

10

12

14

16

18

20

22

24

26

Perc

enta

ge w

ithou

t high

scho

ol dip

loma

(b) Using the CubicReg function on a TI-84 Plus calculator yields

P (x) = 0.00026x3 0.0303x2 + 1.0877x 0.5071

as the cubic function that best fits these data.

(c) Based on these data, the approximate percentage of 35-year olds who do not have a high schooldiploma is

P (35) = 0.00026(35)3 0.0303(35)2 + 1.0877(35) 0.5071 11.59 .

16

P.3: Operations on Functions; Graphing Techniques


1. If the domain of function f is {x|0 x 7} and the domain of function g is {x| 2 x 5}, thenthe domain of the sum function f + g is {x|0 x 5} ; that is, the intersection of the domains of fand g.

2. False . If f and g are functions, then the domain off

gconsists of all numbers x that are in the domains

of both f and g, but excluding the numbers x for which g(x) = 0.

3. True . The domain of (f g)(x) consists of the numbers x that are in the domains of both f and g.

4. False . The domain of the composite function (f g)(x) is the set of all numbers x in the domain ofg for which g(x) is in the domain of f .

5. True . The graph of y = f(x) is the reflection of the graph of y = f(x) about the x-axis.

6. False . To obtain the graph of y = f(x+ 2) 3, shift the graph of y = f(x) horizontally to the left 2units and vertically down 3 units.

7. True . Suppose the x-intercepts of the graph of the function f are 3 and 2. Then the x-interceptsof the graph of y = 2f(x) are 3 and 2.

8. Suppose that the graph of a function f is known. Then the graph of the function y = f(x 2) can beobtained by a horizontal shift of the graph of f to the right a distance of 2 units.

9. Suppose that the graph of a function f is known. Then the graph of the function y = f(x) can beobtained by a reflection about the y -axis of the graph of f .

10. Suppose the x-intercepts of the graph of the function f are2, 1, and 5. The x-intercepts of y = f(x+3)are 5, 2, and 2 .

Practice Problems

11. Let f(x) = 3x+ 4 and g(x) = 2x 3. Note the domain of both f and g is the set of all real numbers.

(a) (f + g)(x) = f(x) + g(x) = 3x+ 4 + 2x 3 = 5x+ 1 . The domain of this function is the set ofall real numbers or the interval (,) .

(b) (f g)(x) = f(x) g(x) = 3x+ 4 (2x 3) = x+ 7 . The domain of this function is the set ofall real numbers or the interval (,) .

(c) (f g)(x) = f(x)g(x) = (3x+ 4)(2x 3) = 6x2 x 12 . The domain of this function is the setof all real numbers or the interval (,) .

(d)

(f

g

)(x) =

f(x)

g(x)=

3x+ 4

2x 3 . The domain of this function is the set {x|x 6=32} .

12. Let f(x) = 1 +1

xand g(x) =

1

x. Note the domain of both f and g is the set {x|x 6= 0}.

(a) (f+g)(x) = f(x)+g(x) = 1+1

x+

1

x= 1 +

2

x. The domain of this function is the set {x|x 6= 0} .

17

(b) (f g)(x) = f(x) g(x) = 1 + 1x 1x

= 1 . The domain of this function is the set {x|x 6= 0} .

(c) (f g)(x) = f(x)g(x) =(

1 +1

x

)1

x=

1

x+

1

x2. The domain of this function is the set {x|x 6= 0} .

(d)

(f

g

)(x) =

f(x)

g(x)=

1 + 1x1x

= x+ 1 . The domain of this function is the set {x|x 6= 0} .

13. Let f(x) =x+ 1 and g(x) =

2

x. Note the domain of f is the set {x|x 1} and the domain of g is

the set {x|x 6= 0}.

(a) (f + g)(x) = f(x) + g(x) =x+ 1 +

2

x. The domain of this function is the set

{x| 1 x < 0} {x|x > 0} .

(b) (f g)(x) = f(x) g(x) = x+ 1 2x

. The domain of this function is the set

{x| 1 x < 0} {x|x > 0} .

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

=2x+ 1

x. The domain of this function is the set

{x| 1 x < 0} {x|x > 0} .

(d)

(f

g

)(x) =

f(x)

g(x)=

x+ 12x

=x

2

x+ 1 . The domain of this function is the set

{x| 1 x < 0} {x|x > 0} .

14. Let f(x) = |x| and g(x) = x. Note the domain of both f and g is the set of all real numbers.

(a) (f + g)(x) = f(x) + g(x) = |x|+ x . The domain of this function is the set of all real numbersor the interval (,) .

(b) (f g)(x) = f(x) g(x) = |x| x . The domain of this function is the set of all real numbersor the interval (,) .

(c) (f g)(x) = f(x)g(x) = |x|x . The domain of this function is the set of all real numbers or theinterval (,) .

(d)

(f

g

)(x) =

f(x)

g(x)=|x|x

. The domain of this function is the set {x|x 6= 0} .

15. Let f(x) = 2x and g(x) = 3x2 + 1.

(a) (f g)(4) = f(g(4)) = f(49) = 98 .(b) (g f)(2) = g(f(2)) = g(4) = 49 .(c) (f f)(1) = f(f(1)) = f(2) = 4 .(d) (g g)(0) = g(g(0)) = g(1) = 4 .

16. Let f(x) =3

x+ 1and g(x) =

x.

(a) (f g)(4) = f(g(4)) = f(2) = 1 .

18

(b) (g f)(2) = g(f(2)) = g(1) = 1 .

(c) (f f)(1) = f(f(1)) = f(

3

2

)=

6

5.

(d) (g g)(0) = g(g(0)) = g(0) = 0 .

17. (a) (f g)(1) = f(g(1)) = f(0) = 1 .(b) (f g)(1) = f(g(1)) = f(0) = 1 .(c) (g f)(1) = g(f(1)) = g(3) = 8 .(d) (g f)(1) = g(f(1)) = g(3) = 8 .(e) (g g)(2) = g(g(2)) = g(3) = 8 .(f) (f f)(1) = f(f(1)) = f(3) = 7 .

18. (a) (f g)(1) = f(g(1)) = f(0) = 5 .(b) (f g)(2) = f(g(2)) = f(3) = 11 .(c) (g f)(2) = g(f(2)) = g(1) = 0 .(d) (g f)(3) = g(f(3)) = g(1) = 0 .(e) (g g)(1) = g(g(1)) = g(0) = 1 .(f) (f f)(3) = f(f(3)) = f(1) = 7 .

19. (a) (g f)(1) = g(f(1)) = g(1) = 4 .(b) (g f)(6) = g(f(6)) = g(2) = 2 .(c) (f g)(6) = f(g(6)) = f(5) = 1 .(d) (f g)(4) = f(g(4)) = f(2) = 2 .

20. (a) (g f)(1) = g(f(1)) = g(1) = 3 .(b) (g f)(5) = g(f(5)) = g(1) = 4 .(c) (f g)(7) = f(g(7)) = f(5) = 1 .(d) (f g)(2) = f(g(2)) = f(2) = 2 .

21. Let f(x) = 3x+ 1 and g(x) = 8x. Note the domain of both f and g is the set of all real numbers.

(a) (f g)(x) = f(g(x)) = f(8x) = 3(8x) + 1 = 24x+ 1 . The domain of this function is the set ofall real numbers or the interval (,) .

(b) (g f)(x) = g(f(x)) = g(3x+ 1) = 8(3x+ 1) = 24x+ 8 . The domain of this function is the setof all real numbers or the interval (,) .

