Unit 5 trigonometric functions answer key all things algebra

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1.1 Functions and Function Notation

1.

1. ⓐ yes
2. ⓑ yes. (Note: If two players had been tied for, say, 4th place, then the name would not have been a function of rank.)

6.

y=f(x)= x32y=f(x)=x32

9.

1. ⓐ yes, because each bank account has a single balance at any given time
2. ⓑ no, because several bank account numbers may have the same balance
3. ⓒ no, because the same output may correspond to more than one input.

10.

1. ⓐ Yes, letter grade is a function of percent grade;
2. ⓑ No, it is not one-to-one. There are 100 different percent numbers we could get but only about five possible letter grades, so there cannot be only one percent number that corresponds to each letter grade.

12.

No, because it does not pass the horizontal line test.

1.2 Domain and Range

1.

{−5,0,5,10,15}{−5,0,5,10,15}

3.

(−∞,12) ∪(12,∞)(−∞,12)∪(12,∞)

4.

[−52,∞) [−52,∞)

5.

1. ⓐ values that are less than or equal to –2, or values that are greater than or equal to –1 and less than 3;
2. ⓑ{x |x≤−2or−1≤x<3}{x|x≤−2or−1≤x<3 } ;
3. ⓒ(−∞,−2]∪[−1,3)(−∞,−2]∪[−1,3)

6.

domain =[1950,2002] range = [47,000,000,89,000,000]

7.

domain: (− ∞,2];(−∞,2]; range: (−∞,0](−∞,0]

1.3 Rates of Change and Behavior of Graphs

1.

$2.84−$2.315 years= $0.535 years=$0.106$2.84−$2.315 years=$0.535 years=$0.106 per year.

4.

The local maximum appears to occur at (−1,28),(−1,28), and the local minimum occurs at (5,−80).(5,−80). The function is increasing on (−∞,−1)∪(5,∞ )(−∞,−1)∪(5,∞) and decreasing on (−1,5) .(−1,5).

1.4 Composition of Functions

1.

(fg)(x)=f (x)g(x)=(x−1)(x2−1)=x3−x2−x+1(f−g) (x)=f(x)−g(x)=(x−1)−(x2−1)=x−x2(fg)(x)=f(x)g(x)=(x−1)(x2−1)=x 3−x2−x+1(f−g)(x)=f(x)−g(x)=(x−1)−(x 2−1)=x−x2

No, the functions are not the same.

2.

A gravitational force is still a force, so a(G(r)) a(G(r)) makes sense as the acceleration of a planet at a distance r from the Sun (due to gravity), but G(a( F))G(a(F)) does not make sense.

3.

f(g(1))=f(3)=3f(g(1))=f(3)=3 and g (f(4))=g(1)=3g(f(4))=g(1)=3

4.

g (f(2))=g(5)=3g(f(2))=g(5)=3

6.

[−4,0)∪(0,∞)[−4,0)∪(0,∞)

7.

Possible answer:

g(x) =4+x2g(x)=4+x2
h(x)=4 3−xh(x)=43−x
f=h∘gf=h∘ g

1.5 Transformation of Functions

1.

b(t)=h(t)+10= −4.9t2+30t+10b(t)=h(t)+10=−4.9t2+30t+10

2.

The graphs of f(x )f(x) and g(x)g(x) are shown below. The transformation is a horizontal shift. The function is shifted to the left by 2 units.

4.

g(x)=1x-1+1 g(x)=1x-1+1

6.

1. ⓐ

   g(x)=− f(x)g(x)=−f(x)

   xx	-2	0	2	4
   g(x)g(x)	−5−5	−10−10	−15−15	−20−20

2. ⓑ

   h(x)=f(−x)h(x)=f(−x )

   xx	-2	0	2	4
   h(x)h(x)	15	10	5	unknown

7.

Notice: g(x)=f(−x)g(x)=f(−x) looks the same as f(x)f(x ).

9.

