4-2 Inverse Relations and Functions
For each polygon, find the inverse of the relation. Then, graph both the original relation and its inverse.
1. △MNP with vertices at {(–8, 6), (6, –2), (4, –6)}
SOLUTION:
Graph the ordered pairs and connect the points to form a triangle.
To find the inverse, exchange the coordinates of the ordered pairs.
{(6, –8), (–2, 6), (–6, 4)}
Graph the inverse.
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4-2 Inverse Relations and Functions
2. △XYZ with vertices at {(7, 7), (4, 9), (3, –7)}
SOLUTION:
Graph the ordered pairs and connect the points to form a triangle.
To find the inverse, exchange the coordinates of the ordered pairs.
{(7, 7), (9, 4), (–7, 3)}
Graph the inverse.
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4-2 Inverse Relations and Functions
3. trapezoid QRST with vertices at {(8, –1), (–8, –1), (–2, –8), (2, –8)}
SOLUTION:
Graph the ordered pairs and connect the points to form a trapezoid.
To find the inverse, exchange the coordinates of the ordered pairs.
{(–1, 8), (–1, –8), (–8, –2), (–8, 2)}
Graph the inverse.
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4-2 Inverse Relations and Functions
4. quadrilateral FGHJ with vertices at {(4, 3), (–4, –4), (–3, –5), (5, 2)}
SOLUTION:
Graph the ordered pairs and connect the points to form a quadrilateral.
To find the inverse, exchange the coordinates of the ordered pairs.
{(3, 4), (–4, –4), (–5, –3), (2, 5)}
Graph the inverse.
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4-2 Inverse Relations and Functions
Find the inverse of each function. Then graph the function and its inverse. If necessary, restrict the
domain of the inverse so that it is a function.
5. f(x) = x + 2
SOLUTION:
Rewrite the function as an equation relating x and y.
f(x) = x + 2 → y = x + 2
Exchange x and y.
x=y+2
Solve for y.
Replace y with f –1(x) in the equation.
y = x – 2 → f –1(x) = x – 2
The inverse of f(x) = x + 2 is f –1(x) = x – 2.
Graph f(x) and f –1(x).
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4-2 Inverse Relations and Functions
6. g(x) = 5x
SOLUTION:
Rewrite the function as an equation relating x and y.
g(x) = 5x → y = 5x
Exchange x and y.
x = 5y
Solve for y.
Replace y with g–1(x) in the equation.
The inverse of g(x) = 5x is
.
Graph g(x) and g–1(x).
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4-2 Inverse Relations and Functions
7. f(x) = –2x + 1
SOLUTION:
Rewrite the function as an equation relating x and y.
f(x) = –2x + 1 → y = –2x + 1
Exchange x and y.
x = –2y + 1
Solve for y.
Replace y with f –1(x) in the equation.
The inverse of f(x) = –2x + 1 is
.
Graph f(x) and f –1(x).
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4-2 Inverse Relations and Functions
8.
SOLUTION:
Rewrite the function as an equation relating x and y.
Exchange x and y.
Solve for y.
Replace y with h–1(x) in the equation.
y = 3x + 4 → h–1(x) = 3x + 4
The inverse of
is h–1(x) = 3x + 4.
Graph h(x) and h–1(x).
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4-2 Inverse Relations and Functions
9.
SOLUTION:
Rewrite the function as an equation relating x and y.
Exchange x and y.
Solve for y.
Replace y with f –1(x) in the equation.
The inverse of
is
.
Graph f(x) and f –1(x).
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4-2 Inverse Relations and Functions
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4-2 Inverse Relations and Functions
10. g(x) = x + 4
SOLUTION:
Rewrite the function as an equation relating x and y.
g(x) = x + 4 → y = x + 4
Exchange x and y.
x=y+4
Solve for y.
Replace y with g–1(x) in the equation.
y = x – 4 → g–1(x) = x – 4
The inverse of g(x) = x + 4 is g–1(x) = x – 4.
Graph g(x) and g–1(x).
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4-2 Inverse Relations and Functions
11. f(x) = 4x
SOLUTION:
Rewrite the function as an equation relating x and y.
f(x) = 4x → y = 4x
Exchange x and y.
x = 4y
Solve for y.