(c) (f f)(x) = f(f(x)) = f(3x+ 1) = 3(3x+ 1) + 1 = 9x+ 4 . The domain of this function is theset of all real numbers or the interval (,) .

(d) (g g)(x) = g(g(x)) = g(8x) = 8(8x) = 64x . The domain of this function is the set ofall real numbers or the interval (,) .

22. Let f(x) = x and g(x) = 2x 4. Note the domain of both f and g is the set of all real numbers.

19

(a) (f g)(x) = f(g(x)) = f(2x 4) = (2x 4) = 4 2x . The domain of this function is the setof all real numbers or the interval (,) .

(b) (g f)(x) = g(f(x)) = g(x) = 2(x) 4 = 2x 4 . The domain of this function is the set ofall real numbers or the interval (,) .

(c) (f f)(x) = f(f(x)) = f(x) = (x) = x . The domain of this function is the set ofall real numbers or the interval (,) .

(d) (g g)(x) = g(g(x)) = g(2x 4) = 2(2x 4) 4 = 4x 12 . The domain of this function is theset of all real numbers or the interval (,) .

23. Let f(x) = x2 + 1 and g(x) =x 1. Note the domain of f is the set of all real numbers and the

domain of g is the set {x|x 1}.(a) (f g)(x) = f(g(x)) = f(x 1) = (x 1)2 + 1 = x . The domain of f g begins with the

domain of g: {x|x 1}. Because the domain of f is the set of all real numbers, any value of g(x)will be in the domain of f . Therefore, the domain of f g is the set {x|x 1} .

(b) (g f)(x) = g(f(x)) = g(x2 + 1) = x2 + 1 1 = |x| . The domain of g f begins with thedomain of f : the set of all real numbers. Because the domain of g is the set {x|x 1} and thevalue of f(x) is always a number greater than or equal to one, it follows that the value of f(x) is

always in the domain of g. Therefore, the domain of g f is the set of all real numbers or theinterval (,) .

(c) (f f)(x) = f(f(x)) = f(x2 + 1) = (x2 + 1)2 + 1 = x4 + 2x2 + 2 . The domain of this functionis the set of all real numbers or the interval (,) .

(d) (gg)(x) = g(g(x)) = g(x 1) =

x 1 1 . The domain of gg begins with the domain ofg: {x|x 1}. Next, the value of g(x) = x 1 must be in the domain of g; that is, the inequalityx 1 1 must be satisfied. Solving this inequality yields x 1 1, or x 2. Therefore, the

domain of g g is the set {x|x 2} .

24. Let f(x) = 2x+ 3 and g(x) =x. Note the domain of f is the set of all real numbers and the domain

of g is the set {x|x 0}.

(a) (f g)(x) = f(g(x)) = f(x) = 2x+ 3 . The domain of f g begins with the domain of g:{x|x 0}. Because the domain of f is the set of all real numbers, any value of g(x) will be in thedomain of f . Therefore, the domain of f g is the set {x|x 0} .

(b) (g f)(x) = g(f(x)) = g(2x+ 3) = 2x+ 3 . The domain of g f begins with the domain of f :the set of all real numbers. Next, the value of f(x) = 2x+ 3 must be in the domain of g; that is,the inequality 2x+ 3 0 must be satisfied. Solving this inequality yields x 32 . Therefore, thedomain of g f is the set {x|x 32} .

(c) (f f)(x) = f(f(x)) = f(2x+ 3) = 2(2x+ 3) + 3 = 4x+ 10 . The domain of this function is theset of all real numbers or the interval (,) .

(d) (g g)(x) = g(g(x)) = g(x) =

x = 4x . The domain of g g begins with the domain of

g: {x|x 0}. Next, the value of g(x) = x must be in the domain of g; that is, the inequalityx 0 must be satisfied. Solving this inequality yields x 0. Therefore, the domain of g g is

the set {x|x 0} .

20

25. Let f(x) =x

x 1 and g(x) =2

x. Note the domain of f is the set {x|x 6= 1} and the domain of g is the

set {x|x 6= 0}.

(a) (f g)(x) = f(g(x)) = f(

2

x

)=

2x

2x 1

=2

2 x . The domain of f g begins with the domain ofg: {x|x 6= 0}. Next, the value of g(x) must be in the domain of f ; that is, the value of 2x cannotbe equal to 1. This means that x cannot be equal to 2. The domain of f g is therefore the set{x|x 6= 0, x 6= 2} .

(b) (g f)(x) = g(f(x)) = g(

x

x 1)

=2xx1

=2(x 1)

x. The domain of g f begins with the

domain of f : {x|x 6= 1}. Next, the value of f(x) must be in the domain of g; that is, the valueof xx1 cannot be equal to 0. This means that x cannot be equal to 0. The domain of g f istherefore the set {x|x 6= 0, x 6= 1} .

(c) (f f)(x) = f(f(x)) = f(

x

x 1)

=xx1xx1 1

=x

x (x 1) = x . The domain of f f beginswith the domain of f : {x|x 6= 1}. Next, the value of f(x) must be in the domain of f ; that is,the value of xx1 cannot be equal to 1. The equation

x

x 1 = 1,

however, has no solution. The domain of f f is therefore the set {x|x 6= 1} .

(d) (g g)(x) = g(g(x)) = g(

2

x

)=

22x

= x . The domain of g g begins with the domain of g:{x|x 6= 0}. Next, the value of g(x) must be in the domain of g; that is, the value of 2x cannot beequal to 0. The equation

2

x= 0,

however, has no solution. The domain of g g is therefore the set {x|x 6= 0} .

26. Let f(x) =1

x+ 3and g(x) = 2

x. Note the domain of f is the set {x|x 6= 3} and the domain of g is

the set {x|x 6= 0}.

(a) (f g)(x) = f(g(x)) = f( 2x

)=

1

2x + 3=

x

2 + 3x . The domain of f g begins with thedomain of g: {x|x 6= 0}. Next, the value of g(x) must be in the domain of f ; that is, the valueof 2x cannot be equal to -3. This means that x cannot be equal to 23 . The domain of f g istherefore the set {x|x 6= 0, x 6= 23} .

(b) (g f)(x) = g(f(x)) = g(

1

x+ 3

)= 21

x+3

= 2(x+ 3) . The domain of g f begins with thedomain of f : {x|x 6= 3}. Next, the value of f(x) must be in the domain of g; that is, the valueof 1x+3 cannot be equal to 0. The equation

1

x+ 3= 0,

however, has no solution. The domain of g f is therefore the set {x|x 6= 3} .

21

(c) (f f)(x) = f(f(x)) = f(

1

x+ 3

)=

11

x+3 + 3=

x+ 3

1 + 3(x+ 3)=

x+ 3

3x+ 10. The domain of f f

begins with the domain of f : {x|x 6= 3}. Next, the value of f(x) must be in the domain of f ;that is, the value of 1x+3 cannot be equal to 3. This means that x cannot be equal to 103 . Thedomain of f f is therefore the set {x|x 6= 3, x 6= 103 } .

(d) (g g)(x) = g(g(x)) = g( 2x

)= 2 2x

= x . The domain of g g begins with the domain of g:{x|x 6= 0}. Next, the value of g(x) must be in the domain of g; that is, the value of 2x cannotbe equal to 0. The equation

2x

= 0,

however, has no solution. The domain of g g is therefore the set {x|x 6= 0} .

27. Let f(x) = x4 and g(x) = 2x+ 3 . Then

(f g)(x) = f(g(x)) = f(2x+ 3) = (2x+ 3)4 = F (x).