11.

g(x)=f(13x)g(x)=f(13x) so using the square root function we get g(x)=13xg(x)=13x

1.6 Absolute Value Functions

2.

using the variable pp for passing, |p−80|≤20|p−80|≤20

3.

f(x )=−|x+2|+3f(x)=−|x+2|+3

5.

f(0)=1,f(0)=1, so the graph intersects the vertical axis at (0,1).(0,1). f(x)=0f(x)=0 when x =−5x=−5 and x=1x=1 so the graph intersects the horizontal axis at (−5,0)(−5,0) and (1,0).(1,0).

7.

k≤1k≤1 or k≥7;k≥7; in interval notation, this would be (−∞,1]∪[7,∞)(−∞,1]∪[7,∞)

1.7 Inverse Functions

4.

The domain of function f−1f−1 is (−∞,−2)(−∞,−2) and the range of function f−1f−1 is (1,∞).( 1,∞).

5.

1. f(60)=50.f(60)=50. In 60 minutes, 50 miles are traveled.
2. f−1( 60)=70.f−1(60)=70. To travel 60 miles, it will take 70 minutes.

8.

f−1(x)=(2− x)2;domainoff:[0,∞);domainoff− 1:(−∞,2]f−1(x)=(2−x)2; domainoff:[0,∞);domainoff−1:(− ∞,2]

1.1 Section Exercises

1.

A relation is a set of ordered pairs. A function is a special kind of relation in which no two ordered pairs have the same first coordinate.

3.

When a vertical line intersects the graph of a relation more than once, that indicates that for that input there is more than one output. At any particular input value, there can be only one output if the relation is to be a function.

5.

When a horizontal line intersects the graph of a function more than once, that indicates that for that output there is more than one input. A function is one-to-one if each output corresponds to only one input.

27.

f(−3)=−11;f(−3)=−11;
f(2)=−1;f(2)=−1;
f(−a)=−2a−5;f(−a)=− 2a−5;
−f(a)=−2a+5;−f(a)=−2a +5;
f(a+h)=2a+2h−5f(a+h)=2a +2h−5

29.

f(−3)=5+5;f(−3)=5+5;
f (2)=5;f(2)=5;
f(−a)=2+a +5;f(−a)=2+a+5;
−f(a)=−2 −a−5;−f(a)=−2−a−5;
f(a+h) =2−a−h+5f(a+h)=2−a−h+5

31.

f(−3)=2;f(−3)=2; f(2)=1−3=−2;f(2)=1−3=−2;
f(−a)=|−a−1|−|−a+1|;f(−a)=|−a−1|−|−a+1|;
−f(a)=−| a−1|+|a+1|;−f(a)=−|a−1|+|a+1|;
f(a+h)=|a+h−1|−|a+h+1|f(a+h)=|a+h−1|− |a+h+1|

33.

g(x)−g(a)x−a=x+a+2,x≠ag(x)−g(a)x−a=x+a+2,x≠a

35.

1. ⓐf(−2)=14; f(−2)=14;
2. ⓑx=3x=3

37.

1. ⓐf(5)=10;f(5)=10;
2. ⓑx=−1x=−1 or x=4x=4

39.

1. ⓐf(t)=6−2 3t;f(t)=6−23t;
2. ⓑf(−3)=8; f(−3)=8;
3. ⓒt=6t=6

53.

1. ⓐf(0)= 1;f(0)=1;
2. ⓑf(x)=−3,x=−2 f(x)=−3,x=−2 or x=2x=2

55.

not a function so it is also not a one-to-one function

59.

function, but not one-to-one

67.

f(x)=1,x =2f(x)=1,x=2

69.

f(−2)=14; f(−1)=11;f(0)=8;f(1)=5;f(2)= 2f(−2)=14;f(−1)=11;f(0)=8; f(1)=5;f(2)=2

71.

f(−2)=4; f(−1)=4.414;f(0)=4.732;f(1)=5;f(2)=5.236f(−2)=4;f(−1)=4.414;f(0 )=4.732;f(1)=5;f(2)=5.236

73.

f(−2)=19 ;f(−1)=13;f(0)=1;f(1)=3;f( 2)=9f(−2)=19;f(−1)=13; f(0)=1;f(1)=3;f(2)=9

77.

[0, 100][0, 100]

79.

[−0.001, 0.001 ][−0.001, 0.001]

81.

