Download 1-7 Inverse Relations and Functions

1-7 Inverse Relations and Functions
Graph each function using a graphing calculator, and apply the
horizontal line test to determine whether its inverse function
exists. Write yes or no.
2
3. f (x) = x2 – 10x + 25
SOLUTION: 2
1. f (x) = x + 6x + 9
SOLUTION: 2
The graph of f (x) = x – 10x + 25 below shows that it is possible to find
a horizontal line that intersects the graph of f (x) more than once.
Therefore, you can conclude that an inverse function does not exist.
The graph of f (x) = x + 6x + 9 below shows that it is possible to find a
horizontal line that intersects the graph of f (x) more than once.
Therefore, you can conclude that an inverse function does not exist.
4. f (x) = 3x − 8
SOLUTION: It appears from the portion of the graph of f (x) = 3x − 8 shown below
that there is no horizontal line that intersects the graph of f (x) more than
once. Therefore, you can conclude that an inverse function does exist.
2. f (x) = x2 – 16x + 64
SOLUTION: 2
The graph of f (x) = x – 16x + 64below shows that it is possible to find
a horizontal line that intersects the graph of f (x) more than once.
Therefore, you can conclude that an inverse function does not exist.
5. f (x) = SOLUTION: It appears from the portion of the graph of f (x) =
shown below
that there is no horizontal line that intersects the graph of f (x) more than
once. Therefore, you can conclude that an inverse function does exist.
2
3. f (x) = x – 10x + 25
SOLUTION: 2
The
graph
of f (x) by
= xCognero
– 10x
eSolutions
Manual
- Powered
+ 25 below shows that it is possible to find
a horizontal line that intersects the graph of f (x) more than once.
Therefore, you can conclude that an inverse function does not exist.
Page 1
1-7 Inverse Relations and Functions
5. f (x) = 7. f (x) = SOLUTION: SOLUTION: It appears from the portion of the graph of f (x) =
shown below
that there is no horizontal line that intersects the graph of f (x) more than
once. Therefore, you can conclude that an inverse function does exist.
It appears from the portion of the graph of f (x) =
shown below
that there is no horizontal line that intersects the graph of f (x) more than
once. Therefore, you can conclude that an inverse function does exist.
6. f (x) = 4
SOLUTION: The graph of f (x) = 4 below shows that it is possible to find a horizontal
line that intersects the graph of f (x) more than once. That horizontal line
is the same as the graph of f (x) itself. Therefore, you can conclude that
an inverse function does not exist.
7. f (x) = SOLUTION: It appears from the portion of the graph of f (x) =
shown below
that there is no horizontal line that intersects the graph of f (x) more than
once. Therefore, you can conclude that an inverse function does exist.
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8. f (x) = −4x2 + 8
SOLUTION: 2
The graph of f (x) = −4x + 8 below shows that it is possible to find a
horizontal line that intersects the graph of f (x) more than once.
Therefore, you can conclude that an inverse function does not exist.
9. f (x) =
SOLUTION: It appears from the portion of the graph of f (x) =
shown below that there is no horizontal line that intersects the graph of f (x) morePage
than2
once. Therefore, you can conclude that an inverse function does exist.
1-7 Inverse Relations and Functions
11. f (x) = x3 – 9
9. f (x) =
SOLUTION: SOLUTION: 3
shown below It appears from the portion of the graph of f (x) =
It appears from the portion of the graph of f (x) = x – 9 shown below
that there is no horizontal line that intersects the graph of f (x) more than
once. Therefore, you can conclude that an inverse function does exist.
that there is no horizontal line that intersects the graph of f (x) more than
once. Therefore, you can conclude that an inverse function does exist.
12. f (x) =
x
3
SOLUTION: 10. f (x) =
It appears from the portion of the graph of f (x) =
SOLUTION: shown below It appears from the portion of the graph of f (x) =
3
x shown below
that there is no horizontal line that intersects the graph of f (x) more than
once. Therefore, you can conclude that an inverse function does exist.
that there is no horizontal line that intersects the graph of f (x) more than
once. Therefore, you can conclude that an inverse function does exist.
Determine whether f has an inverse function. If it does, find the
inverse function and state any restrictions on its domain.
13. g(x) = −3x4 + 6x2 – x
11. f (x) = x3 – 9
SOLUTION: SOLUTION: eSolutions Manual - Powered by Cognero
4
3
It appears from the portion of the graph of f (x) = x – 9 shown below
that there is no horizontal line that intersects the graph of f (x) more than
once. Therefore, you can conclude that an inverse function does exist.
2
Page 3
The graph of g(x) = −3x + 6x – x below shows that it is possible to
find a horizontal line that intersects the graph of g(x) more than once.
Therefore, you can conclude that an inverse function does not exist.
1-7 Inverse Relations and Functions
Determine whether f has an inverse function. If it does, find the
inverse function and state any restrictions on its domain.
15. h(x) = x7 + 2x3 − 10x2
SOLUTION: 13. g(x) = −3x4 + 6x2 – x
7
SOLUTION: 4
2
The graph of g(x) = −3x + 6x – x below shows that it is possible to
find a horizontal line that intersects the graph of g(x) more than once.
Therefore, you can conclude that an inverse function does not exist.
3
2
The graph of h(x) = x + 2x − 10x below shows that it is possible to
find a horizontal line that intersects the graph of h(x) more than once.
Therefore, you can conclude that an inverse function does not exist.
16. f (x) =
5
14. f (x) = 4x – 8x
4
SOLUTION: SOLUTION: 5
4
The graph of f (x) = 4x – 8x below shows that it is possible to find a
horizontal line that intersects the graph of f (x) more than once.
Therefore, you can conclude that an inverse function does not exist.
It appears from the portion of the graph of f (x) =
shown below
that there is no horizontal line that intersects the graph of f (x) more than
once. Therefore, you can conclude that an inverse function does exist.
The function f has domain [–8,
) and range [0,
).
15. h(x) = x7 + 2x3 − 10x2
SOLUTION: 7
3
2
The graph of h(x) = x + 2x − 10x below shows that it is possible to
find
a horizontal
that intersects the graph of h(x) more than once.
eSolutions
Manual
- Poweredline
by Cognero
Therefore, you can conclude that an inverse function does not exist.
−1
2
Page 4
f (x) = x − 8
2
From the graph y = x – 8 below, you can see that the inverse relation
has domain (– , ) and range [8, ).
1-7 Inverse Relations and Functions
−1
2
f (x) = x − 8
2
From the graph y = x – 8 below, you can see that the inverse relation
has domain (– , ) and range [8, ).
By restricting the domain of the inverse relation to [0,
and range of f are equal to the range and domain of f
−1
2
Therefore, f (x) = x − 8 for x ≥ 0.