28. Let f(x) = x3 and g(x) = 1 + x2 . Then

(f g)(x) = f(g(x)) = f(1 + x2) = (1 + x2)3 = F (x).

29. Let f(x) =x and g(x) = x2 + 1 . Then

(f g)(x) = f(g(x)) = f(x2 + 1) =x2 + 1 = F (x).

30. Let f(x) =x and g(x) = 1 x2 . Then

(f g)(x) = f(g(x)) = f(1 x2) =

1 x2 = F (x).

31. Let f(x) = |x| and g(x) = 2x+ 1 . Then(f g)(x) = f(g(x)) = f(2x+ 1) = |2x+ 1| = F (x).

32. Let f(x) = |x| and g(x) = 2x2 + 3 . Then

(f g)(x) = f(g(x)) = f(2x2 + 3) = |2x2 + 3| = F (x).

33. The graph of f(x) = x3+2 can be obtained from the graph of y = x3 by shifting vertically up 2 units .

In the figure below, the graph of y = x3 is represented by the dashed curve, and the graph of y = x3+2is represented by the solid curve.

-2 -1 1 2

-10

-5

5

10

22

34. The graph of g(x) = x31 can be obtained from the graph of y = x3 by shifting vertically down 1 unit .In the figure below, the graph of y = x3 is represented by the dashed curve, and the graph of y = x31is represented by the solid curve.

-2 -1 1 2

-8

-6

-4

-2

2

4

6

8

35. The graph of h(x) =x 2 can be obtained from the graph of y = x by

shifting horizontally right 2 units . In the figure below, the graph of y =x is represented by the

dashed curve, and the graph of y =x 2 is represented by the solid curve.

1 2 3 4 5 6

0.5

1

1.5

2

2.5

36. The graph of f(x) =x+ 1 can be obtained from the graph of y =

x by

shifting horizontally left 1 unit . In the figure below, the graph of y =x is represented by the

dashed curve, and the graph of y =x+ 1 is represented by the solid curve.

-1 1 2 3 4

0.5

1

1.5

2

37. The graph of g(x) = 4x can be obtained from the graph of y =

x by

stretching vertically by a factor of 4 . In the figure below, the graph of y =x is represented by

the dashed curve, and the graph of y = 4x is represented by the solid curve.

23

1 2 3 4

2

4

6

8

38. The graph of f(x) = 12x can be obtained from the graph of y =

x by

compressing vertically by a factor of 12 . In the figure below, the graph of y =x is represented

by the dashed curve, and the graph of y = 12x is represented by the solid curve.

1 2 3 4

0.5

1

1.5

2

39. The graph of f(x) = (x 1)3 + 2 can be obtained from the graph of y = x3 byshifting horizontally right 1 unit and then shifting vertically up 2 units . In the figure below at the

left, the graph of y = x3 is represented by the dashed curve, and the graph of y = (x1)3 is representedby the dash-dot curve. In the figure below at the right, the graph of y = (x 1)3 is represented by thedash-dot curve, and the graph of y = (x 1)3 + 2 is represented by the solid curve.

-1 1 2

-6

-4

-2

2

4

6

-1 1 2 3

-6

-4

-2

2

4

6

8

40. The graph of g(x) = 3(x 2)3 + 1 can be obtained from the graph of y = x3 byshifting horizontally right 2 units , then stretching vertically by a factor of 3 and finally

shifting vertically up 1 unit . In the figure below at the top left, the graph of y = x3 is represented

24

by the dashed curve, and the graph of y = (x 2)3 is represented by the dash-dot curve. In the figurebelow at the top right, the graph of y = (x 2)3 is represented by the dash-dot curve, and the graphof y = 3(x 2)3 is represented by the dotted curve. In the bottom figure, the graph of y = 3(x 2)3 isrepresented by the dotted curve, and the graph of y = 3(x 2)3 + 1 is represented by the solid curve.

-2 -1 1 2 3 4

-8

-6

-4

-2

2

4

6

8

1 2 3 4

-24

-16

-8

8

16

1 2 3 4

-6

-5

-4

-3

-2

-1

1

2

3

4

5

6

7

41. The graph of h(x) = 12x can be obtained from the graph of y =1x by

compressing vertically by a factor of 12 . In the figure below, the graph of y =1x is represented by

the dashed curve, and the graph of y = 12x is represented by the solid curve.

-2 -1 1 2

-4

-2

2

4

(1/2, 1)

(1/2, 2)

(1/2, 2)

(1/2, 1)

42. The graph of f(x) = 4x + 2 can be obtained from the graph of y =1x by

stretching vertically by a factor of 4 and then shifting vertically up 2 units . In the figure below at

the left, the graph of y = 1x is represented by the dashed curve, and the graph of y =4x is represented

25

by the dash-dot curve. In the figure below at the right, the graph of y = 4x is represented by thedash-dot curve, and the graph of y = 4x + 2 is represented by the solid curve.

-2 -1 1 2

-8

-6

-4

-2

2

4

6

8

10

-2 -1 1 2

-8

-6

-4

-2

2

4

6

8

43. First note that because |t| = |t|, the function G can be written as G(x) = 2|(x1)| = 2|x1|. Thegraph of G can therefore be obtained from the graph of y = |x| by shifting horizontally right 1 unitand then stretching vertically by a factor of 2 . In the figure below at the left, the graph of y = |x| isrepresented by the dashed curve, and the graph of y = |x 1| is represented by the dash-dot curve.In the figure below at the right, the graph of y = |x 1| is represented by the dash-dot curve, and thegraph of y = 2|x 1| is represented by the solid curve.

-2 -1 1 2 3

0.5

1

1.5

2

-1 1 2 3

1

2

3

4

44. The graph of g(x) = (x + 1)3 1 can be obtained from the graph of y = x3 byshifting horizontally left 1 unit , then reflecting about the x-axis and finally

shifting vertically down 1 unit . In the figure below at the top left, the graph of y = x3 is repre-

sented by the dashed curve, and the graph of y = (x + 1)3 is represented by the dash-dot curve. Inthe figure below at the top right, the graph of y = (x+ 1)3 is represented by the dash-dot curve, andthe graph of y = (x + 1)3 is represented by the dotted curve. In the bottom figure, the graph ofy = (x + 1)3 is represented by the dotted curve, and the graph of y = (x + 1)3 1 is representedby the solid curve.

26

-3 -2 -1 1 2

-8

-6

-4

-2

2

4

6

8

-3 -2 -1 1

-8

-6

-4

-2

2

4

6

8

-3 -2 -1 1

-8

-6

-4

-2

2

4

6

8

45. The graph of g(x) = 4x 1 can be obtained from the graph of y = x byshifting horizontally right 1 unit , then stretching vertically by a factor of 4 , and finally

reflecting about the x-axis . In the figure below at the top left, the graph of y =x is represented

by the dashed curve, and the graph of y =x 1 is represented by the dash-dot curve. In the figure

below at the top right, the graph of y =x 1 is represented by the dash-dot curve, and the graph

of y = 4x 1 is represented by the dotted curve. In the bottom figure, the graph of y = 4x 1 is

represented by the dotted curve, and the graph of y = 4x 1 is represented by the solid curve.

1 2 3 4 5

0.5

1

1.5

2

1 2 3 4 5

2

4

6

8

27

1 2 3 4 5

-8

-6

-4

-2

2

4

6

8

46. The graph of f(x) = 4

2 x = 4x+ 2 can be obtained from the graph of y = x byshifting horizontally left 2 units , then reflecting about the y-axis , and finally

stretching vertically by a factor of 4 . In the figure below at the top left, the graph of y =x is

represented by the dashed curve, and the graph of y =x+ 2 is represented by the dash-dot curve.