[−1,000,000, 1,000,000][−1,000,000, 1,000,000]

83.

[0, 10][0, 10]

85.

[−0.1,0.1][−0.1,0.1]

87.

[−100, 100 ][−100, 100]

89.

1. ⓐ g(5000)=50;g(5000)=50;
2. ⓑ The number of cubic yards of dirt required for a garden of 100 square feet is 1.

91.

1. ⓐ The height of a rocket above ground after 1 second is 200 ft.
2. ⓑ the height of a rocket above ground after 2 seconds is 350 ft.

1.2 Section Exercises

1.

The domain of a function depends upon what values of the independent variable make the function undefined or imaginary.

3.

There is no restriction on xx for f(x)=x3f(x)=x3 because you can take the cube root of any real number. So the domain is all real numbers, (−∞,∞).(−∞,∞). When dealing with the set of real numbers, you cannot take the square root of negative numbers. So xx-values are restricted for f(x)=x f(x)=x to nonnegative numbers and the domain is [0,∞).[0,∞).

5.

Graph each formula of the piecewise function over its corresponding domain. Use the same scale for the xx-axis and yy-axis for each graph. Indicate inclusive endpoints with a solid circle and exclusive endpoints with an open circle. Use an arrow to indicate −∞−∞ or ∞.∞. Combine the graphs to find the graph of the piecewise function.

15.

(−∞,−12)∪(−12,∞)(−∞,−12)∪(− 12,∞)

17.

(−∞,−11)∪(−11,2)∪(2,∞)(−∞,−11)∪(−11,2)∪(2,∞)

19.

(−∞,−3)∪(−3, 5)∪(5,∞)(−∞,−3)∪(−3,5)∪(5,∞)

25.

(−∞, −9)∪(−9,9)∪(9,∞)(−∞,−9)∪(−9,9)∪( 9,∞)

27.

domain: (2,8],(2,8], range [6,8)[6,8)

29.

domain: [−4, 4],[−4, 4], range: [0, 2] [0, 2]

31.

domain: [−5,3),[−5,3), range: [0,2][0,2]

33.

domain: (−∞,1], (−∞,1], range: [0,∞)[0,∞)

35.

domain: [−6,−16 ]∪[16,6];[−6,−16]∪[16,6]; range: [−6,−16]∪[16,6][−6,−16]∪[ 16,6]

37.

domain: [−3,∞);[−3,∞); range: [0,∞)[0,∞)

39.

domain: (−∞,∞)(−∞,∞)

41.

domain: (−∞,∞)(−∞,∞)

43.

domain: (−∞,∞)(−∞,∞)

45.

domain: (−∞,∞)(−∞,∞)

47.

f(−3)=1;f (−2)=0;f(−1)=0;f(0)=0f(−3)=1;f(−2)=0;f(−1)=0;f(0)=0

49.

f(−1)=−4;f(0)=6;f(2)=20;f(4)=34 f(−1)=−4;f(0)=6;f(2)=20; f(4)=34

51.

f(−1)=−5; f(0)=3;f(2)=3;f(4)=16f( −1)=−5;f(0)=3;f(2)=3;f(4)=16

53.

domain: (−∞,1 )∪(1,∞)(−∞,1)∪(1,∞)

55.

window: [−0.5,−0.1];[ −0.5,−0.1]; range: [4,100][4,100]

window: [0.1, 0.5];[0.1,0.5]; range: [4,100] [4,100]

59.

Many answers. One function is f(x)=1x−2 .f(x)=1x−2.

1.3 Section Exercises

1.

Yes, the average rate of change of all linear functions is constant.

3.

The absolute maximum and minimum relate to the entire graph, whereas the local extrema relate only to a specific region around an open interval.

11.

−113(13 +h)−113(13+h)

13.

3h2+9h+93h2+9h+9

19.

increasing on (−∞,−2.5)∪ (1,∞),(−∞,−2.5)∪(1,∞), decreasing on (−2.5 ,1)(−2.5,1)

21.

increasing on (−∞,1)∪( 3,4),(−∞,1)∪(3,4), decreasing on (1,3) ∪(4,∞)(1,3)∪(4,∞)

23.

local maximum: (−3,50),(−3,50), local minimum: (3,−50) (3,−50)

25.

absolute maximum at approximately (7,150),(7,150), absolute minimum at approximately (−7.5,−220) (−7.5,−220)

35.

