ACCUPLACER Advanced Algebra & Functions Quiz: Inverse Functions

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Inverse FunctionsQuestion 1 of 20

Which of the following functions has an inverse function over its entire domain, $(-\infty, \infty)$ ?

f(x) = x^2

f(x) = |x|

f(x) = x^3

f(x) = 5

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ACCUPLACER Advanced Algebra & Functions Quiz

ACCUPLACER Advanced Algebra & Functions Quiz: Inverse Functions

Practice Inverse Functions in ACCUPLACER Advanced Algebra & Functions with focused quiz questions that help you check what you know, review explanations, and build confidence with test-style prompts.

What this quiz covers

This quiz focuses on Inverse Functions, giving you a quick way to practice the rules, question types, and explanations that matter most for ACCUPLACER Advanced Algebra & Functions.

How to use this quiz

Try each quiz question before looking at the correct answer. Use the explanations to review missed ideas, then come back to similar questions until the pattern feels familiar.

All questions

Question 1

Which of the following functions has an inverse function over its entire domain, $(-\infty, \infty)$ ?

$f(x) = x^2$
$f(x) = |x|$
$f(x) = x^3$ (correct answer)
$f(x) = 5$

Explanation: A function has an inverse over its domain if and only if it is one-to-one, meaning each output value corresponds to exactly one input value. This can be verified with the horizontal line test. The function

f(x) = x^3

is strictly increasing, so any horizontal line will intersect its graph only once. The other functions are not one-to-one:

f(x)=x^2

(e.g.,

f(2)=f(-2)=4

f(x)=|x|

(e.g.,

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

), and

f(x)=5

(a horizontal line).

Question 2

The graph of an invertible function $f(x)$ contains the point $(3, -5)$ . Which of the following points must be on the graph of $f^{-1}(x)$ ?

$(-5, 3)$ (correct answer)
$(-3, 5)$
$(3, -5)$
$(\frac{1}{3}, -\frac{1}{5})$

Explanation: If a point

(a, b)

is on the graph of

f(x)

, this means

f(a) = b

. By the definition of an inverse,

f^{-1}(b) = a

, which means the point

(b, a)

must be on the graph of

f^{-1}(x)

. Since

(3, -5)

is on the graph of

f(x)

, the point

(-5, 3)

must be on the graph of

f^{-1}(x)

Question 3

The function $C(t) = 20t + 50$ gives the total cost in dollars, $C$ , to rent a tool for $t$ hours, for $t > 0$ . Which of the following best describes the meaning of its inverse function, $C^{-1}(x)$ ?

The number of hours the tool is rented for a total cost of $x$ dollars. (correct answer)
The total cost to rent the tool for $x$ hours.
The initial fee, in dollars, to rent the tool for $x$ hours.
The rate of change of the cost, in dollars per hour.

Explanation: The original function

C(t)

takes hours (

t

) as input and gives total cost (

C

) as output. The inverse function swaps the roles of the input and output. Therefore,

C^{-1}(x)

must take a total cost (represented by

x

) as input and give the corresponding number of hours as output.

Question 4

The function $h(x) = \frac{2x + 5}{x - 1}$ has domain $x \neq 1$ . What is the domain of $h^{-1}(x)$ ?

$x \neq 1$
$x \neq 2$ (correct answer)
$x \neq -5$
All real numbers

Explanation: The domain of

h^{-1}

equals the range of

h

. To find the range of

h(x) = \frac{2x + 5}{x - 1}

, we solve

y = \frac{2x + 5}{x - 1}

for

x

y(x - 1) = 2x + 5

, so

yx - y = 2x + 5

, giving

x(y - 2) = y + 5

, thus

x = \frac{y + 5}{y - 2}

. This is undefined when

y = 2

, so the range of

h

is all real numbers except

y = 2

. Therefore, the domain of

h^{-1}

x \neq 2

. Choice A incorrectly uses the domain restriction of

h

. Choices C and D don't reflect the correct range analysis.

Question 5

Let $f(x)$ be an invertible function. What is the value of $f(f^{-1}(10))$ ?

$-10$
$\frac{1}{10}$
$10$ (correct answer)
The value cannot be determined without knowing $f(x)$ .

Explanation: By the definition of inverse functions, the composition of a function with its inverse yields the input value. That is,

f(f^{-1}(x)) = x

for all

x

in the domain of

f^{-1}

. In this case, the input to the composite function is 10, so

f(f^{-1}(10)) = 10

. The specific formula for

f(x)

is not needed.