In the figure below at the top right, the graph of y =x+ 2 is represented by the dash-dot curve,

and the graph of y =x+ 2 is represented by the dotted curve. In the bottom figure, the graph of

y =x+ 2 is represented by the dotted curve, and the graph of y = 4x+ 2 is represented by the

solid curve.

-2 -1 1 2 3 4

0.5

1

1.5

2

-2 -1 1 2

0.5

1

1.5

2

-2 -1 1 2

2

4

6

8

47. (a) The graph of F (x) = f(x) + 3 can be obtained from the graph of f by

shifting vertically up 3 units . In the figure below, the graph of f is represented by the dashed

curve, and the graph of F is represented by the solid curve.

28

-4 -2 2 4

-2

-1

1

2

3

4

5

(b) The graph of G(x) = f(x + 2) can be obtained from the graph of f by

shifting horizontally left 2 units . In the figure below, the graph of f is represented by the dashed

curve, and the graph of G is represented by the solid curve.

-6 -4 -2 2 4

-2

-1

1

2

(c) The graph of P (x) = f(x) can be obtained from the graph of f by reflecting about the x-axis .In the figure below, the graph of f is represented by the dashed curve, and the graph of P isrepresented by the solid curve.

-6 -4 -2 2 4

-2

-1

1

2

(d) The graph of H(x) = f(x + 1) 2 can be obtained from the graph of f byshifting horizontally left 1 unit and vertically down 2 units . In the figure below, the graph of

f is represented by the dashed curve, and the graph of H is represented by the solid curve.

29

-6 -4 -2 2 4

-4

-3

-2

-1

1

2

(e) The graph of Q(x) = 12f(x) can be obtained from the graph of f by

compressing vertically by a factor of 12 . In the figure below, the graph of f is represented by

the dashed curve, and the graph of Q is represented by the solid curve.

-4 -2 2 4

-2

-1

1

2

(f) The graph of g(x) = f(x) can be obtained from the graph of f by reflecting about the y-axis .In the figure below, the graph of f is represented by the dashed curve, and the graph of g isrepresented by the solid curve.

-4 -2 2 4

-2

-1

1

2

(g) The graph of h(x) = f(2x) can be obtained from the graph of f by

compressing horizontally by a factor of 12 . In the figure below, the graph of f is represented

by the dashed curve, and the graph of h is represented by the solid curve.

30

-4 -2 2 4

-2

-1

1

2

48. (a) The graph of F (x) = f(x) + 3 can be obtained from the graph of f by

shifting vertically up 3 units . In the figure below, the graph of f is represented by the dashed

curve, and the graph of F is represented by the solid curve.

-3 -2 -1 1 2 3

-1

1

2

3

4

(b) The graph of G(x) = f(x + 2) can be obtained from the graph of f by

shifting horizontally left 2 units . In the figure below, the graph of f is represented by the dashed

curve, and the graph of G is represented by the solid curve.

-4 -2 2

-1

-0.5

0.5

1

(c) The graph of P (x) = f(x) can be obtained from the graph of f by reflecting about the x-axis .In the figure below, the graph of f is represented by the dashed curve, and the graph of P isrepresented by the solid curve.

31

-3 -2 -1 1 2 3

-1

-0.5

0.5

1

(d) The graph of H(x) = f(x + 1) 2 can be obtained from the graph of f byshifting horizontally left 1 unit and vertically down 2 units . In the figure below, the graph of

f is represented by the dashed curve, and the graph of H is represented by the solid curve.

-4 -3 -2 -1 1 2 3

-3

-2

-1

1

(e) The graph of Q(x) = 12f(x) can be obtained from the graph of f by

compressing vertically by a factor of 12 . In the figure below, the graph of f is represented by

the dashed curve, and the graph of Q is represented by the solid curve.

-3 -2 -1 1 2 3

-1

-0.5

0.5

1

(f) The graph of g(x) = f(x) can be obtained from the graph of f by reflecting about the y-axis .Because the graph of f is symmetric with respect to the y-axis, the graphs of f and g coincide.

32

-3 -2 -1 1 2 3

-1

-0.5

0.5

1

(g) The graph of h(x) = f(2x) can be obtained from the graph of f by

compressing horizontally by a factor of 12 . In the figure below, the graph of f is represented

by the dashed curve, and the graph of h is represented by the solid curve.

-3 -2 -1 1 2 3

-1

-0.5

0.5

1

49. (a) The graph of T (l) is shown below:

2 4 6 8 10

Length (meters)

1

2

3

4

5

6

Perio

d (s

econ

ds)

(b) The graphs of T (l + 1), T (l + 2), and T (l + 3) are shown below:

33

2 4 6 8Length (meters)

1

2

3

4

5

6

7

Perio

d (s

econ

ds)

T(l+1)T(l+2)T(l+3)

(c) Answers will vary, but one interpretation follows. As seen in the graphs for parts (a) and (b),

the period T is an increasing function of the length l . Adding to the length of the pendulum

therefore increases the period. Algebraically, observe that

l + 1 = l

(1 +

1

l

),

so that

T (l + 1) = 2pi

l + 1

g=

1 +

1

l 2pi

l

g=

1 +

1

lT (l).

The factor

1 +

1

lis always larger than 1 (indicating that the period increases) but becomes

smaller as l becomes larger.

(d) The graphs of T (2l), T (3l), and T (4l) are shown below:

2 4 6 8 10

Length (meters)

2

4

6

8

10

12

Perio

d (s

econ

ds)

T(2l)T(3l)T(4l)

(e) Answers will vary, but one interpretation follows. Note that

T (2l) = 2pi

2l

g=

2 2pil

g=

2T (l).

Similarly, T (3l) =

3T (l), and T (4l) =

4T (l) = 2T (l). Thus, multiplying the length of thependulum by 2, 3, and 4, multiplies the period of the pendulum by

2,

3, and

4 = 2,respectively.

34

50. (a) The graph of y = g(x+ 3)5 can be obtained from the graph of y = g(x) by shifting horizontallyleft 3 units and vertically down 5 units. The graph of y = g(x + 3) 5 therefore contains thepoint (1 3, 3 5) = (2,2) .

(b) The graph of y = 2g(x 2) + 1 can be obtained from the graph of y = g(x) by shiftinghorizontally right 2 units, stretching vertically by a factor of 2, reflecting about the x-axis, andfinally shifting vertically up 1 unit. The graph of y = 2g(x 2) + 1 therefore contains the point(1 + 2,2 3 + 1) = (3,5) .

(c) The graph of y = g(2x + 3) can be obtained from the graph of y = g(x) by shifting horizontallyleft 3 units and then compressing horizontally by a factor of 12 . The graph of y = g(2x + 3)

therefore contains the point

(1

2(1 3), 3

)= (1, 3) .

35

P.4: Inverse Functions


1. False . If every horizontal line intersects the graph of a function f at no more than one point, f isa one-to-one function.

2. If the domain of a one-to-one function f is [4,), the range of its inverse f1 is [4,) .

3. False . If f and g are inverse functions, the domain of f does not need to be the same as thedomain of g. However, the domain of f will always be the same as the range of g.