Local minimum at (3,−22), (3,−22), decreasing on (−∞,3),( −∞,3), increasing on (3,∞)(3,∞)

37.

Local minimum at (−2,−2),(−2,−2), decreasing on (−3, −2),(−3,−2), increasing on (−2,∞) (−2,∞)

39.

Local maximum at (−0.5,6),(−0.5,6), local minima at (−3.25,−47)(−3.25,−47) and (2.1,−32),(2.1,−32), decreasing on (−∞,−3.25)(−∞,−3.25) and (−0.5,2.1),(−0.5,2.1), increasing on (−3.25,−0.5)(−3.25,−0.5) and (2.1, ∞)(2.1,∞)

45.

2.7 gallons per minute

47.

approximately –0.6 milligrams per day

1.4 Section Exercises

1.

Find the numbers that make the function in the denominator gg equal to zero, and check for any other domain restrictions on ff and g,g, such as an even-indexed root or zeros in the denominator.

3.

Yes. Sample answer: Let f(x)=x+1 and g(x)=x−1.f(x)=x+1 and g(x )=x−1. Then f(g(x))=f(x−1)=(x−1)+1=xf(g(x))=f(x−1)=(x−1)+1=x and g(f(x))=g(x+1)=(x+1)−1=x. g(f(x))=g(x+1)=(x+1)−1=x. So f∘g=g∘f.f∘g=g∘f.

5.

(f+g)(x)=2x +6,(f+g)(x)=2x+6, domain: (−∞,∞) (−∞,∞)

(f−g)(x)=2x2+2x−6 ,(f−g)(x)=2x2+2x−6, domain: (−∞,∞)(−∞,∞)

(fg)(x)=−x4−2 x3+6x2+12x,(fg)(x)=−x4−2x3+6x2+12x, domain: (−∞,∞)(−∞,∞)

(fg)(x)=x2+2x6−x2,(fg)(x)= x2+2x6−x2, domain: (−∞,−6)∪(−6,6 )∪(6,∞)(−∞,−6)∪(−6,6)∪( 6,∞)

7.

(f+g)(x)=4x3+8x2+12x,(f+g)(x)=4x3+8x2+12x, domain: (−∞,0)∪( 0,∞)(−∞,0)∪(0,∞)

(f−g )(x)=4x3+8x2−12x,(f−g)(x)=4x 3+8x2−12x, domain: (−∞,0)∪(0,∞)(−∞,0)∪(0,∞)

(fg)(x)=x+ 2,(fg)(x)=x+2, domain: (−∞,0)∪(0 ,∞)(−∞,0)∪(0,∞)

(fg) (x)=4x3+8x2,(fg)(x)=4x3+8x2, domain: (−∞,0)∪(0,∞)(−∞,0)∪(0,∞ )

9.

(f+g)(x)=3x2+x−5,(f+g)(x)= 3x2+x−5, domain: [5,∞)[5,∞)

(f−g)(x)=3x2−x−5,(f− g)(x)=3x2−x−5, domain: [5,∞)[5,∞)

(fg)(x)=3x2x−5, (fg)(x)=3x2x−5, domain: [5,∞) [5,∞)

(fg)(x)=3x2 x−5,(fg)(x)=3x2x−5, domain: (5,∞)(5,∞)

11.

1. ⓐ 3
2. ⓑf(g(x)) =2(3x−5)2+1;f(g(x))=2(3x−5)2+ 1;
3. ⓒg(f)(x))=6x2−2;g(f)(x) )=6x2−2;
4. ⓓ(g∘g)(x)=3(3x−5)−5=9 x−20;(g∘g)(x)=3(3x−5)−5=9x−20;
5. ⓔ (f∘f)(−2)=163(f∘f)(−2)=163

13.

f(g(x))=x2+3+2,g(f(x))=x+4x+7f(g(x))=x2+3+2,g(f(x))=x+4x+7

15.

f( g(x))=x+1x33=x+13x,g(f(x))=x3+1xf(g(x))=x+1x33 =x+13x,g(f(x))=x3+1x

17.