18. f (x) = | x – 6 |
SOLUTION: The graph of f (x) = | x – 6 | below shows that it is possible to find a
horizontal line that intersects the graph of f (x) more than once.
Therefore, you can conclude that an inverse function does not exist.
), the domain
–1
, respectively.
17. f (x) =
19. f (x) =
SOLUTION: SOLUTION: The graph of f (x) =
below shows that it is possible to find a horizontal line that intersects the graph of f (x) more than once.
Therefore, you can conclude that an inverse function does not exist.
It appears from the portion of the graph of f (x) =
18. f (x) = | x – 6 |
SOLUTION: The graph of f (x) = | x – 6 | below shows that it is possible to find a
horizontal line that intersects the graph of f (x) more than once.
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Therefore, you can conclude that an inverse function does not exist.
shown below that there is no horizontal line that intersects the graph of f (x) more than
once. Therefore, you can conclude that an inverse function does exist.
The function f has domain (–
(–1, ).
, 0)
(0, ) and range [–
, –1)
Page 5
20. g(x) =
1-7 Inverse
Relations and Functions
The function f has domain (– , 0) (0, (–1,
) and range [–
, –1)
SOLUTION: ).
It appears from the portion of the graph of g(x) =
shown below that there is no horizontal line that intersects the graph of g(x) more
than once. Therefore, you can conclude that an inverse function does
exist.
f
–−1
(x) =
From the graph y =
has domain (–
, –1)
below, you can see that the inverse relation (–1,
) and range (–
, 0)
(0, The function g has domain (–
(1, ).
, 0)
(0, ) and range [–
, 1)
).
–1
The domain and range of f are equal to the range and domain of f ,
–−1
respectively. Therefore, no further restrictions are necessary. f (x)
=
g−1(x) =
From the graph y =
for x ≠ −1.
has domain [–
, 1)
below, you can see that the inverse relation (1, ) and range (–
, 0)
(0, ).
20. g(x) =
SOLUTION: It appears from the portion of the graph of g(x) =
eSolutions Manual - Powered by Cognero
shown below that there is no horizontal line that intersects the graph of g(x) more
than once. Therefore, you can conclude that an inverse function does
exist.
Page 6
–1
g−1(x) =
From the graph
y=
below, you can see that the inverse relation 1-7 Inverse
Relations
and
Functions
has domain [–
, 1)
(1, ) and range (–
(0, , 0)
).
The function f has domain (–
, 8) and range (0,
).
–1
The domain and range of g are equal to the range and domain of g ,
–−1
respectively. Therefore, no further restrictions are necessary. g (x)
=
for x ≠ 1.
21. f (x) =
f
SOLUTION: –−1
(x) = 8 −
.
From the graph y =
shown below
It appears from the portion of the graph of f (x) =
has domain [–
, 0)
below, you can see that the inverse relation (0, ) and range (–
, 8).
that there is no horizontal line that intersects the graph of f (x) more than
once. Therefore, you can conclude that an inverse function does exist.
By restricting the domain of the inverse relation to (0, ∞), the domain
and range of f are equal to the range and domain of f
The function f has domain (–
, 8) and range (0,
).
Therefore, f
−1
(x) = 8 −
–1
, respectively.
for x > 0.
22. g(x) =
SOLUTION: eSolutions Manual - Powered by Cognero
Page 7
It appears from the portion of the graph of g(x) =
shown Therefore, f
−1
(x) = 8 −
for x > 0.
g−1(x) = −3 +
1-7
22. Inverse
g(x) = Relations and Functions
From the graph y = −3 +
relation has domain (–
below, you can see that the inverse , 0)
(0, ) and range (–3,
).
SOLUTION: shown It appears from the portion of the graph of g(x) =
below that there is no horizontal line that intersects the graph of g(x)
more than once. Therefore, you can conclude that an inverse function
does exist.
By restricting the domain of the inverse relation to (0, ), the domain
–1
and range of g are equal to the range and domain of g , respectively.
Therefore, g
–−1
(x) = −3 +
for x > 0.
23. f (x) =
The function g has domain (–3,
) and range (0,
).
SOLUTION: It appears from the portion of the graph of f (x) =
shown below that there is no horizontal line that intersects the graph of f (x) more than
once. Therefore, you can conclude that an inverse function does exist.
The function f has domain (–
(6, ).
g−1(x) = −3 +
From the graph y = −3 +
relation has domain (–
, 8)
(8, ) and range (–
, 6)
below, you can see that the inverse , 0)
eSolutions Manual - Powered by Cognero
(0, ) and range (–3,
).
Page 8
24. h(x) =
SOLUTION: 1-7 Inverse
Relations and Functions
The function f has domain (– , 8) (8, (6, ) and range (–
, 6)
It appears from the portion of the graph of h(x) =
).
shown below
that there is no horizontal line that intersects the graph of h(x) more
than once. Therefore, you can conclude that an inverse function does
exist.
f
–−1
(x) =
The function h has domain
has domain (–
, 6)
.
below, you can see that the inverse relation
From the graph y =
(6, ) and range (–
, 8)
(8, and range
).
–1
The domain and range of f are equal to the range and domain of f ,
–−1
respectively. Therefore, no further restrictions are necessary. f (x) =
for x
−1
h (x) =
6.
From the graph y =
24. h(x) =
has domain
below, you can see that the inverse relation
and range
.
SOLUTION: It appears from the portion of the graph of h(x) =
shown below
that there is no horizontal line that intersects the graph of h(x) more
than once. Therefore, you can conclude that an inverse function does
exist.
eSolutions Manual - Powered by Cognero
Page 9
h (x) =
From the graph y =
below, you can see that the inverse relation
1-7 Inverse Relations and Functions
has domain
and range
.
26. SPEED The speed of an object in kilometers per hour y is y = 1.6x,
where x is the speed of the object in miles per hour.
a. Find an equation for the inverse of the function. What does each
variable represent?
b. Graph each equation on the same coordinate plane.
SOLUTION: a.
–1
The domain and range of h are equal to the range and domain of h ,
−1
respectively. Therefore, no further restrictions are necessary. h (x) =
for x ≠ .
25. g(x) = | x + 1 | + | x – 4 |
y = speed in mi/h, x = speed in km/h
b.
SOLUTION: The graph of g(x) = | x + 1 | + | x – 4 || below shows that it is possible
to find a horizontal line that intersects the graph of g(x) more than once.
Therefore, you can conclude that an inverse function does not exist.
Show algebraically that f and g are inverse functions.
27. f (x) = −6x + 3
g(x) =
26. SPEED The speed of an object in kilometers per hour y is y = 1.6x,
where x is the speed of the object in miles per hour.
a. Find an equation for the inverse of the function. What does each
variable represent?
b. Graph each equation on the same coordinate plane.
SOLUTION: SOLUTION: a.