4. True . If f and g are inverse functions, their graphs are symmetric with respect to the line y = x.

5. False . If f and g are inverse functions, then (f g)(x) = (gf)(x) = x; that is, f(g(x)) = g(f(x)) = x.

6. False . If a function f is one-to-one, then f(f1(x)) = x, where x is in the domain of f1.

7. First, examine the x-coordinates of the ordered pairs. If each value of x appears in one and only oneordered pair, then the collection of points represents a function y = f(x); otherwise, the collectionof points is not a function. If the collection of points does represent a function, next check the y-coordinates. If each value of y appears in one and only one ordered pair, then the function representedby the collection of points is one-to-one; otherwise, the collection of points represents a function thatis not one-to-one.

8. Given the graph of the one-to-one function y = f(x), the graph of the inverse function f1 can beobtained by reflecting the graph of f about the line y = x.

Practice Problems

9. The graph of f passes the horizontal line test, so this function is one-to-one .


11. The graph of f does not pass the horizontal line test (for example, the horizontal line y = 2 intersects

the graph of f in two points), so this function is not one-to-one .

12. The graph of f does not pass the horizontal line test (for example, the horizontal line y = 2 intersects

the graph of f in two points), so this function is not one-to-one .


14. The graph of f does not pass the horizontal line test (the horizontal line y = 2 intersects the graph in

infinitely many points), so this function is not one-to-one .

15. Let f(x) = 3x+ 4 and g(x) =1

3(x 4). Then

f(g(x)) = f

(1

3(x 4)

)= 3 1

3(x 4) + 4 = x 4 + 4 = x,

and

g(f(x)) = g(3x+ 4) =1

3(3x+ 4 4) = 1

3(3x) = x,

so that f and g are inverses of each other.

36

16. Let f(x) = x3 8 and g(x) = 3x+ 8. Then

f(g(x)) = f( 3x+ 8) =

(3x+ 8

)3 8 = x+ 8 8 = x,and

g(f(x)) = g(x3 8) = 3x3 8 + 8 = 3

x3 = x,


17. Let f(x) = g(x) =1

x. Then

f(g(x)) = f

(1

x

)=

11x

= x,

and

g(f(x)) = g

(1

x

)=

11x

= x,


18. Let f(x) =2x+ 3

x+ 4and g(x) =

4x 32 x . Then

f(g(x)) = f

(4x 32 x

)=

2 4x32x + 34x32x + 4

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

4x 3 + 4(2 x) =8x 6 + 6 3x4x 3 + 8 4x =

5x

5= x,

and

g(f(x)) = g

(2x+ 3

x+ 4

)=

4 2x+3x+4 32 2x+3x+4

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

8x+ 12 3x 122x+ 8 2x 3 =

5x

5= x,


19. (a) Because each value of the second coordinate appears in one and only one ordered pair, this is a

one-to-one function.

(b) Because the function is one-to-one, the function has an inverse. To obtain the inverse, interchangethe coordinates in each ordered pair; thus, the inverse f1 is the set

{(5,3), (9,2), (2,1), (11, 0), (5, 1)} .

(c) The domain of f is the set {3,2,1, 0, 1} , and the range of f is the set {5, 2, 5, 9, 11} . Thedomain of f1 is the set {5, 2, 5, 9, 11} , and the range of f1 is the set {3,2,1, 0, 1} .

20. (a) Because the number 8 appears as the second coordinate in two ordered pairs with different first

coordinates, (0, 8) and (2, 8), this is not a one-to-one function.

(b) Because the function is not one-to-one, the function does not have an inverse .

(c) The domain of f is the set {2,1, 0, 1, 2} , and the range of f is the set {3, 2, 6, 8} . Thereis no inverse function.

21. (a) Because the number 1 appears as the second coordinate in two ordered pairs with different first

coordinates, (2, 1) and (2, 1), this is not a one-to-one function.(b) Because the function is not one-to-one, the function does not have an inverse .

(c) The domain of f is the set {10,3,2, 1, 2} , and the range of f is the set {0, 1, 2, 9} . Thereis no inverse function.

37

22. (a) Because each value of the second coordinate appears in one and only one ordered pair, this is a

one-to-one function.

(b) Because the function is one-to-one, the function has an inverse. To obtain the inverse, interchangethe coordinates in each ordered pair; thus, the inverse f1 is the set

{(8,2), (1,1), (0, 0), (1, 1), (8, 2)} .

(c) The domain of f is the set {2,1, 0, 1, 2} , and the range of f is the set {8,1, 0, 1, 8} . Thedomain of f1 is the set {8,1, 0, 1, 8} , and the range of f1 is the set {2,1, 0, 1, 2} .

23. In the figure below, the dashed curve is the graph of y = f(x) and the solid curve is the graph ofy = f1(x). The graph of y = x is included for clarity. Observe how the graph of y = f1(x) is thereflection of the graph of y = f(x) about the line y = x.

-2 -1 1 2

-2

-1

1

2y = x


-2 -1 1 2

-2

-1

1

2y = x


38

-2 -1 1 2

-2

-1

1

2y = x


-2 -1 1

-2

-1

1 y =x


-3 -2 -1 1 2 3

-3

-2

-1

1

2

3 y = x

(1, 1/3)

(1, 3)

(1/3, 1)

(3, 1)


39

-3 -2 -1 1 2 3 4 5

-3

-2

-1

1

2

3

4

5 y = x

(1, 1/5)

(1, 5)

(1/5, 1)

(5, 1)

29. (a) Start by writing y = f(x) = 4x+ 2. Next, interchange the x and y variables to obtain an implicitdefinition of the inverse function: x = 4y + 2. To obtain the inverse function explicitly, solve theimplicit equation for y:

4y = x 2, so that y = 14

(x 2).

Thus, f1(x) = 14 (x 2) . To check that this is the correct formula for the inverse function,show that f(f1(x)) = x and f1(f(x)) = x:

f(f1(x)) = f(

1

4(x 2)

)= 4

(1

4(x 2)

)+ 2 = x 2 + 2 = x ,

f1(f(x)) = f1(4x+ 2) =1

4(4x+ 2 2) = 1

4(4x) = x .

(b) Both the domain and range of f are the set of all real numbers or the interval (,) .The same is true for the inverse function. Both the domain and range of f1 are the set ofall real numbers or the interval (,) .

30. (a) Start by writing y = f(x) = 1 3x. Next, interchange the x and y variables to obtain an implicitdefinition of the inverse function: x = 1 3y. To obtain the inverse function explicitly, solve theimplicit equation for y:

3y = 1 x, so that y = 13

(1 x).

Thus, f1(x) = 13 (1 x) . To check that this is the correct formula for the inverse function,show that f(f1(x)) = x and f1(f(x)) = x:

f(f1(x)) = f(

1

3(1 x)

)= 1 3

(1

3(1 x)

)= 1 (1 x) = x ,

f1(f(x)) = f1(1 3x) = 13

(1 (1 3x)) = 13

(3x) = x .


31. (a) Start by writing y = f(x) = 3x+ 10. Next, interchange the x and y variables to obtain an

implicit definition of the inverse function: x = 3y + 10. To obtain the inverse function explicitly,

solve the implicit equation for y:

y + 10 = x3, so that y = x3 10.

40

Thus, f1(x) = x3 10 . To check that this is the correct formula for the inverse function, showthat f(f1(x)) = x and f1(f(x)) = x:

f(f1(x)) = f(x3 10) = 3

(x3 10) + 10 = 3x3 = x ,

f1(f(x)) = f1( 3x+ 10) = ( 3

x+ 10)3 10 = x+ 10 10 = x .