Replace y with f –1(x) in the equation.
The inverse of f(x) = 4x is
.
Graph f(x) and f –1(x).
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4-2 Inverse Relations and Functions
12. f(x) = –8x + 9
SOLUTION:
Rewrite the function as an equation relating x and y.
f(x) = –8x + 9 → y = –8x + 9
Exchange x and y.
x = –8y + 9
Solve for y.
Replace y with f –1(x) in the equation.
The inverse of f(x) = –8x + 9 is
.
Graph f(x) and f –1(x).
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4-2 Inverse Relations and Functions
13. f(x) = 5x 2
SOLUTION:
Find the inverse of f(x).
Because f(x) fails the horizontal line test, f –1(x) is not a function. Find the restricted domain of f(x) so that f –1(x)
will be a function. Look for a portion of the graph that is one-to-one. If the domain of f(x) is restricted to (–∞, 0],
then the inverse is
. If the domain of f(x) is restricted to [0, ∞), then the inverse is
.
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4-2 Inverse Relations and Functions
14. h(x) = x 2 + 4
SOLUTION:
Find the inverse of h(x).
Because h(x) fails the horizontal line test, h–1(x) is not a function. Find the restricted domain of h(x) so that h–1(x)
will be a function. Look for a portion of the graph that is one-to-one. If the domain of h(x) is restricted to (–∞, 0],
then the inverse is
. If the domain of h(x) is restricted to [0, ∞), then the inverse is
.
15. WEIGHT The formula to convert weight in pounds to stones is
, where x is the weight in pounds.
a. Find the inverse of p(x), and describe its meaning.
b. Graph p(x) and p–1(x). Use the graph to find the weight in pounds of a dog that weighs about 2.5 stones.
SOLUTION:
a. Find the inverse of p(x), and describe its meaning.
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4-2 Inverse Relations and Functions
The inverse converts stones to pounds, where x is the weight in stones.
b. Graph p(x) and p–1(x).
Look at p–1(x). When x = 2.5, p–1(x) = 35. So, the weight in pounds of a dog that weighs about 2.5 stones is
about 35 pounds.
16. CRYPTOGRAPHY DeAndre is designing a code to send secret messages. He assigns each letter of the
alphabet to a number, where A = 1, B = 2, C = 3, and so on. Then he uses c(x) = 4x – 9 to create the secret
code.
a. Find the inverse of c(x), and describe its meaning.
b. Make tables of c(x) and c–1(x). Use the table to decipher the message: 15, 75, 47, 3, 71, 27, 51, 47, 67.
SOLUTION:
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a. Find the inverse of c(x), and describe its meaning.
The inverse can be used to convert the secret code to the original message.
b. Make tables of c(x) and c–1(x).
When A = 1, c(x) = c(1) = 4(1) – 9 = –5.
When B = 2, c(x) = c(2) = 4(2) – 9 = –1.
When C = 3, c(x) = c(3) = 4(3) – 9 = 3.
When D = 4, c(x) = c(4) = 4(4) – 9 = 7.
When E = 5, c(x) = c(5) = 4(5) – 9 = 11.
When F = 6, c(x) = c(6) = 4(6) – 9 = 15.
When G = 7, c(x) = c(7) = 4(7) – 9 = 19.
When H = 8, c(x) = c(8) = 4(8) – 9 = 23.
When I = 9, c(x) = c(9) = 4(9) – 9 = 27.
When J = 10, c(x) = c(10) = 4(10) – 9 = 31.
When K = 11, c(x) = c(11) = 4(11) – 9 = 35.
When L = 12, c(x) = c(12) = 4(12) – 9 = 39.
When M = 13, c(x) = c(13) = 4(13) – 9 = 43.
When N = 14, c(x) = c(14) = 4(14) – 9 = 47.
When O = 15, c(x) = c(15) = 4(15) – 9 = 51.
When P = 16, c(x) = c(16) = 4(16) – 9 = 55.
When Q = 17, c(x) = c(17) = 4(17) – 9 = 59.
When R = 18, c(x) = c(18) = 4(18) – 9 = 63.