(f∘g)(x)=12x+4−4=x2,(g∘f)(x)=2x−4( f∘g)(x)=12x+4−4=x2,(g∘f)(x)=2x −4

19.

f(g(h(x)))=(1x+3)2+1f(g( h(x)))=(1x+3)2+1

21.

• ⓐ Text (g∘f)(x)=−32−4x;(g∘f)(x)=−32−4x;
• ⓑ(−∞,12)(−∞,12)

23.

1. ⓐ(0,2)∪ (2,∞);(0,2)∪(2,∞);
2. ⓑ(−∞ ,−2)∪(2,∞);(−∞,−2)∪(2,∞); c. (0,∞)(0,∞)

27.

sample: f(x)=x3 g(x)=x−5f(x)=x3g(x)=x−5

29.

sample: f(x)=4xg(x)=(x+2)2 f(x)=4xg(x)=(x+2)2

31.

sample: f(x)=x3g(x)=12x−3f (x)=x3g(x)=12x−3

33.

sample: f(x)=x4 g(x)=3x−2x+5f(x)=x4g( x)=3x−2x+5

35.

sample: f(x)=xf(x)=x
g(x)=2x+6g(x)=2x+6

37.

sample: f(x)=x3f(x)=x3
g(x)=(x−1)g(x)=(x−1)

39.

sample: f(x)=x3f(x)=x3
g(x)=1x−2g(x)=1x−2

41.

sample: f(x)=xf(x)=x
g(x)=2x−13x+4g(x)=2x−13x+4

73.

f(g(0))=27 ,g(f(0))=−94f(g(0))=27,g(f (0))=−94

75.

f(g(0))=15,g(f(0))=5 f(g(0))=15,g(f(0))=5

77.

18x2+60x+51 18x2+60x+51

79.

g∘g(x)=9x+20 g∘g(x)=9x+20

87.

(f∘g)(6)= 6(f∘g)(6)=6; (g∘f)(6)= 6(g∘f)(6)=6

89.

(f∘g)(11)= 11,(g∘f)(11)=11(f∘g)(11)=11,(g∘f)(11)=11

93.

A(t)=π(25t +2)2A(t)=π(25t+2)2 and A(2)=π(254)2=2500πA(2)=π(254) 2=2500π square inches

95.

A(5)=π(2(5)+1)2=121πA(5)=π (2(5)+1)2=121π square units

97.

• ⓐN(T(t)) =23(5t+1.5)2−56(5t+1.5)+1;N(T(t)) =23(5t+1.5)2−56(5t+1.5)+1;
• ⓑ 3.38 hours

1.5 Section Exercises

1.

A horizontal shift results when a constant is added to or subtracted from the input. A vertical shifts results when a constant is added to or subtracted from the output.

3.

A horizontal compression results when a constant greater than 1 is multiplied by the input. A vertical compression results when a constant between 0 and 1 is multiplied by the output.

5.

For a function f,f, substitute (−x)(−x) for (x)(x) in f(x).f(x). Simplify. If the resulting function is the same as the original function, f(−x)=f(x),f(−x)=f(x), then the function is even. If the resulting function is the opposite of the original function, f(−x)=−f( x),f(−x)=−f(x), then the original function is odd. If the function is not the same or the opposite, then the function is neither odd nor even.

7.

g(x)=| x-1|−3g(x)=|x-1|−3

9.

g(x)=1(x +4)2+2g(x)=1(x+4)2+2

11.

The graph of f(x+43)f(x+43) is a horizontal shift to the left 43 units of the graph of f.f.

13.

The graph of f(x-4 )f(x-4) is a horizontal shift to the right 4 units of the graph of f.f.

15.

The graph of f(x)+8f(x)+8 is a vertical shift up 8 units of the graph of f.f.

17.

The graph of f(x)−7f(x)−7 is a vertical shift down 7 units of the graph of f.f.

19.

The graph of f(x+4 )−1f(x+4)−1 is a horizontal shift to the left 4 units and a vertical shift down 1 unit of the graph of f. f.