32. (a) Start by writing y = f(x) = 2x3 +4. Next, interchange the x and y variables to obtain an implicitdefinition of the inverse function: x = 2y3 + 4. To obtain the inverse function explicitly, solve theimplicit equation for y:

2y3 = x 4, so that y = 3x 4

2.

Thus, f1(x) = 3x 4

2. To check that this is the correct formula for the inverse function, show

that f(f1(x)) = x and f1(f(x)) = x:

f(f1(x)) = f

(3

x 4

2

)= 2

(3

x 4

2

)3+ 4 = 2

(x 4

2

)+ 4 = x 4 + 4 = x ,

f1(f(x)) = f1(2x3 + 4) = 3

2x3 + 4 42

=3x3 = x .


33. (a) Start by writing y = f(x) =1

x 2 . Next, interchange the x and y variables to obtain an implicit

definition of the inverse function: x =1

y 2 . To obtain the inverse function explicitly, solve theimplicit equation for y:

y 2 = 1x, so that y =

1

x+ 2.

Thus, f1(x) =1

x+ 2 . To check that this is the correct formula for the inverse function, show

that f(f1(x)) = x and f1(f(x)) = x:

f(f1(x)) = f(

1

x+ 2

)=

11x + 2 2

=11x

= x ,

f1(f(x)) = f1(

1

x 2)

=11

x2+ 2 = x 2 + 2 = x .

(b) The domain of f is the set {x|x 6= 2} , and the range of f is the set {y|y 6= 0} . The domain off1 is the set {x|x 6= 0} , and the range of f1 is the set {y|y 6= 2} .

41

34. (a) Start by writing y = f(x) =2x

3x 1 . Next, interchange the x and y variables to obtain an implicit

definition of the inverse function: x =2y

3y 1 . To obtain the inverse function explicitly, solve theimplicit equation for y:

x(3y 1) = 2y3xy x = 2y

3xy 2y = xy(3x 2) = x

y =x

3x 2 = f1(x) .

To check that this is the correct formula for the inverse function, show that f(f1(x)) = x andf1(f(x)) = x:

f(f1(x)) = f(

x

3x 2)

=2 x3x2

3 x3x2 1=

2x

3x (3x 2) =2x

2= x ,

f1(f(x)) = f1(

2x

3x 1)

=2x

3x13 2x3x1 2

=2x

6x 2(3x 1) =2x

2= x .

(b) The domain of f is the set {x|x 6= 13} , and the range of f is the set {y|y 6= 23} . The domain off1 is the set {x|x 6= 23} , and the range of f1 is the set {y|y 6= 13} .

35. (a) Start by writing y = f(x) =2x+ 3

x+ 2. Next, interchange the x and y variables to obtain an implicit

definition of the inverse function: x =2y + 3

y + 2. To obtain the inverse function explicitly, solve the

implicit equation for y:

x(y + 2) = 2y + 3

xy + 2x = 2y + 3

xy 2y = 3 2xy(x 2) = 3 2x

y =3 2xx 2 = f

1(x) .


f(f1(x)) = f(

3 2xx 2

)=

2 32xx2 + 332xx2 + 2

=2(3 2x) + 3(x 2)

3 2x+ 2(x 2) =x1 = x ,

f1(f(x)) = f1(

2x+ 3

x+ 2

)=

3 2 2x+3x+22x+3x+2 2

=3(x+ 2) 2(2x+ 3)

2x+ 3 2(x+ 2) =x1 = x .

(b) The domain of f is the set {x|x 6= 2} , and the range of f is the set {y|y 6= 2} . The domainof f1 is the set {x|x 6= 2} , and the range of f1 is the set {y|y 6= 2} .

36. (a) Start by writing y = f(x) =3x 4x 2 . Next, interchange the x and y variables to obtain an

implicit definition of the inverse function: x =3y 4y 2 . To obtain the inverse function explicitly,

42

solve the implicit equation for y:

x(y 2) = 3y 4xy 2x = 3y 4xy + 3y = 2x 4y(x+ 3) = 2x 4

y =2x 4x+ 3

= f1(x) .


f(f1(x)) = f(

2x 4x+ 3

)=3 2x4x+3 42x4x+3 2

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

2x 4 2(x+ 3) =10x10 = x ,

f1(f(x)) = f1(3x 4

x 2)

=23x4x2 43x4x2 + 3

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

10x10 = x .

(b) The domain of f is the set {x|x 6= 2} , and the range of f is the set {y|y 6= 3} . The domainof f1 is the set {x|x 6= 3} , and the range of f1 is the set {y|y 6= 2} .

37. (a) Start by writing y = f(x) = x2 + 4, where x 0. Next interchange the x and y variables toobtain an implicit definition of the inverse function: x = y2 + 4, where y 0. To obtain theinverse function explicitly, solve the implicit equation for y:

y2 = x 4, so that y = x 4.

Because y 0, only the positive square root is appropriate; thus, f1(x) = x 4 . To checkthat this is the correct formula for the inverse function, show that f(f1(x)) = x and f1(f(x)) =x:

f(f1(x)) = f(x 4) = (x 4)2 + 4 = x 4 + 4 = x ,

f1(f(x)) = f1(x2 + 4) =x2 + 4 4 =

x2 = |x| = x ,

where, in both calculations, attention is restricted to x 0.(b) The domain of f is the set {x|x 0} , and the range of f is the set {y|y 4} . The domain of

f1 is the set {x|x 4} , and the range of f1 is the set {y|y 0} .

38. (a) Start by writing y = f(x) = (x 2)2 + 4, where x 2. Next interchange the x and y variables toobtain an implicit definition of the inverse function: x = (y 2)2 + 4, where y 2. To obtain theinverse function explicitly, solve the implicit equation for y:

(y 2)2 = x 4, so that y = 2x 4.

Because y 2, only the negative square root is appropriate; thus, f1(x) = 2x 4 . Tocheck that this is the correct formula for the inverse function, show that f(f1(x)) = x andf1(f(x)) = x:

f(f1(x)) = f(2x 4) = (2x 4 2)2 + 4= (x 4)2 + 4 = x 4 + 4 = x ,

f1(f(x)) = f1((x 2)2 + 4) = 2

(x 2)2 + 4 4= 2

(x 2)2 = 2 |x 2| = 2 (2 x) = x ,

where, in both calculations, attention is restricted to x 2.

43

(b) The domain of f is the set {x|x 2} , and the range of f is the set {y|y 4} . The domain off1 is the set {x|x 4} , and the range of f1 is the set {y|y 2} .

44

P.5: Exponential and Logarithmic Functions


1. The graph of every exponential function f(x) = ax, a > 0 and a 6= 1, passes through three points:(1, 1a ), (0, 1), and (1, a) .

2. False . The graph of the exponential function f(x) =(32

)xis increasing, because the base a = 32 > 1.

3. If 3x = 34, then x = 4 .

4. If 4x = 82, then x = 3 .

5. False . The graphs of y = 3x and y =(13

)x= 3x are symmetric with respect to the y-axis. The

graphs of y = 3x and its inverse function y = log3 x are symmetric with respect to the line y = x.

6. False . The range of the exponential function f(x) = ax, a > 0 and a 6= 1, is the set of positive realnumbers.

7. The number e is defined as the base of the exponential function f whose tangent line to the graph off at the point (0, 1) has slope 1 .

8. The domain of the logarithmic function f(x) = loga x is the set {x|x > 0} .

9. The graph of every logarithmic function f(x) = loga x, a > 0 and a 6= 1, passes through three points:( 1a ,1), (1, 0), and (a, 1) .