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Page 10
1-7 Inverse Relations and Functions
Show algebraically that f and g are inverse functions.
27. f (x) = −6x + 3
29. f (x) = −3x2 + 5; x ≥ 0
g(x) =
g(x) =
SOLUTION: SOLUTION: 28. f (x) = 4x + 9
g(x) =
30. f (x) =
SOLUTION: + 8; x ≥ 0
g(x) =
SOLUTION: 2
29. f (x)Manual
= −3x- Powered
+ 5; x ≥ 0
eSolutions
by Cognero
g(x) =
Page 11
1-7 Inverse Relations and Functions
30. f (x) =
+ 8; x ≥ 0
31. f (x) = 2x3 – 6
g(x) =
SOLUTION: SOLUTION: 31. f (x) = 2x3 – 6
32. f (x) =
g(x) =
SOLUTION: eSolutions Manual - Powered by Cognero
− 8; x ≥ 0
SOLUTION: Page 12
1-7 Inverse Relations and Functions
34. g(x) =
32. f (x) =
+ 5
2
f (x) = x – 10x + 33; x ≥ 5
− 8; x ≥ 0
g(x) =
SOLUTION: SOLUTION: 35. f (x) =
33. g(x) =
− 4
2
f (x) = x + 8x + 8; x ≥ −4
g(x) =
SOLUTION: SOLUTION: eSolutions
Manual
34. g(x)
= - Powered
+ 5by Cognero
2
f (x) = x – 10x + 33; x ≥ 5
Page 13
1-7 Inverse Relations and Functions
35. f (x) =
36. f (x) =
g(x) =
g(x) =
SOLUTION: SOLUTION: eSolutions
36. f (x)Manual
= - Powered by Cognero
g(x) =
Page 14
36. f (x) =
g(x) =
1-7 Inverse Relations and Functions
SOLUTION: 37. PHYSICS The kinetic energy of an object in motion in joules can be
2
described by f (x) = 0.5mx , where m is the mass of the object in
kilograms and x is the velocity of the object in meters per second.
a. Find the inverse of the function. What does each variable represent?
b. Show that f (x) and the function you found in part a are inverses.
c. Graph f (x) and f –1(x) on the same graphing calculator screen if the
mass of the object is 1 kilogram.
SOLUTION: a.
g(x) = velocity in m/s, x = kinetic energy in joules, m = mass in kg
b.
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Page 15
37. PHYSICS The kinetic energy of an object in motion in joules can be
2
described by f (x) = 0.5mx , where m is the mass of the object in
Because
= = x, the functions are inverse when the domain
g(x) = velocity
in m/s, xand
= kinetic
energy in joules, m = mass in kg
1-7 Inverse
Relations
Functions
b.
Use the graph of each function to graph its inverse function.
38. SOLUTION: Graph the line y = x and reflect the points. Then connect the points with
a smooth curve that resembles the original graph.
Because
= is restricted to [0, ).
= x, the functions are inverse when the domain
c.
Use the graph of each function to graph its inverse function.
39. SOLUTION: Graph the line y = x and reflect the points. Then connect the points with
a smooth curve that resembles the original graph.
38. eSolutions
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SOLUTION: Graph the line y = x and reflect the points. Then connect the points with
a smooth curve that resembles the original graph.
Page 16
1-7 Inverse Relations and Functions
39. 40. SOLUTION: SOLUTION: Graph the line y = x and reflect the points. Then connect the points with
a smooth curve that resembles the original graph.
Graph the line y = x and reflect the points. Then connect the points with
a smooth curve that resembles the original graph.
40. SOLUTION: Graph the line y = x and reflect the points. Then connect the points with
a smooth curve that resembles the original graph.
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41. SOLUTION: Graph the line y = x and reflect the points. Then connect the points with
a smooth curve that resembles the original graph.
Page 17
1-7 Inverse Relations and Functions
42. 41. SOLUTION: SOLUTION: Graph the line y = x and reflect the points. Then connect the points with
a smooth curve that resembles the original graph.
Graph the line y = x and reflect the points. Then connect the points with
a smooth curve that resembles the original graph.
43. 42. SOLUTION: SOLUTION: Graph the line y = x and reflect the points. Then connect the points with
a smooth curve that resembles the original graph.
Graph the line y = x and reflect the points. Then connect the points with
a smooth curve that resembles the original graph.
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Page 18
1-7 Inverse Relations and Functions
44. JOBS Jamie sells shoes at a department store after school. Her base
salary each week is $140, and she earns a 10% commission on each
pair of shoes that she sells. Her total earnings f (x) for a week in which
she sold x dollars worth of shoes is f (x) = 140 + 0.1x.
−1
−1
a. Explain why the inverse function f (x) exists. Then find f (x).
b. What do f −1(x) and x represent in the inverse function?
c. What restrictions, if any, should be placed on the domains of f (x) and
43. −1
SOLUTION: Graph the line y = x and reflect the points. Then connect the points with
a smooth curve that resembles the original graph.
f (x)? Explain.
d. Find Jamie’s total sales last week if her earnings for that week were
$220.
SOLUTION: a. Sample answer: The graph of the function is linear, so it passes the
horizontal line test. Therefore, it is a one-to-one function and it has an
inverse.
44. JOBS Jamie sells shoes at a department store after school. Her base
salary each week is $140, and she earns a 10% commission on each
pair of shoes that she sells. Her total earnings f (x) for a week in which
she sold x dollars worth of shoes is f (x) = 140 + 0.1x.
−1
−1
a. Explain why the inverse function f (x) exists. Then find f (x).
b. What do f −1(x) and x represent in the inverse function?
c. What restrictions, if any, should be placed on the domains of f (x) and
−1
f (x) = 10x − 1400
−1
b. x represents Jamie’s earnings for a week, and f (x) represents her
sales.
c. x ≥ 0; Jamie cannot have negative sales.
d.
−1
f (x)? Explain.
d. Find Jamie’s total sales last week if her earnings for that week were
$220.
SOLUTION: a. Sample answer: The graph of the function is linear, so it passes the
horizontal line test. Therefore, it is a one-to-one function and it has an
inverse.
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45. CURRENCY The average exchange rate from Euros to U.S. dollars
for a recent four-month period can be described by f (x) = 0.66x, where
x is the currency value in Euros.
a. Explain why the inverse function f −1(x) exists. Then find f −1(x).
−1
b. What do f (x) and x represent in the inverse function?
Page 19
c. What restrictions, if any, should be placed on the domains of f (x) and
−1
f (x)? Explain.
c. x ≥ 0; You cannot exchange negative money.
d.
d.
1-7 Inverse Relations and Functions
Determine whether each function has an inverse function.