21.

decreasing on (−∞,−3)(−∞,−3) and increasing on (−3,∞) (−3,∞)

23.

decreasing on [0,∞) [0,∞)

31.

g(x)=f(x-1 ),h(x)=f(x)+1g(x)=f(x-1), h(x)=f(x)+1

33.

f(x)=|x-3|−2 f(x)=|x-3|−2

35.

f(x)=x+3− 1f(x)=x+3−1

37.

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

39.

f(x)=|x+3|−2 f(x)=|x+3|−2

43.

f(x)=−(x+ 1)2+2f(x)=−(x+1)2+2

45.

f(x)=−x+1 f(x)=−x+1

53.

The graph of gg is a vertical reflection (across the xx-axis) of the graph of f.f.

55.

The graph of gg is a vertical stretch by a factor of 4 of the graph of f.f.

57.

The graph of gg is a horizontal compression by a factor of 1515 of the graph of f.f.

59.

The graph of g g is a horizontal stretch by a factor of 3 of the graph of f.f.

61.

The graph of gg is a horizontal reflection across the yy-axis and a vertical stretch by a factor of 3 of the graph of f.f.

63.

g(x)=| −4x|g(x)=|−4x|

65.

g(x)=13( x+2)2−3g(x)=13(x+2)2−3

67.

g(x) =12(x-5)2+1g(x)=12(x-5)2+1

69.

The graph of the function f(x)=x2f(x)=x2 is shifted to the left 1 unit, stretched vertically by a factor of 4, and shifted down 5 units.

71.

The graph of f(x)=|x| f(x)=|x| is stretched vertically by a factor of 2, shifted horizontally 4 units to the right, reflected across the horizontal axis, and then shifted vertically 3 units up.

73.

The graph of the function f(x)=x3 f(x)=x3 is compressed vertically by a factor of 12.12.

75.

The graph of the function is stretched horizontally by a factor of 3 and then shifted vertically downward by 3 units.

77.

The graph of f(x)=x f(x)=x is shifted right 4 units and then reflected across the vertical line x=4.x=4.

1.6 Section Exercises

1.

Isolate the absolute value term so that the equation is of the form |A|=B.|A|=B. Form one equation by setting the expression inside the absolute value symbol, A,A, equal to the expression on the other side of the equation, B.B. Form a second equation by setting AA equal to the opposite of the expression on the other side of the equation, −B.−B. Solve each equation for the variable.

3.

The graph of the absolute value function does not cross the xx -axis, so the graph is either completely above or completely below the xx-axis.

5.

First determine the boundary points by finding the solution(s) of the equation. Use the boundary points to form possible solution intervals. Choose a test value in each interval to determine which values satisfy the inequality.

7.

|x+4| =12|x+4|=12

9.

|f(x)−8|<0.03 |f(x)−8|<0.03

13.

{-94,134 }{-94,134}

15.

{103,203 }{103,203}

17.

{115,295 }{115,295}

19.

{52,72}{52,72}

23.

{−57,27} {−57,27}

25.

(0,−8);(−6,0),(4,0)(0,−8 );(−6,0),(4,0)

27.

(0,−7); (0,−7); no xx-intercepts

29.

(−∞,−8)∪(12, ∞)(−∞,−8)∪(12,∞)

33.

(−∞,−83]∪[6 ,∞)(−∞,−83]∪[6,∞)

35.

(−∞,−83]∪[16 ,∞)(−∞,−83]∪[16,∞)

53.

range: [0,20] [0,20]

55.

x-x- intercepts:

59.

There is no solution for aa that will keep the function from having a yy-intercept. The absolute value function always crosses the yy-intercept when x=0.x=0.

61.

|p−0.08|≤0.015|p−0.08|≤0.015

63.

|x−5.0|≤0.01 |x−5.0|≤0.01

1.7 Section Exercises

1.

Each output of a function must have exactly one output for the function to be one-to-one. If any horizontal line crosses the graph of a function more than once, that means that yy-values repeat and the function is not one-to-one. If no horizontal line crosses the graph of the function more than once, then no yy-values repeat and the function is one-to-one.

3.