10. The graph of f(x) = log2 x is (a) increasing , because the base a = 2 > 1.

11. False . If y = loga x, then x = ay.

12. True . The graph of f(x) = loga x, a > 0 and a 6= 1, has an x-intercept equal to 1 and no y-intercept.

13. True . ln ex = x for all real numbers.

14. ln e = 1 .

15. The number e is defined as the base of the exponential function f whose tangent line to the graph off at the point (0, 1) has slope 1.

16. The x-intercept of the function f(x) = lnx is 1. Because the graph of h(x) = ln(x+1) can be obtainedfrom the graph of f(x) = lnx by shifting horizontally left 1 unit, the x-intercept of the function

h(x) = ln(x+ 1) is 1 1 = 0 .

Practice Problems

17. Let g(x) = 4x + 2.

(a) g(1) = 41 + 2 = 14 + 2 = 94 . The corresponding point on the graph of g is (1, 94 ) .(b) If g(x) = 4x + 2 = 66, then 4x = 64 = 43. It follows that x = 3 . The corresponding point on the

graph of g is (3, 66) .

18. Let g(x) = 5x 3.

(a) g(1) = 51 3 = 15 3 = 145 . The corresponding point on the graph of g is (1, 145 ) .

45

(b) If g(x) = 5x 3 = 122, then 5x = 125 = 53. It follows that x = 3 . The corresponding point onthe graph of g is (3, 122) .

19. The graph appears to be the reflection of y = 3x about the y-axis. The function would therefore be

(a): y = 3x .

20. The graph appears to be y = 3x shifted horizontally right 1 unit. The function would therefore be

(e): y = 3x1 .

21. The graph appears to be the reflection of y = 3x about both the x-axis and the y-axis. The function

would therefore be (c): y = 3x .

22. The graph appears to be the reflection of y = 3x about the x-axis and then shifted vertically up 1 unit.

The function would therefore be (f): y = 1 3x .

23. The graph appears to be the reflection of y = 3x about the x-axis. The function would therefore be

(b): y = 3x .

24. The graph appears to be y = 3x shifted vertically down 1 unit. The function would therefore be

(d): y = 3x 1 .

25. The graph of y = 2x+2 can be obtained from the graph of y = 2x by shifting horizontally left 2 units .

In the figure below, the dashed curve represents the graph of y = 2x, and the solid curve represents

the graph of y = 2x+2. The domain of f is the set of all real numbers or the interval (,) , andthe range is the set {y|y > 0} or the interval (0,) .

-3 -2 -1 1 2 3

2

4

6

8

26. The graph of y = 12x/3 can be obtained from the graph of y = 2x through the following sequence oftransformations: reflect about the y-axis , stretch horizontally by a factor of 3 ,

reflect about the x-axis , and shift vertically up 1 unit . In the figures below, the dashed curve rep-

resents the graph of y = 2x, the dash-dot curve represents the graph of y = 2x, the dash-double dotcurve represents the graph of y = 2x/3, the dotted curve represents the graph of y = 2x/3, and thesolid curve represents the graph of y = 1 2x/3. The domain of f is the set of all real numbers orthe interval (,) , and the range is the set {y|y < 1} or the interval (, 1) .

46

-3 -2 -1 1 2 3

1

2

3

4

5

6

7

8

-8 -6 -4 -2 2 4 6 8

1

2

3

4

5

6

7

-8 -6 -4 -2 2 4 6 8

-7-6-5-4-3-2-1

1234567

-8 -6 -4 -2 2 4 6 8

-7

-6

-5

-4

-3

-2

-1

1

2

27. The graph of y = 4

(1

3

)xcan be obtained from the graph of y =

(1

3

)xby

stretching vertically by a factor of 4 . In the figure below, the dashed curve represents the graph

of y =

(1

3

)x, and the solid curve represents the graph of y = 4

(1

3

)x. The domain of f is the set

of all real numbers or the interval (,) , and the range is the set {y|y > 0} or the interval(0,) .

-3 -2 -1 1 2 3

2

4

6

8

10

28. The graph of f(x) =

(1

2

)x+ 1 can be obtained from the graph of f(x) =

(1

2

)xby

reflecting about the y-axis and then shifting vertically up 1 unit . In the figure below at the left,

the graph of f(x) =

(1

2

)xis represented by the dashed curve, and the graph of f(x) =

(1

2

)xis

47

represented by the dash-dot curve. In the figure below at the right, the graph of f(x) =

(1

2

)xis

represented by the dash-dot curve, and the graph of f(x) =

(1

2

)x+ 1 is represented by the solid

curve. The domain of f is the set of all real numbers or the interval (,) , and the range is theset {y|y > 1} or the interval (1,) .

-3 -2 -1 1 2 3

2

4

6

8

-3 -2 -1 1 2 3

2

4

6

8

29. The graph of y = ex can be obtained from the graph of y = ex by reflecting about the y-axi s. Inthe figure below, the dashed curve represents the graph of y = ex, and the solid curve represents the

graph of y = ex. The domain of f is the set of all real numbers or the interval (,) , and therange is the set {y|y > 0} or the interval (0,) .

-3 -2 -1 1 2 3

5

10

15

20

30. The graph of y = 5 ex can be obtained from the graph of y = ex by reflecting about the x-axis andthen shifting vertically up 5 units . In the figure below at the left, the graph of y = ex is represented

by the dashed curve, and the graph of y = ex is represented by the dash-dot curve. In the figurebelow at the right, the graph of y = ex is represented by the dash-dot curve, and the graph ofy = 5 ex is represented by the solid curve. The domain of f is the set of all real numbers or theinterval (,) , and the range is the set {y|y < 5} or the interval (, 5) .

48

-3 -2 -1 1 2 3

-8

-6

-4

-2

2

4

6

8

-3 -2 -1 1 2 3

-7

-6

-5

-4

-3

-2

-1

1

2

3

4

5

31. The argument of a logarithmic function must be positive, so the domain of F (x) = log2 x2 is the

solution of the inequality x2 > 0. The domain of F is therefore the set {x|x 6= 0} .

32. The argument of a logarithmic function must be positive, so the domain of g(x) = 8 + 5 ln(2x + 3) is

the solution of the inequality 2x+ 3 > 0. The domain of g is therefore the set {x|x > 32} .

33. The argument of a logarithmic function must be positive, so the domain of f(x) = ln(x 1) is thesolution of the inequality x 1 > 0. The domain of f is therefore the set {x|x > 1} .

34. The square root function is only defined for nonnegative real numbers, so the domain of g(x) =

lnxis the solution of the inequality lnx 0. Because ln 1 = 0 and the natural logarithm function isincreasing for all real numbers, it follows that the domain of g is the set {x|x 1} .

35. The graph appears to be the reflection of y = log3 x about the y-axis. The function would therefore

be (b): y = log3(x) .

36. The graph appears to be y = log3 x shifted horizontally right 1 unit. The function would therefore be

(e): y = log3(x 1) .

37. The graph appears to be the reflection of y = log3 x about the x-axis and then shifted vertically up 1

unit. The function would therefore be (f): y = 1 log3 x .

38. The graph appears to be y = log3 x. The function would therefore be (a): y = log3 x .

39. The graph appears to be the reflection of y = log3 x about the x-axis. The function would therefore

be (c): y = log3 x .

40. The graph appears to be y = log3 x shifted vertically down 1 unit. The function would therefore be

(d): y = log3 x 1 .

41. (a) The argument of a logarithmic function must be positive, so the domain of f(x) = ln(x + 4) is

the solution of the inequality x+ 4 > 0. The domain of f is therefore the set {x|x > 4} .