45. CURRENCY The average exchange rate from Euros to U.S. dollars
for a recent four-month period can be described by f (x) = 0.66x, where
x is the currency value in Euros.
a. Explain why the inverse function f −1(x) exists. Then find f −1(x).
−1
b. What do f (x) and x represent in the inverse function?
c. What restrictions, if any, should be placed on the domains of f (x) and
−1
f (x)? Explain.
d. What is the value in Euros of 100 U.S. dollars?
46. SOLUTION: SOLUTION: a. Sample answer: The graph of the function is linear, so it passes the
horizontal line test. Therefore, it is a one-to-one function and it has an
inverse.
The graph does not pass the Horizontal Line Test.
b. x represents the value of the currency in U.S. dollars, and f
−1
(x)
represents the value of the currency in Euros.
c. x ≥ 0; You cannot exchange negative money.
d.
Determine whether each function has an inverse function.
47. SOLUTION: The graph passes the Horizontal Line Test.
46. SOLUTION: The graph does not pass the Horizontal Line Test.
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Page 20
48. 49. 47. SOLUTION: SOLUTION: The graph passes the Horizontal Line Test.
The graphRelations
passes the Horizontal
Line Test.
1-7 Inverse
and Functions
Determine if f −1 exists. If so, complete a table for f −1.
50. SOLUTION: No output value corresponds with more than one input value, so an
inverse exists.
48. SOLUTION: The graph does not pass the Horizontal Line Test.
51. SOLUTION: An output value corresponds with more than one input value, so an
inverse does not exist.
52. SOLUTION: An output value corresponds with more than one input value, so an
inverse does not exist.
49. SOLUTION: The graph passes the Horizontal Line Test.
Determine if f −1 exists. If so, complete a table for f −1.
53. SOLUTION: No output value corresponds with more than one input value, so an
inverse exists.
50. SOLUTION: NoManual
output- Powered
value corresponds
eSolutions
by Cognero
inverse exists.
with more than one input value, so an
54. TEMPERATURE The formula f (x) =
Page 21
x + 32 is used to convert x
degrees Celsius to degrees Fahrenheit. To convert x degrees
No output value corresponds with more than one input value, so an
inverse exists.
1-7 Inverse Relations and Functions
54. TEMPERATURE The formula f (x) =
−1
f represents the formula used to convert degrees Fahrenheit to
degrees Celsius.
b.
x + 32 is used to convert x
degrees Celsius to degrees Fahrenheit. To convert x degrees
Fahrenheit to Kelvin, the formula k(x) =
(x + 459.67) is used.
−1
a. Find f . What does this function represent?
b. Show that f and f −1 are inverse functions. Graph each function on
the same graphing calculator screen.
c. Find (k o f )(x).What does this function represent?
d. If the temperature is 60°C, what would the temperature be in
Kelvin?
SOLUTION: a.
c.
−1
f represents the formula used to convert degrees Fahrenheit to
degrees Celsius.
b.
[k o f ](x) = x + 273.15; represents the formula used to convert degrees
Celsius to degrees Kelvin.
d.
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Restrict the domain of each function so that the resulting
function is one-to-one. Then determine the inverse of the
function.
Page 22
Celsius to degrees Kelvin.
d.
1-7 Inverse Relations and Functions
f
−1
(x) =
+ 5
Restrict the domain of each function so that the resulting
function is one-to-one. Then determine the inverse of the
function.
56. SOLUTION: The domain needs to be restricted so that the graph passes the
Horizontal Line Test. There is more than one way to accomplish this.
Sample answer: x ≤ −9
55. SOLUTION: The domain needs to be restricted so that the graph passes the
Horizontal Line Test. There is more than one way to accomplish this.
Sample answer: x ≥ 5
We are selecting the negative side of the domain, so we will replace
the absolute value symbols with the opposite of the expression.
f
−1
(x) =
+ 5
−1
f (x) = x – 11
Since the range of the restricted function f is y ≤ –2, we must restrict
the inverse relation so that the domain is x ≤ 2. Therefore, f
11.
−1
(x) = x –
56. SOLUTION: The domain needs to be restricted so that the graph passes the
Horizontal Line Test. There is more than one way to accomplish this.
Sample answer: x ≤ −9
57. eSolutions Manual - Powered by Cognero
SOLUTION: The domain needs to be restricted so that the graph passes the
Page 23
−1
f (x) = x – 11
Since the range of the restricted function f is y ≤ –2, we must restrict
the inverseRelations
relation so that
theFunctions
domain is x ≤ 2. Therefore, f
1-7 Inverse
and
−1
We are using the positive side of the domain, so we only need the
positive value of y .
(x) = x –
−1
f (x) =
11.
57. 58. SOLUTION: SOLUTION: The domain needs to be restricted so that the graph passes the
Horizontal Line Test. There is more than one way to accomplish this.
Sample answer: x ≥ 0
The domain needs to be restricted so that the graph passes the
Horizontal Line Test. There is more than one way to accomplish this.
Sample answer: x ≥ −5
We are using the positive side of the domain, so we only need the
positive value of y .
We are selecting the positive side of the domain, so we will replace
the absolute value symbols with the expression.
−1
f (x) =
−1
f (x) = x – 1
Since the range of the restricted function f is y ≥ –4, we must restrict
the inverse relation so that the domain is x ≥ –4. Therefore, f
– 1, x ≥ –4.
−1
(x) = x
State the domain and range of f and f −1, if f −1 exists.
58. 59. f (x) =
SOLUTION: The domain needs to be restricted so that the graph passes the
Horizontal Line Test. There is more than one way to accomplish this.
Sample answer: x ≥ −5
We are selecting the positive side of the domain, so we will replace
the absolute value symbols with the expression.
eSolutions Manual - Powered by Cognero
SOLUTION: For f , the square root must be at least 0, so x must be greater than or
equal to 6.
f: D = {x | x ≥ 6, x R}, R = {y| y ≥ 0, y R}
Graph f .
Page 24
−1
f (x) = x – 1
Since the range of the restricted function f is y ≥ –4, we must restrict
the inverseRelations
relation so that
theFunctions
domain is x ≥ –4. Therefore, f
1-7 Inverse
and
−1
(x) = x
– 1, x ≥ –4.
State the domain and range of f and f −1, if f −1 exists.
59. f (x) =
−1
By restricting the domain of f to [0, ∞], the range becomes [6, ∞].
−1
f : D = {x| x ≥ 0, y R}, R = {y | y ≥ 6, x R}
60. f (x) = x2 + 9 SOLUTION: SOLUTION: For f , the square root must be at least 0, so x must be greater than or
equal to 6.
f: D = {x | x ≥ 6, x R}, R = {y| y ≥ 0, y R}
Graph f .
f
−1
does not exist
61. f (x) =
SOLUTION: The graph of f passes the Horizontal Line Test, so f −1 exists.
−1
Graph f .