Yes. For example, f(x)= 1xf(x)=1x is its own inverse.

5.

Given a function y=f(x),y=f(x), solve for xx in terms of y.y. Interchange the xx and y.y. Solve the new equation for y.y. The expression for yy is the inverse, y=f−1(x).y=f−1(x).

7.

f−1(x)=x −3f−1(x)=x−3

9.

f−1(x)=2−xf−1( x)=2−x

11.

f−1(x)=−2xx−1f−1(x)=−2x x−1

13.

domain of f(x):[−7,∞);f−1(x)=x−7f(x):[−7,∞);f−1(x)=x−7

15.

domain of f(x): [0,∞);f−1(x)=x+5f(x):[0,∞);f−1(x)=x+5

16.

• ⓐf(g(x))=x f(g(x))=x and g(f(x))=x. g(f(x))=x.
• ⓑ This tells us that ff and gg are inverse functions

17.

f(g(x))=x, g(f(x))=xf(g(x))=x,g(f(x))=x

41.

43.

f−1(x)=(1+x)1/3f−1(x)=(1+x)1/3

45.

f−1(x)= 59(x−32).f−1(x)=59(x−32). Given the Fahrenheit temperature, x,x, this formula allows you to calculate the Celsius temperature.

47.

t(d)=d50,t(d)=d50, t(180)=18050. t(180)=18050. The time for the car to travel 180 miles is 3.6 hours.

Review Exercises

5.

f(−3)=−27;f(−3)=−27; f(2)=−2;f(2)=−2; f(−a)=−2a2−3a;f(−a) =−2a2−3a;
−f(a)=2a2−3a;−f (a)=2a2−3a; f(a+h)=−2a2+3a−4ah+ 3h−2h2f(a+h)=−2a2+3a−4ah+3h−2h2

17.

x=−1.8 x=−1.8 or  or x=1.8 or x=1.8

19.

−64+80a−16a2−1+a =−16a+64−64+80a−16a2−1+a=−16a+64

21.

(−∞,−2)∪(− 2,6)∪(6,∞)(−∞,−2)∪(−2,6)∪(6,∞)

27.

increasing (2,∞) ;(2,∞); decreasing (−∞,2)(−∞,2 )

29.

increasing ( −3,1);(−3,1); constant (−∞,−3)∪(1, ∞)(−∞,−3)∪(1,∞)

31.

local minimum (−2,−3); (−2,−3); local maximum (1,3)(1,3)

35.

(f∘g)(x )=17−18x;(g∘f)(x)=−7−18x(f∘g)(x )=17−18x;(g∘f)(x)=−7−18x

37.

(f∘g)(x)=1 x+2;(f∘g)(x)=1x+2;(g∘ f)(x)=1x+2(g∘f)(x)=1x+2

39.

(f∘g )(x)=1+x1+4x,x≠0,x≠−14(f∘ g)(x)=1+x1+4x,x≠0,x≠−14

41.

(f∘g)(x) =1x,x>0(f∘g)(x)=1x,x>0

43.

sample: g(x )=2x−13x+4;f(x)=xg(x)=2x− 13x+4;f(x)=x

55.

f(x)=|x−3| f(x)=|x−3|

63.

f(x)=12|x+2 |+1f(x)=12|x+2|+1

65.

f(x)=−3|x−3| +3f(x)=−3|x−3|+3

69.

x=−22,x=14 x=−22,x=14

71.

(−53,3)(−53,3)

73.

f−1(x)=x- 1f−1(x)=x-1

77.

The function is one-to-one.

78.

The function is not one-to-one.

Practice Test

1.

The relation is a function.

5.

The graph is a parabola and the graph fails the horizontal line test.

19.

x=−7x=−7 and x=10x=10

21.

f−1(x)=x+53 f−1(x)=x+53

23.

(−∞,−1.1) and (1.1, ∞)(−∞,−1.1) and (1.1,∞)

25.

(1.1,−0.9)(1.1 ,−0.9)

29.

f(x)={|x|ifx≤23ifx>2 f(x)={|x|ifx≤23ifx>2

35.

f− 1(x)=−x−112f−1(x)=−x−112