Question 6

Let $h(x) = x^3 + x$ . Since $h(x)$ is one-to-one, it has an inverse $h^{-1}(x)$ . What is the value of $h^{-1}(h(2))$ ?

$2$ (correct answer)
$8$
$10$
The value cannot be determined without knowing $h^{-1}(x)$ .

Explanation: By the definition of inverse functions,

h^{-1}(h(x)) = x

for all

x

in the domain of

h

. Here, the input is 2. Therefore,

h^{-1}(h(2)) = 2

. Calculating

h(2) = 2^3 + 2 = 10

is an intermediate step that is immediately undone by the inverse function.

Question 7

Let $h$ be a one-to-one function such that $h(4) = -7$ . What is the value of $h^{-1}(-7)$ ?

$-7$
$-\frac{1}{7}$
$4$ (correct answer)
$\frac{1}{4}$

Explanation: The definition of an inverse function states that if a point

(a, b)

is on the graph of

h(x)

, meaning

h(a) = b

, then the point

(b, a)

is on the graph of

h^{-1}(x)

, meaning

h^{-1}(b) = a

. Given

h(4) = -7

, we can identify

a=4

and

b=-7

. Therefore,

h^{-1}(-7) = 4

Question 8

The function $f(x)$ has an inverse, $f^{-1}(x)$ . If $f^{-1}(5) = 2$ , what is the value of $f(2)$ ?

$\frac{1}{5}$
$\frac{1}{2}$
$2$
$5$ (correct answer)

Explanation: The relationship between a function and its inverse is that if

f^{-1}(b) = a

, then

f(a) = b

. In this case, we are given

f^{-1}(5) = 2

, where

b=5

and

a=2

. Therefore, it must be true that

f(2) = 5

Question 9

A one-to-one function $g(x)$ is defined by the set of points $\{(-2, 3), (0, 1), (1, 5), (3, -4)\}$ . What is the value of $g^{-1}(3)$ ?

$-2$ (correct answer)
$\frac{1}{3}$
$1$
$5$

Explanation: To find

g^{-1}(3)

, we are looking for the input to the function

g(x)

that gives an output of 3. We examine the given points. The point

(-2, 3)

indicates that

g(-2) = 3

. By the definition of an inverse function, this means

g^{-1}(3) = -2

Question 10

A one-to-one function $g(x)$ has a domain of $(-\infty, 3]$ and a range of $[-4, \infty)$ . What is the range of $g^{-1}(x)$ ?

$[-4, \infty)$
$(-\infty, -4]$
$[3, \infty)$
$(-\infty, 3]$ (correct answer)

Explanation: The range of an inverse function

g^{-1}(x)

is the domain of the original function

g(x)

. The domain of

g(x)

is given as

(-\infty, 3]

. Therefore, the range of

g^{-1}(x)

is also

(-\infty, 3]

Question 11

If $g(x)$ is a one-to-one function, which statement describes the relationship between the graph of $y = g(x)$ and the graph of $y = g^{-1}(x)$ ?

They are reflections of each other across the $y$ -axis.
They are reflections of each other across the $x$ -axis.
They are reflections of each other across the line $y = x$ . (correct answer)
They are reflections of each other across the line $y = -x$ .

Explanation: A key property of inverse functions is that their graphs are reflections of each other across the line

y=x

. This is because if a point

(a, b)

is on the graph of

g(x)

, the point

(b, a)

is on the graph of

g^{-1}(x)

, and these two points are symmetric with respect to the line

y=x

Question 12

Let $f(x) = x^3 - 7$ . What is the value of $f^{-1}(20)$ ?

$3$ (correct answer)
$27$
(7,993)
The value cannot be determined.

Explanation: To find

f^{-1}(20)

, we need to find the input value

x

such that the output of

f(x)

is 20. We set up the equation

f(x) = 20

, which is

x^3 - 7 = 20

. Adding 7 to both sides gives

x^3 = 27

. Taking the cube root of both sides yields

x = 3

Question 13

Two functions, $f(x)$ and $g(x)$ , are inverses of each other. If $f(x) = \frac{x-1}{3}$ , which of the following is $g(x)$ ?

$g(x) = 3x + 1$ (correct answer)
$g(x) = 3(x+1)$
$g(x) = 3x - 1$
$g(x) = \frac{3}{x-1}$

Explanation: To find the inverse of

f(x)

, we set

y = f(x)

, swap

x

and

y

, and solve for the new

y

. Start with

y = \frac{x-1}{3}

. Swap variables:

x = \frac{y-1}{3}

. Now, solve for

y

. Multiply by 3:

3x = y - 1

. Add 1:

y = 3x + 1

. So, the inverse function is

g(x) = 3x + 1

Question 14

Let $f(x)$ be an invertible function. If $f(5) = 1$ and $f(2) = 5$ , what is the value of $f^{-1}(f^{-1}(1))$ ?