When S = 19, c(x) = c(19) = 4(19) – 9 = 67.
When T = 20, c(x) = c(20) = 4(20) – 9 = 71.
When U = 21, c(x) = c(21) = 4(21) – 9 = 75.
When V = 22, c(x) = c(22) = 4(22) – 9 = 79.
When W = 23, c(x) = c(23) = 4(23) – 9 = 83.
When X = 24, c(x) = c(24) = 4(24) – 9 = 87.
When Y = 25, c(x) = c(25) = 4(25) – 9 = 91.
When Z = 26, c(x) = c(26) = 4(26) – 9 = 95.
Letter
c(x)
Letter
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c(x)
Letter
c(x)
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4-2 Inverse Relations and Functions
A=1
B=2
C=3
D=4
E=5
F=6
G=7
H=8
I=8
–5
–1
3
7
11
15
19
23
27
J = 10
K = 11
L = 12
M = 13
N = 14
O = 15
P = 16
Q = 17
R = 18
31
35
39
43
47
51
55
59
63
S = 19
T = 20
U = 21
V = 22
W = 23
X = 24
Y = 25
Z = 26
67
71
75
79
83
87
91
95
When x = –5,
and A = 1.
When x = –1,
and B = 2.
When x = 3,
and C = 3.
When x = 7,
and D = 4.
When x = 11,
and E = 5.
When x = 15,
and F = 6.
When x = 19,
and G = 7.
When x = 23,
and H = 8.
When x = 27,
and I = 9.
When x = 31,
and J = 10.
When x = 35,
and K = 11.
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4-2 Inverse Relations and Functions
When x = 39,
and L = 12.
When x = 43,
and M = 13.
When x = 47,
and N = 14.
When x = 53,
and O = 15.
When x = 55,
and P = 16.
When x = 59,
and Q = 17.
When x = 63,
and R = 18.
When x = 67,
and S = 19.
When x = 71,
and T = 20.
When x = 75,
and U = 21.
When x = 79,
and V = 22.
When x = 83,
and W = 23.
When x = 87,
and X = 24.
When x = 91,
and Y = 25.
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4-2 Inverse Relations and Functions
When x = 95,
c–1(x)
–5
–1
3
7
11
15
19
23
27
Letter c–1(x)
A=1
31
B=2
35
C=3
39
D=4
43
E=5
47
F=6
51
G=7
55
H=8
59
I=9
63
and Z = 26.
Letter c–1(x)
J = 10
67
K = 11 71
L = 12 75
M = 13 79
N = 15 83
O = 15 87
P = 16 91
Q = 17 95
R = 18
Letter
S = 19
T = 20
U = 21
V = 22
W = 23
X = 24
X = 25
Z = 26
Using the table, c–1(15) = F, c–1(75) = U, c–1(47) = N, c–1(3) = C, c–1(21) = T, c–1(27) = I, c–1(51) = O, c–
1(47) = N, c–1(67) = S, so the message is FUNCTIONS.
Determine whether each pair of functions are inverse functions. Write yes or no.
17. f(x) = x – 1
g(x) = 1 – x
SOLUTION:
Find [f ◦ g](x).
Because [f ◦ g](x) is not the identity function, f(x) and g(x) are not inverses.
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4-2 Inverse Relations and Functions
18. f(x) = 2x + 3
SOLUTION:
Find [f ◦ g](x).
Because [f ◦ g](x) is the identity function, f(x) and g(x) are inverses.
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4-2 Inverse Relations and Functions
19. f(x) = 5x – 5
SOLUTION:
Find [f ◦ g](x).
Find [g ◦ f](x).
Because [f ◦ g](x) and [g ◦ f](x) is the identity function, f(x) and g(x) are inverses.
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4-2 Inverse Relations and Functions
20. f(x) = 2x
SOLUTION:
Find [f ◦ g](x).
Because [f ◦ g](x) is the identity function, f(x) and g(x) are inverses.
21. h(x) = 6x – 2
SOLUTION:
Find [h ◦ g](x).
Because [h ◦ g](x) is not the identity function, h(x) and g(x) are not inverses.
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4-2 Inverse Relations and Functions
22. f(x) = 8x – 10
SOLUTION:
Find [f ◦ g](x).