There appears to be asymptotes at x = 4 and y = 3.
f: D = {x | x 4, x R}, R = {y | y 3, x R}
The graph of f passes the Horizontal Line Test, so f −1 exists.
By restricting the domain of f to [0, ∞], the range becomes [6, ∞].
−1
f : D = {x| x ≥ 0, y R}, R = {y | y ≥ 6, x R}
60. f (x) = x2 + 9 SOLUTION: eSolutions Manual - Powered by Cognero
Page 25
1-7
There appears to be asymptotes at x = 4 and y = 3.
f: D = {x | x 4, x R}, R = {y | y 3, x R}
Inverse Relations and Functions
The graph of f passes the Horizontal Line Test, so f −1 exists.
There appears to be asymptotes at x = 3 and y = 4.
f: D = {x | x 3, x R}, R = {y | y 4, x R}
The domain and range restrictions of f correspond with the range and
−1
domain restrictions of f . Therefore, no further restrictions are needed.
−1
f : D = {x | x 3, x R}, R = {y | y 4, x R}
62. f (x) =
The domain and range restrictions of f correspond with the range and
−1
domain restrictions of f . Therefore, no further restrictions are needed.
−1
f : D = {x | x 4, x R}, R = {y | y 3, x R}
63. ENVIRONMENT Once an endangered species, the bald eagle was
SOLUTION: Graph f .
There appears to be asymptotes at x = 3 and y = 4.
f: D = {x | x 3, x R}, R = {y | y 4, x R}
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eSolutions
downlisted to threatened status in 1995. The table shows the number of
nesting pairs each year.
a. Use the table to approximate a linear function that relates the number
of nesting pairs to the year. Let 0 represent 1984.
Page 26
b. Find the inverse of the function you generated in part a. What does
each variable represent?
c. Using the inverse function, in approximately what year was the
An inverse function is f
−1
(x) =
, where x represents the
−1
1-7 Inverse Relations and Functions
a. Use the table to approximate a linear function that relates the number
of nesting pairs to the year. Let 0 represent 1984.
b. Find the inverse of the function you generated in part a. What does
each variable represent?
c. Using the inverse function, in approximately what year was the
number of nesting pairs 5094?
SOLUTION: a. Use two points from the table to write the related equation of the
function. Because x = 0 corresponds to 1984, two points that can be
used are (0, 1757) and (21, 7066).
First, find the slope of the line.
One of the points chosen is (0, 1757), so the y-intercept is 1757. So, the
related equation is y = 252.81x + 1757. Therefore, a linear function that
can be used to relate the number of nesting pairs to the year is f (x) =
252.81x + 1757.
b.
number of nesting pairs and f (x) represents the number of years
after 1984.
c. Substitute 5094 for x in f ––1(x).
Because x = 0 corresponds to 1984, 13.20 corresponds to 1984 + 13.2
or about 1997.2. Therefore, the number of nesting pairs was 5094 in
1997.
64. FLOWERS Bonny needs to purchase 75 flower stems for banquet
decorations. She can choose between lilies and hydrangea, which cost
$5.00 per stem and $3.50 per stem, respectively.
a. Write a function for the total cost of the flowers.
b. Find the inverse of the cost function. What does each variable
represent?
c. Find the domain of the cost function and its inverse.
d. If the total cost for the flowers was $307.50, how many lilies did
Bonny purchase?
SOLUTION: a. Let x represent the number of stems of hydrangea;
b.
c(x) = 3.5x + 5(75 − x).
An inverse function is f
−1
(x) =
, where x represents the
−1
number of nesting pairs and f (x) represents the number of years
after 1984.
c. Substitute 5094 for x in f ––1(x).
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c
−1
(x) = 250 −
; x represents the total cost and c
−1
(x) represents
the number of stems of hydrangea
c. Bonny is not purchasing more than 75 stems. Therefore, the domain
Page 27
of c(x) is {x | 0 ≤ x ≤ 75, x ∈ R}.
The range of c(x) is from c(0) to c(75).
c
−1
(x) = 250 −
; x represents the total cost and c
−1
(x) represents
Remember that x is the number of hydrangea. The number of lilies is
1-7 Inverse Relations and Functions
the number of stems of hydrangea
c. Bonny is not purchasing more than 75 stems. Therefore, the domain
of c(x) is {x | 0 ≤ x ≤ 75, x ∈ R}.
The range of c(x) is from c(0) to c(75).
75 − 45 or 30.
Find an equation for the inverse of each function, if it exists.
Then graph the equations on the same coordinate plane. Include
any domain restrictions.
65. SOLUTION: The range of the first equation is y ≥ 16, which is equivalent to the domain of f
The domain of c −1(x) is equal to the range of c(x), or {x | 262.5 ≤ x ≤
375, x ∈R}
d.
−1
(x).
The domain of f (x) is for negative values of x, so the inverse of f (x) =
2
−1
x is f (x) = −
.
The range of the second equation is y < 13, which is equivalent to the
−1
domain of f (x).
Remember that x is the number of hydrangea. The number of lilies is
75 − 45 or 30.
Find an equation for the inverse of each function, if it exists.
Then graph the equations on the same coordinate plane. Include
any domain restrictions.
65. SOLUTION: The range of the first equation is y ≥ 16, which is equivalent to the domain of f
−1
(x).
66. eSolutions Manual - Powered by Cognero
The domain of f (x) is for negative values of x, so the inverse of f (x) =
2
x is f
−1
(x) = −
.
Page 28
SOLUTION: The graph of f (x) does not pass the Horizontal Line Test. Therefore, f
1-7 Inverse Relations and Functions
−1
does not exist.
67. FLOW RATE The flow rate of a gas is the volume of gas that passes
66. through an area during a given period of time. The velocity v of air
flowing through a vent can be found using v(r) =
SOLUTION: , where r is the
flow rate in cubic feet per second and A is the cross-sectional area of
the vent in square feet.
The graph of f (x) does not pass the Horizontal Line Test. Therefore, f
−1
does not exist.
a. Find v−1 of the vent shown. What does this function represent?
b. Determine the velocity of air flowing through the vent in feet per
second if the flow rate is 15,000
67. FLOW RATE The flow rate of a gas is the volume of gas that passes
through an area during a given period of time. The velocity v of air
flowing through a vent can be found using v(r) =
.
c. Determine the gas flow rate of a circular vent that has a diameter of
5 feet with a gas stream that is moving at 1.8
.
, where r is the
flow rate in cubic feet per second and A is the cross-sectional area of
the vent in square feet.
SOLUTION: a. The cross section is a rectangle 2 ft by 1.5 ft. So, v(r) =
and v
−1
(x) = 3x. This represents the formula for the flow rate of the gas.
b. The velocity of air flowing through the vent in feet per second
a. Find v−1 of the vent shown. What does this function represent?
b. Determine the velocity of air flowing through the vent in feet per
second if the flow rate is 15,000
.