$1$
$2$ (correct answer)
$5$
The value cannot be determined.

Explanation: This is a two-step problem that must be solved from the inside out. First, we evaluate the inner part,

f^{-1}(1)

. Since we are given

f(5) = 1

, the definition of an inverse tells us that

f^{-1}(1) = 5

. Now we substitute this result back into the expression: we need to find

f^{-1}(5)

. We are given that

f(2) = 5

, so

f^{-1}(5) = 2

. Therefore,

f^{-1}(f^{-1}(1)) = 2

Question 15

A function $f(x)$ is defined as $f(x) = \sqrt{x-2}$ for $x \ge 2$ . What is the range of the inverse function, $f^{-1}(x)$ ?

$[0, \infty)$
$(-\infty, 0]$
$[2, \infty)$ (correct answer)
$(-\infty, 2]$

Explanation: The range of an inverse function

f^{-1}(x)

is identical to the domain of the original function

f(x)

. The domain of

f(x) = \sqrt{x-2}

is restricted to values of

x

for which the expression under the square root is non-negative, which is

x-2 \ge 0

, or

x \ge 2

. This is given as the domain. Therefore, the range of

f^{-1}(x)

[2, \infty)

Question 16

If $f(x) = 4x - 3$ , for what value of $c$ is $f^{-1}(c) = 2$ ?

$2$
$5$ (correct answer)
$8$
$11$

Explanation: The statement

f^{-1}(c) = 2

is equivalent to the statement

f(2) = c

by the definition of an inverse function. We can find the value of

c

by evaluating

f(2)

. Substitute

x=2

into the function:

f(2) = 4(2) - 3 = 8 - 3 = 5

. Thus,

c = 5

Question 17

A one-to-one function $f(x)$ has a domain of $\{x | x \ge 5\}$ and a range of $\{y | y \le -2\}$ . What is the domain of its inverse, $f^{-1}(x)$ ?

$\{x | x \ge 5\}$
$\{x | x \le 5\}$
$\{x | x \ge -2\}$
$\{x | x \le -2\}$ (correct answer)

Explanation: The domain of an inverse function

f^{-1}(x)

is the range of the original function

f(x)

. The range of

f(x)

is given as

\{y | y \le -2\}

. Therefore, the domain of

f^{-1}(x)

\{x | x \le -2\}

Question 18

The function $f(x) = 2x + 5$ has an inverse $f^{-1}(x)$ . What is the value of $3 \cdot f^{-1}(11)$ ?

$3$
$9$ (correct answer)
$27$
$33$

Explanation: First, we must find the value of

f^{-1}(11)

. This is the value of

x

such that

f(x) = 11

. We set up the equation

2x + 5 = 11

. Subtracting 5 from both sides gives

2x = 6

, and dividing by 2 gives

x = 3

. So,

f^{-1}(11) = 3

. The question asks for

3 \cdot f^{-1}(11)

, which is

3 \cdot 3 = 9

Question 19

The graphs of a function $f(x)$ and its inverse $f^{-1}(x)$ intersect at a point $P$ . Which of the following must be true about the coordinates of $P$ ?

The coordinates of $P$ must be $(0, 0)$ .
The point $P$ must lie on the line $y = x$ . (correct answer)
The point $P$ must lie on the $x$ -axis.
The point $P$ must lie on the $y$ -axis.

Explanation: The graphs of

y=f(x)

and

y=f^{-1}(x)

are symmetric with respect to the line

y=x

. If a point

(x,y)

is an intersection point, it lies on both graphs. Because of the symmetry, the point must lie on the line of symmetry,

y=x

Question 20

The function $f(x) = (x-2)^2$ is not one-to-one. However, it has an inverse function if its domain is restricted. On which of the following domains does $f(x)$ have an inverse?

$(-\infty, \infty)$
$[0, 4]$
$[2, \infty)$ (correct answer)
$[1, 3]$

Explanation: A function has an inverse on a domain where it is one-to-one. The graph of

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

is a parabola with its vertex at

x=2

. The function is strictly decreasing for

x<2

and strictly increasing for

x>2

. To be one-to-one, the domain must be restricted to one side of the vertex. The domain

[2, \infty)

includes the vertex and all points to its right, where the function is strictly increasing, so it is one-to-one on this domain. On the interval

[1, 3]

f(1)=1

and

f(3)=1

, so it is not one-to-one.