Because [f ◦ g](x) is the identity function, f(x) and g(x) are inverses.
23. GEOMETRY The formula for the volume of a right circular cone with a height of 2 feet is
whether
. Determine
is the inverse of the original function.
SOLUTION:
Find V ◦ r.
Find r ◦ V.
is the inverse of
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.
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4-2 Inverse Relations and Functions
24. GEOMETRY The formula for the area of a trapezoid is
. Determine whether h = 2A – (a + b) is
the inverse of the original function.
SOLUTION:
Find A ◦ h.
h = 2A – (a + b) is not the inverse of
Find h ◦ A.
.
Find the inverse of each function. Then graph the function and its inverse. If necessary, restrict the
domain of the inverse so that it is a function.
25. y = 4
SOLUTION:
Exchange x and y.
x=4
The inverse of y = 4 is x = 4.
Graph y = 4 and x = 4.
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4-2 Inverse Relations and Functions
26. f(x) = 3x
SOLUTION:
Rewrite the function as an equation relating x and y.
f(x) = 3x → y = 3x
Exchange x and y.
x = 3y
Solve for y.
Replace y with f –1(x) in the equation.
The inverse of f(x) = 3x is
.
Graph f(x) and f –1(x).
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4-2 Inverse Relations and Functions
27. f(x) = x + 2
SOLUTION:
Rewrite the function as an equation relating x and y.
f(x) = x + 2 → y = x + 2
Exchange x and y.
x=y+2
Solve for y.
Replace y with f –1(x) in the equation.
y = x – 2 → f –1(x) = x – 2
The inverse of f(x) = x + 4 is f –1(x) = x – 2.
Graph f(x) and f –1(x).
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4-2 Inverse Relations and Functions
28. g(x) = 2x – 1
SOLUTION:
Rewrite the function as an equation relating x and y.
g(x) = 2x – 1 → y = 2x – 1
Exchange x and y.
x = 2y – 1
Solve for y.
Replace y with g–1(x) in the equation.
The inverse of g(x) = 2x – 1 is
.
Graph g(x) and g–1(x).
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4-2 Inverse Relations and Functions
29.
SOLUTION:
Find the inverse of f(x).
Because f(x) fails the horizontal line test, f –1(x) is not a function. Find the restricted domain of f(x) so that f –1(x)
will be a function. Look for a portion of the graph that is one-to-one. If the domain of f(x) is restricted to (–∞, 0],
then the inverse is
. If the domain of f(x) is restricted to [0, ∞), then the inverse is
.
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30. f(x) = (x + 1)2 + 3
SOLUTION:
Find the inverse of f(x).
Because f(x) fails the horizontal line test, f –1(x) is not a function. Find the restricted domain of f(x) so that f –1(x)
will be a function. Look for a portion of the graph that is one-to-one. If the domain of f(x) is restricted to (–∞, –1],
then the inverse is
. If the domain of f(x) is restricted to [–1, ∞), then the inverse is
.
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4-2 Inverse Relations and Functions
Determine whether each pair of functions are inverse functions. Write yes or no.
31. f(x) = 4x 2
SOLUTION:
Find [f ◦ g](x).
Because [f ◦ g](x) is the identity function, f(x) and g(x) are inverses.
32.
SOLUTION:
Find [f ◦ g](x).
Because [f ◦ g](x) is the identity function, f(x) and g(x) are inverses.
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4-2 Inverse Relations and Functions
33. f(x) = x 2 – 9
g(x) = x + 3
SOLUTION:
Find [f ◦ g](x).
Because [f ◦ g](x) is not the identity function, f(x) and g(x) are not inverses.
34.
SOLUTION:
Find [f ◦ g](x).
Because [f ◦ g](x) is not the identity function, f(x) and g(x) are not inverses.
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4-2 Inverse Relations and Functions
35. f(x) = (x + 6)2
SOLUTION:
Find [f ◦ g](x).
Because [f ◦ g](x) is the identity function, f(x) and g(x) are inverses.
36.
SOLUTION:
Find [f ◦ g](x).
Because [f ◦ g](x) is the identity function, f(x) and g(x) are inverses.