.
c. The cross sectional area is π(2.5)2 or 6.25π. So, v(r) =
c. Determine the gas flow rate of a circular vent that has a diameter of
−1
and −1
v (x) = 6.25πx. v (1.8) ≈ 35.3.
5 feet with a gas stream that is moving at 1.8
.
68. COMMUNICATION A cellular phone company is having a sale as
shown. Assume the $50 rebate only after the 10% discount is given.
SOLUTION: a. The cross section is a rectangle 2 ft by 1.5 ft. So, v(r) =
eSolutions Manual - Powered by Cognero
and v
(x) = 3x. This represents the formula for the flow rate of the gas.
−1
Page 29
c. The cross sectional area is π(2.5)2 or 6.25π. So, v(r) =
−1
and −1
v (x) = 6.25πx. v (1.8) ≈ 35.3.
1-7 Inverse Relations and Functions
68. COMMUNICATION A cellular phone company is having a sale as
shown. Assume the $50 rebate only after the 10% discount is given.
a. Write a function r for the price of the phone as a function of the
original price if only the rebate applies.
b. Write a function d for the price of the phone as a function of the
original price if only the discount applies.
c. Find a formula for T(x) = [r ◦ d](x) if both the discount and the
rebate apply.
d. Find T −1 and explain what the inverse represents.
e . If the total cost of the phone after the discount and the rebate was
$49, what was the original price of the phone?
The inverse represents the original price of the phone as a function of
the price of the phone after the rebate and the discount.
e.
Use f (x) = 8x – 4 and g(x) = 2x + 6 to find each of the following.
69. [f –−1o g −1](x)
SOLUTION: SOLUTION: a. r(x) = x − 50
b. d(x) = 0.9x
c. The discount comes before the rebate, so T(x) = 0.9x − 50.
d.
The inverse represents the original price of the phone as a function of
the price of the phone after the rebate and the discount.
e.
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Page 30
1-7 Inverse Relations and Functions
Use f (x) = 8x – 4 and g(x) = 2x + 6 to find each of the following.
69. [f
–−1
−1
o g ](x)
70. [g −1o f −1](x)
SOLUTION: SOLUTION: 71. [f o g]−1(x)
70. [g −1o f −1](x)
SOLUTION: eSolutions Manual - Powered by Cognero
SOLUTION: Page 31
1-7 Inverse Relations and Functions
71. [f o g]−1(x)
SOLUTION: 73. (f · g)−1(x)
SOLUTION: Now, find the inverse.
Now, find the inverse.
72. [g o f ]−1(x)
SOLUTION: Now, find the inverse.
must be greater than or equal to 0. Therefore, x ≥ −1.25.
74. (f −1 · g −1)(x)
SOLUTION: 73. (f · g)−1(x)
eSolutions
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SOLUTION: Page 32
1-7 Inverse must
Relations
and
be greater
thanFunctions
or equal to 0. Therefore, x ≥ −1.25.
74. (f −1 · g −1)(x)
Use f (x) = x 2 + 1 with domain [0, ∞) and g(x) =
each of the following.
SOLUTION: to find 75. [f −1o g −1](x)
SOLUTION: 76. [g −1o f −1](x)
Use f (x) = x 2 + 1 with domain [0, ∞) and g(x) =
each of the following.
to find SOLUTION: 75. [f −1o g −1](x)
SOLUTION: eSolutions Manual - Powered by Cognero
Page 33
1-7 Inverse Relations and Functions
76. [g −1o f −1](x)
SOLUTION: 77. [f o g]−1(x)
SOLUTION: Now, find the inverse.
78. [g o f ]−1(x)
SOLUTION: 77. [f o g]−1(x)
SOLUTION: Next, find the inverse.
Now, find the inverse.
79. (f · g −1) (x)
SOLUTION: eSolutions Manual - Powered by Cognero
−1
78. [g o f ] (x)
SOLUTION: Page 34
1-7 Inverse Relations and Functions
79. (f · g −1) (x)
SOLUTION: 80. (f
−1
· g)(x)
SOLUTION: 81. COPIES Karen’s Copies charges users $0.40 for every minute or part
of a minute to use their computer scanner. Suppose you use the
scanner for x minutes, where x is any real number greater than 0.
a. Sketch the graph of the function, C(x), that gives the cost of using
the scanner for x minutes.
b. What are the domain and range of C(x)?
c. Sketch the graph of the inverse of C(x).
d. What are the domain and range of the inverse?
e . What real-world situation is modeled by the inverse?
SOLUTION: eSolutions
Manual - Powered by Cognero
a. The function is C(x) = 0.4x.
the scanner for x minutes.
b. What are the domain and range of C(x)?
c. Sketch the graph of the inverse of C(x).
d. What are the domain and range of the inverse?
e . What real-world situation is modeled by the inverse?
SOLUTION: a. The function is C(x) = 0.4x.
b. The minutes are rounded up, so the domain will consist of whole
numbers. D={x | x R}, R={ y| y positive multiples of 0.4}
c. To graph the inverse, interchange the axes.
d. D={ x| x positive multiples of 0.4} R = {y | y R}
e . The inverse gives the number of possible minutes spent using the
scanner that costs x dollars.
82. MULTIPLE REPRESENTATIONS In this problem, you will
investigate inverses of even and odd functions.
a. GRAPHICAL Sketch the graphs of three different even functions.
Do the graphs pass the horizontal line test?
b. ANALYTICAL What pattern can you discern regarding the
inverses of even functions? Confirm or deny the pattern algebraically.
Page 35
c. GRAPHICAL Sketch the graphs of three different odd functions.
Do the graphs pass the horizontal line test?
d. ANALYTICAL What pattern can you discern regarding the
82. MULTIPLE REPRESENTATIONS In this problem, you will
1-7
investigate inverses of even and odd functions.
a. GRAPHICAL Sketch the graphs of three different even functions.
Inverse
Relations and Functions
Do the graphs pass the horizontal line test?
b. ANALYTICAL What pattern can you discern regarding the
inverses of even functions? Confirm or deny the pattern algebraically.
c. GRAPHICAL Sketch the graphs of three different odd functions.
Do the graphs pass the horizontal line test?
d. ANALYTICAL What pattern can you discern regarding the
inverses of odd functions? Confirm or deny the pattern algebraically.
d. Sample answer: The pattern indicates that all odd functions have
inverses. While the pattern of the three graphs presented indicates that
some odd functions have inverses, this is in fact a false statement. If a
function is odd, then f (−x) = −f (x) for all x. An example of an odd
function is
. Notice from the graph of this function
that, while it is odd, it fails the horizontal line test and, therefore, does
not have an inverse function.