(b) The graph of f(x) = ln(x+ 4) is shown below:

49

-5 -4 -3 -2 -1 1 2 3 4 5

-2

-1

1

2

(c) From the graph of f , the range of f is the set of all real numbers or the interval (,) .(d) Start by writing y = f(x) = ln(x + 4). Next, interchange the variables x and y to obtain

x = ln(y + 4). Finally, solve for y:

y + 4 = ex, so that y = f1(x) = ex 4 .

(e) Recall that the range of f is the domain of its inverse function f1. Because the domain of f1 isthe set of all real numbers, the range of f is the set of all real numbers or the interval (,) .

(f) The graph of f1(x) = ex 4 is shown below:

-5 -4 -3 -2 -1 1 2 3

-4

-3

-2

-1

1

2

3

4

5

6

42. (a) The argument of a logarithmic function must be positive, so the domain of f(x) = 12 ln(2x) is the

solution of the inequality 2x > 0. The domain of f is therefore the set {x|x > 0} .(b) The graph of f(x) = 12 ln(2x) is shown below:

0.5 1 1.5 2 2.5 3 3.5 4 4.5 5 5.5 6

-1

-0.5

0.5

1

1.5

50

(c) From the graph of f , the range of f is the set of all real numbers or the interval (,) .(d) Start by writing y = f(x) = 12 ln(2x). Next, interchange the variables x and y to obtain x =

12 ln(2y). Finally, solve for y:

2y = e2x, so that y = f1(x) =1

2e2x .

(e) Recall that the range of f is the domain of its inverse function f1. Because the domain of f1 isthe set of all real numbers, the range of f is the set of all real numbers or the interval (,) .

(f) The graph of f1(x) = 12e2x is shown below:

-2 -1 1 2

0.5

1

1.5

2

2.5

3

3.5

4

4.5

5

5.5

6

43. (a) The domain of f(x) = 3ex + 2 is the set of all real numbers or the interval (,) .(b) The graph of f(x) = 3ex + 2 is shown below:

-2 -1 1 2

5

10

15

20

25

(c) From the graph of f , the range of f is the set {y|y > 2} .(d) Start by writing y = f(x) = 3ex+2. Next, interchange the variables x and y to obtain x = 3ey+2.

Finally, solve for y:

ey =1

3(x 2), so that y = f1(x) = ln

(x 2

3

).

(e) Recall that the range of f is the domain of its inverse function f1. Because the domain of f1

is the set {x|x > 2}, the range of f is the set {y|y > 2} .(f) The graph of f1(x) = ln

(x23

)is shown below:

51

2 4 6 8 10

-5

-4

-3

-2

-1

1

44. (a) The domain of f(x) = 2x/3 + 4 is the set of all real numbers or the interval (,) .(b) The graph of f(x) = 2x/3 + 4 is shown below:

-6 -4 -2 2 4 6

2

4

6

8

(c) From the graph of f , the range of f is the set {y|y > 4} .(d) Start by writing y = f(x) = 2x/3+4. Next, interchange the variables x and y to obtain x = 2y/3+4.

Finally, solve for y:

2y/3 = x 4, so that y = f1(x) = 3 log2(x 4) .

(e) Recall that the range of f is the domain of its inverse function f1. Because the domain of f1

is the set {x|x > 4}, the range of f is the set {y|y > 4} .(f) The graph of f1(x) = 3 log2(x 4) is shown below:

5 10 15 20

-15

-10

-5

5

10

15

45. The transformation y = ln(x + c), c > 0, shifts the graph of f(x) = lnx horiztonally left by c units.

This will move the x-intercept c units to the left .

52

46. The transformation y = ecx, c > 0, horizontally stretches (if 0 < c < 1) or compresses (if c > 1) thegraph of f(x) = ex. Because a horizontal transformation has no effect on the y-intercept of a graph,

the y-intercept of y = ecx is the same as the y-intercept of y = ex .

47. Because 9 = 32, the equation can be written as

3x2

= (32)x = 32x.

By the one-to-one property of exponential functions, it follows that

x2 = 2x, or x2 2x = x(x 2) = 0.Thus, x = 0 or x = 2 , and the solution set is {0, 2}.

48. Because 125 = 53, the equation can be written as

5x2+8 = (53)2x = 56x.

By the one-to-one property of exponential functions, it follows that

x2 + 8 = 6x, or x2 6x+ 8 = (x 4)(x 2) = 0.Thus, x = 4 or x = 2 , and the solution set is {2, 4}.

49. First, rewrite the equation as e3x = e2x. Then, by the one-to-one property of exponential functions,

3x = 2 x, or 4x = 2,so that x = 12 .

50. First, rewrite the equation as e4x+x2

= e12. Then, by the one-to-one property of exponential functions,

4x+ x2 = 12, or x2 + 4x 12 = (x+ 6)(x 2) = 0.Thus, x = 6 or x = 2 , and the solution set is {6, 2}.

51. The numbers e and 4 cannot be expressed with the same base, so apply the natural logarithm functionto both sides of the equation. This yields

ln(e12x) = ln 41 2x = ln 4

2x = 1 ln 4x =

1

2(1 ln 4) .

52. The numbers e and 5 cannot be expressed with the same base, so apply the natural logarithm functionto both sides of the equation. This yields

ln(e1x) = ln 51 x = ln 5

x = 1 ln 5 .

53. First, rewrite the equation as 23x = 95 , and then apply the natural logarithm function to both sides.This yields

ln(23x) = ln

(9

5

)3x ln 2 = ln

(9

5

)

x =ln( 95 )

3 ln 2.

53

54. First, rewrite the equation as 40.2x = 0.20.3 =23 , and then apply the natural logarithm function to both

sides. This yields

ln(40.2x) = ln

(2

3

)0.2x ln 4 = ln

(2

3

)

x =ln( 23 )

0.2 ln 4=

5 ln( 23 )

ln 4.

55. Because the numbers 3 and 4 cannot be expressed with the same base, apply the natural logarithmfunction to both sides of the equation. This yields

ln(312x) = ln(4x)(1 2x) ln 3 = x ln 4ln 3 2x ln 3 = x ln 4

2x ln 3 x ln 4 = ln 3x(2 ln 3 + ln 4) = ln 3

x =ln 3

2 ln 3 + ln 4.

Using properties of logarithms, the answer can be simplified further as

x =ln 3

2 ln 3 + ln 4=

ln 3

ln 32 + ln 4=

ln 3

ln(32 4) =ln 3

ln 36.

56. Because the numbers 2 and 5 cannot be expressed with the same base, apply the natural logarithmfunction to both sides of the equation. This yields

ln(2x+1) = ln(512x)(x+ 1) ln 2 = (1 2x) ln 5x ln 2 + ln 2 = ln 5 2x ln 5

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

x =ln 5 ln 2

ln 2 + 2 ln 5.

Using properties of logarithms, the answer can be simplified further as

x ==ln 5 ln 2

ln 2 + 2 ln 5=

ln(52

)ln 2 + ln 52

=ln(52

)ln(2 52) =

ln(52

)ln 50

.

57. Change the logarithmic equation log2(2x + 1) = 3 into the equivalent exponential equation 2x + 1 =

23 = 8. Then 2x = 7, or x = 72 .

58. Change the logarithmic equation log3(3x 2) = 2 into the equivalent exponential equation 3x 2 =32 = 9. Then 3x = 11, or x = 113 .

59. Change the logarithmic equation logx

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