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4-2 Inverse Relations and Functions
Restrict the domain of f(x) so that its inverse is also a function. State the restricted domain of f(x) and the
domain of f–1(x).
37. f(x) = x 2 + 5
SOLUTION:
Find the inverse of f(x).
Graph f(x) and f –1(x).
Because f(x) fails the horizontal line test, f –1(x) is not a function. Find the restricted domain of f(x) so that f –1(x)
will be a function. Look for a portion of the graph that is one-to-one. If the domain of f(x) is restricted to (–∞, 0],
then the inverse is
. If the domain of f(x) is restricted to [0, ∞), then the inverse is
. So, the domain of f(x) is {x | x ≥ 0} or {x | x ≤ 0}. The range of f(x) is {y | y ≥ 5}. The range
of f(x) is the domain of f –1(x). So, the domain of f –1(x) is {x | x ≥ 5}.
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38. f(x) = 3x 2
SOLUTION:
Find the inverse of f(x).
Graph f(x) and f –1(x).
Because f(x) fails the horizontal line test, f –1(x) is not a function. Find the restricted domain of f(x) so that f –1(x)
will be a function. Look for a portion of the graph that is one-to-one. If the domain of f(x) is restricted to (–∞, 0],
then the inverse is
. If the domain of f(x) is restricted to [0, ∞), then the inverse is
. So, the domain of f(x) is {x | x ≥ 0} or {x | x ≤ 0}. The range of f(x) is {y | y ≥ 0}. The range
of f(x) is the domain of f –1(x). So, domain of f –1(x) is {x | x ≥ 0}.
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4-2 Inverse Relations and Functions
39.
SOLUTION:
Find the inverse of f(x).
Graph f(x) and f –1(x).
Because f(x) passes the horizontal line test, f –1(x) is a function. Look at the graph of f(x). The domain of f(x) is
{x | x ≥ –6}. The range of f(x) is {y | y ≥ 0}. The range of f(x) is the domain of f –1(x). So, the domain of f –1(x) is
{x | x ≥ 0}.
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40.
SOLUTION:
Find the inverse of f(x).
Graph f(x) and f –1(x).
Because f(x) passes the horizontal line test, f –1(x) is a function. Look at the graph of f(x). The domain of f(x) is
{x | x ≥ –3}. The range of f(x) is {y | y ≥ 0}. The range of f(x) is the domain of f –1(x). So, the domain of f –1(x) is
{x | x ≥ 0}.
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Sketch a graph of the inverse of each function. Then state whether the inverse is a function.
41. ​
SOLUTION:
Reflect the graph over the line y = x.
The function fails the horizontal line test, so the inverse is not a function.
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42. ​
SOLUTION:
Reflect the graph over the line y = x.
The function passes the horizontal line test, so the inverse is a function.
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43. ​
SOLUTION:
Reflect the graph over the line y = x.
The function passes the horizontal line test, so the inverse is a function.
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44. ​
SOLUTION:
Reflect the graph over the line y = x.
The function fails the horizontal line test, so the inverse is not a function.
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45. FITNESS Alejandro is a personal trainer. For his clients to gain the maximum benefit from their exercise,
Alejandro calculates their maximum target heart rate using the function f(x) = 0.85(220 – x), where x represents
the age of the client. Find the inverse of this function and interpret its meaning in the context of the situation.
SOLUTION:
Find the inverse of f(x), and describe its meaning.
The inverse function gives the age of the clients based on their maximum target heart rate.
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46. Graph the inverse of the piecewise function shown.
SOLUTION:
Reflect the graph over the line y = x.
47. USE A MODEL The diagram shows a sheet of metal with squares of side length x removed from each corner
which can be used to form an open box.
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a. Write and graph the function V(x) which gives the volume of the open box. Explain how the domain of this
function must be restricted within the context of this scenario.
b. Does restricting the domain in part a allow for the inverse to also be a function? Explain your reasoning.
SOLUTION:
a. The volume of a box is V = lwh. The length of the box is 12 – 2x, the width of the box is 16 – 2x, and the
height of the box is x. So, the function V(x) = x(12 – 2x)(16 – 2x) gives the volume of the open box.