SOLUTION: a. No; sample answer:
83. REASONING If f has an inverse and a zero at 6, what can you
determine about the graph
of f
b. Sample answer: The pattern indicates that no even functions have
inverses. When a function is even, f (x) = f (−x). Two x-values share a
common y-value. This violates the horizontal line test. Therefore, the
statement is true, and no even functions have inverse functions.
c. No; sample answer:
−1
?
SOLUTION: If f has an inverse, then (x, f (x)) = (f (x), x). Therefore, f
intercept at (0, 6).
−1
has a y-
84. Writing in Math Explain what type of restriction on the domain is
needed to determine the inverse of a quadratic function and why a
restriction is needed. Provide an example.
SOLUTION: Sample answer: The domain of a quadratic function needs to be
restricted so that only half of the parabola is shown. The cut-off point
of the restriction will be along the axis of symmetry of the parabola.
This essentially cuts the parabola into two equal halves. The restriction
d. Sample answer: The pattern indicates that all odd functions have
inverses. While the pattern of the three graphs presented indicates that
some odd functions have inverses, this is in fact a false statement. If a
function is odd, then f (−x) = −f (x) for all x. An example of an odd
function is
. Notice from the graph of this function
eSolutions Manual - Powered by Cognero
that, while it is odd, it fails the horizontal line test and, therefore, does
not have an inverse function.
will be
or 2
for f (x) = ax + bx + c.
85. REASONING True or False. Explain your reasoning.
All linear functions have inverse functions.
SOLUTION: Page 36
False; sample answer: Constant functions are linear, but they do not
pass the horizontal line test. Therefore, constant functions are not oneto-one functions and do not have inverse functions.
restricted so that only half of the parabola is shown. The cut-off point
of the restriction will be along the axis of symmetry of the parabola.
This essentially cuts the parabola into two equal halves. The restriction
2
1-7 Inverse
will be Relations
for f (x) = ax
or and Functions
+ bx + c.
85. REASONING True or False. Explain your reasoning.
though both limits approach 0, they do it from opposite sides of 0 and no
x-values ever share a corresponding y-value. Therefore, the function
passes the horizontal line test.
f (x) =
?
SOLUTION: SOLUTION: False; sample answer: Constant functions are linear, but they do not
pass the horizontal line test. Therefore, constant functions are not oneto-one functions and do not have inverse functions.
86. CHALLENGE If f (x) = x3 − ax + 8 and f
−1
(23) = 3, find the value
of a.
Sample answer: If the ± sign is used, then f (x) will no longer be a
function because it violates the vertical line test.
89. Writing in Math Explain how an inverse of f can exist. Give an
example provided that the domain of f is restricted and f does not have
an inverse when the domain is unrestricted.
SOLUTION: SOLUTION: If f
. Even
88. REASONING Why is ± not used when finding the inverse function of
All linear functions have inverse functions.
−1
Yes; sample answer: One function that does this is f (x) =
2
(23) = 3, then (23, 3) is equal to (x, y) in f
Sample answer: If f (x) = x , f does not have an inverse because it is not
one-to-one. If the domain is restricted to x ≥ 0, then the function is now −1
.
one-to-one and f
−1
exists; f
−1
(x) =
.
For each pair of functions, find f o g and g o f . Then state the
domain of each composite function.
90. f (x) = x2 – 9
g(x) = x + 4
Replace (x, y ) with (23, 3).
SOLUTION: 87. REASONING Can f (x) pass the horizontal line test when
and
f (x) = 0
f (x) = 0? Explain.
SOLUTION: Yes; sample answer: One function that does this is f (x) =
. Even
though both limits approach 0, they do it from opposite sides of 0 and no
x-values ever share a corresponding y-value. Therefore, the function
passes the horizontal line test.
There are no restrictions on the domains.
91. f (x) =
x–7
g(x) = x + 6
88. REASONING
Why
is ± not
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Manual - Powered
by Cognero
f (x) =
?
used when finding the inverse function of
SOLUTION: Page 37
1-7 Inverse
Functions
There areRelations
no restrictionsand
on the
domains.
91. f (x) =
x–7
g(x) = x + 6
SOLUTION: There are no restrictions on the domains.
Use the graph of the given parent function to describe the graph
of each related function.
93. f (x) = x2
2
a. g(x) = (0.2x)
b. h(x) = (x – 5)2 – 2
2
c. m(x) = 3x + 6
SOLUTION: a. The graph of g(x) is the graph of f (x) expanded horizontally because
2
g(x) = (0.2x) = f (0.2x) and 0 < 0.2 < 1.
b. This function is of the form h(x) = f (x – 5) – 2. So, the graph of h(x)
is the graph of f (x) translated 5 units to the right and 2 units down.
c. This function is of the form m(x) = 3f (x) + 6. So, the graph of m(x) is
the graph of f (x) expanded vertically because 3 > 1 and translated 6
units up.
94. f (x) = x3
There are no restrictions on the domains.
a. g(x) = |x3 + 3|
3
92. f (x) = x – 4
g(x) = 3x
2
SOLUTION: There are no restrictions on the domains.
Use the graph of the given parent function to describe the graph
of each related function.
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93. f (x) = x2
2
a. g(x) = (0.2x)
b. h(x) = −(2x)
c. m(x) = 0.75(x + 1)3
SOLUTION: a. This function is of the form g(x) = |f (x) + 3|. So, the graph of g(x) is
the graph of f (x) translated 3 units up and the portion of the graph
below the x-axis is reflected in the x-axis.
b. This function is of the form h(x) = –f (2x). So, the graph of h(x) is
the graph of f (x) compressed horizontally because 2 > 1 and reflected
in the x-axis.
c. This function is of the form m(x) = 0.75f (x + 1). So, the graph of m
(x) is the graph of f (x) translated 1 unit to the left and compressed
vertically because 0 < 0.75 < 1.
95. f (x) = |x|
a. g(x) = |2x|
b. h(x) = |x – 5|
c. m(x) = |3x| − 4
SOLUTION: Page 38
1-7
the graph of f (x) compressed horizontally because 2 > 1 and reflected
in the x-axis.
c. This function is of the form m(x) = 0.75f (x + 1). So, the graph of m
(x) is the graph
of f (x) translated
1 unit to the left and compressed
Inverse
Relations
and Functions
vertically because 0 < 0.75 < 1.
Solve each system of equations.
97. x + 2y + 3z = 5
3x + 2y – 2z = −13
5x + 3y – z = −11
SOLUTION: 95. f (x) = |x|
a. g(x) = |2x|
b. h(x) = |x – 5|
c. m(x) = |3x| − 4
Subtract the 2nd equation from the first.