The length of the side of the box (12 – 2x) cannot be less than 0.
The value of x must be between 0 and 6 since the length x must be more than 0, but the length of the side of the
box (12 – 2x) cannot be less than 0.
Graph V(x).
b. Even with the restriction, the function does not pass the horizontal line test, so its inverse will not be a function.
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48. STRUCTURE Use the table to find the relationship between (f + g)–1(x) and f –1(x) + g–1(x).
x
f(x)
g(x)
0
0
1
1
3
0
2
1
4
3
4
3
a. Suppose that functions f(x), g(x), and (f + g)(x) all have inverse functions defined on the domain [0, 3].
Calculate the following
values.
i. f –1(3) + g–1(3)
ii. f –1(1) + g–1(1)
b. Use the value of (f + g)(1) to find (f + g)–1(3). Use the value of (f + g)(0) to find (f + g)–1(1).
c. Joyce claims that (f + g)–1(x) = f –1(x) + g–1(x). Determine whether she is correct. Explain your reasoning.
d. Consider the functions f(x) = 2x + 1 and g(x) = 2x – 1. Compare (f + g)–1(x) and f –1(x) + g–1(x).
SOLUTION:
a. i. f –1(3) = 1 and g–1(3) = 3, so f –1(3) + g–1(3) = 1 + 3 = 4.
a. i. f –1(1) = 2 and g–1(1) = 0, so f –1(1) + g–1(1) = 2 + 0 = 2.
b. (f + g)(1) = f(1) + g(1) = 3 + 0 = 3. Since (f + g)(1) = 3, (f + g)–1(3) = 1.
(f + g)(0) = f(0) + g(0) = 0 + 1 = 1. Since (f + g)(0) = 1, (f + g)–1(1) = 0.
c. Part b provides two counterexamples to the property (f + g)–1(x) = f –1(x) + g–1(x) and (f + g)–1(x) = f –1(x)
+ g–1(x).
(f + g)–1(3) = 1, but f –1(3) + g–1(3) = 1 + 3 = 4.
(f + g)–1(1) = 0, but f –1(1) + g–1(1) = 2 + 0 = 2.
d. Find f –1(x).
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4-2 Inverse Relations and Functions
Find g–1(x).
Find f –1(x) + g–1(x).
Find (f + g)(x).
Find (f + g)–1(x).
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4-2 Inverse Relations and Functions
(f + g)–1(x) and f –1(x) + g–1(x) are not the same.
49. ANALYZE If a relation is not a function, then its inverse is sometimes, always, or never a function. Justify your
argument.
SOLUTION:
is an example of a relation that is not a function because it fails the horizontal line test. Find the inverse.
The inverse of
is y = x 2. The inverse y = x 2 is a function.
A circle is an example of a relation that is not a function because it fails the horizontal line test. The inverse of a
circle is also a circle, so the inverse is not a function because it fails the horizontal line test.
Therefore, the statement is sometimes true.
50. CREATE Give an example of a function and its inverse. Verify that the two functions are inverses.
SOLUTION:
Suppose f(x) = 2x, Find the inverse of f(x).
Find f[f –1(x)] and f –1[f(x)].
f[f –1(x)] = f –1[(x)] = x, so the two functions are inverses.
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4-2 Inverse Relations and Functions
51. PERSEVERE Give an example of a function that is its own inverse.
SOLUTION:
Suppose f(x) = x. Find the inverse of f(x).
Suppose f(x) = –x. Find the inverse of f(x).
52. ANALYZE Can the graphs of a linear function with a slope ≠ 0 and its inverse ever be perpendicular? Justify your
answer.
SOLUTION:
A linear function is defined as f(x) = mx + b. Find the inverse of f(x).
Perpendicular lines have slopes that are opposite reciprocals. Because
, a function with slope ≠ 0 is
not perpendicular to its inverse.
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4-2 Inverse Relations and Functions
53. WRITE Suppose you have a composition of two functions that are inverses. When you put in a value of 5 for x,
why is the result always 5?
SOLUTION:
One of the functions carries out an operation on 5. Then the second function that is an inverse of the first
function reverses the operation on 5. So, the result is 5.
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