SOLUTION: a. The graph of g(x) is the graph of f (x) compressed horizontally
because g(x) = |2x| = f (2x) and 2 > 1.
b. This function is of the form h(x) = f (x – 5). So, the graph of h(x) is
the graph of f (x) translated 5 units to the right.
c. This function is of the form m(x) = f (3x) – 4. So, the graph of m(x) is
the graph of f (x) compressed horizontally because 3 > 1 and translated
4 units down.
rd
nd
Subtract 2 times the 3 original equation from 3 times the 2
equation.
original
96. ADVERTISING A newspaper surveyed companies on the annual
amount of money spent on television commercials and the estimated
number of people who remember seeing those commercials each week.
A soft-drink manufacturer spends $40.1 million a year and estimates
78.6 million people remember the commercials. For a package-delivery
service, the budget is $22.9 million for 21.9 million people. A
telecommunications company reaches 88.9 million people by spending
$154.9 million. Use a matrix to represent these data.
Solve the new system of 2 equations.
Multiply the bottom equation by 2 and add.
SOLUTION: Substitute to find x.
Solve each system of equations.
97. x + 2y + 3z = 5
3x + 2y – 2z = −13
5x + 3y – z = −11
Substitute into one of the original equations to find y .
SOLUTION: eSolutions Manual - Powered by Cognero
Page 39
Add the 1 and 3 equations together.
1-7 Inverse Relations and Functions
Add 2 times the 3rd original equation to the 2nd original equation.
Substitute into one of the original equations to find y .
Substitute all three values into the other two original equations to
confirm your answer.
Solve the new system of 2 equations. Solve for y in the 1s t, then
substitute into the 2nd.
98. 7x + 5y + z = 0
−x + 3y + 2z = 16
x – 6y – z = −18
SOLUTION: Substitute for x and solve for y .
Add the 1s t and 3rd equations together.
Substitute into one of the original equations to find z.
Add 2 times the 3rd original equation to the 2nd original equation.
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Substitute all three values into the other two original equations to
confirm your answer.
Page 40
st
Solve the new system of 2 equations. Solve for y in the 1 , then
substitute into the 2nd.
1-7 Inverse
and
SubstituteRelations
all three values
intoFunctions
the other two original equations to
confirm your answer.
Solve the new system of 2 equations and 2 variables. Solve for x in
the 1s t, then substitute into the 2nd.
99. x – 3z = 7
2x + y – 2z = 11
−x – 2y + 9z = 13
SOLUTION: Substitute into one of the original equations to find y .
Add the 3rd equation to twice the 2nd equation to eliminate the y .
Substitute all three values into the last original equation to confirm your
answer.
Solve the new system of 2 equations and 2 variables. Solve for x in
the 1s t, then substitute into the 2nd.
100. BASEBALL A batter pops up the ball. Suppose the ball was 3.5 feet
above the ground when he hit it straight up with an initial velocity of 80
2
feet per second. The function d(t) = 80t – 16t + 3.5 gives the ball’s
height above the ground in feet as a function of time t in seconds. How
long did the catcher have to get into position to catch the ball after it
was hit?
SOLUTION: eSolutions Manual - Powered by Cognero
Page 41
The ball will hit the ground after about 5.04 seconds, so the catcher
will have about 5 seconds.
1-7 Inverse Relations and Functions
100. BASEBALL A batter pops up the ball. Suppose the ball was 3.5 feet
above the ground when he hit it straight up with an initial velocity of 80
101. SAT/ACT What is the probability that the spinner will land on a
number that is either even or greater than 5?
2
feet per second. The function d(t) = 80t – 16t + 3.5 gives the ball’s
height above the ground in feet as a function of time t in seconds. How
long did the catcher have to get into position to catch the ball after it
was hit?
SOLUTION: A
B
C
D
E
The ball will hit the ground after about 5.04 seconds, so the catcher
will have about 5 seconds.
101. SAT/ACT What is the probability that the spinner will land on a
SOLUTION: The sections of the spinner are all equal and four of the 6 numbers are
either even or greater than 5. Therefore, the probability is
= .
number that is either even or greater than 5?
102. REVIEW If m and n are both odd natural numbers, which of the
following must be true?
2
2
I. m + n is even.
2
A
B
C
eSolutions
D Manual - Powered by Cognero
E
2
II. m + n is divisible by 4.
2
III. (m + n) is divisible by 4.
F none
G I only
H I and II only
J I and III only
SOLUTION: 2
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If m is odd, then m equals an odd number multiplied by an odd number.
2
The sum of an odd number of odd numbers is always odd, so m will
2
SOLUTION: The sections of the spinner are all equal and four of the 6 numbers are
1-7 Inverse
and
Functions
either evenRelations
or greater than
5. Therefore,
the probability is
= .
102. REVIEW If m and n are both odd natural numbers, which of the
following must be true?
2
2
I. m + n is even.
2
2
2
2
where a is an integer, thus (m + n) = (2a) = 4a , a ∈ Z. Since 4a is
divisible by 4, Part III is true.
Part II is not true when m = 1 and n = 3. Therefore, the correct choice
is J.
103. Which of the following is the inverse of f (x) =
?
A g(x) =
2
II. m + n is divisible by 4.
2
III. (m + n) is divisible by 4.
F none
G I only
H I and II only
J I and III only
B g(x) =
C g(x) = 2x + 5
D g(x) =
SOLUTION: SOLUTION: 2
If m is odd, then m equals an odd number multiplied by an odd number.
2
The sum of an odd number of odd numbers is always odd, so m will
2
2
2
always be odd. The same is true for n . Since m and n are odd, the
sum of these two numbers will always be even. Thus, part I is true.
The sum of two odd numbers is always an even number of the form 2a
2
2
2
2
where a is an integer, thus (m + n) = (2a) = 4a , a ∈ Z. Since 4a is
divisible by 4, Part III is true.
Part II is not true when m = 1 and n = 3. Therefore, the correct choice
is J.
103. Which of the following is the inverse of f (x) =
A g(x) =
B g(x) =
?
104. REVIEW A train travels d miles in t hours and arrives at its destination
3 hours late. At what average speed, in miles per hour, should the train
have gone in order to have arrived on time?
Ft–3
G
H
J
− 3
C g(x) = 2x + 5
D g(x) =
SOLUTION: d = rt
SOLUTION: eSolutions Manual - Powered by Cognero
The train’s current rate is r =
. If the train needs to arrive 3 hours
earlier, then replace t with t − 3. Therefore, r =
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.
1-7 Inverse Relations and Functions
104. REVIEW A train travels d miles in t hours and arrives at its destination
3 hours late. At what average speed, in miles per hour, should the train
have gone in order to have arrived on time?
Ft–3
G
H
J
− 3
SOLUTION: d = rt
The train’s current rate is r =
. If the train needs to arrive 3 hours
earlier, then replace t with t − 3. Therefore, r =
eSolutions Manual - Powered by Cognero
.
Page 44