Math 3 Quiz: Finding And Verifying Inverse Functions

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Finding And Verifying Inverse FunctionsQuestion 1 of 16

The function $f(x) = 3x - 7$ has domain $x \geq 2$ . What is the range of the inverse function $f^{-1}(x)$ ?

x \geq 7

x \geq -1

x \geq \frac{7}{3}

x \geq 2

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Math 3 Quiz

Math 3 Quiz: Finding And Verifying Inverse Functions

Practice Finding And Verifying Inverse Functions in Math 3 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 Finding And Verifying Inverse Functions, giving you a quick way to practice the rules, question types, and explanations that matter most for Math 3.

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.

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Question 1

The function $f(x) = 3x - 7$ has domain $x \geq 2$ . What is the range of the inverse function $f^{-1}(x)$ ?

$x \geq 7$
$x \geq -1$
$x \geq \frac{7}{3}$
$x \geq 2$ (correct answer)

Explanation: When you encounter questions about inverse functions and their domains and ranges, remember this key relationship: the domain of a function becomes the range of its inverse, and the range of a function becomes the domain of its inverse. Let's work through this systematically. First, we need to find the range of the original function

f(x) = 3x - 7

with domain

x \geq 2

. Since this is a linear function with a positive slope, it's strictly increasing. When

x = 2

(the minimum input), we get

f(2) = 3(2) - 7 = -1

. As

x

increases beyond 2,

f(x)

increases without bound. Therefore, the range of

f(x)

y \geq -1

. Now, since the range of

f(x)

y \geq -1

, this becomes the domain of

f^{-1}(x)

. The domain of

f(x)

x \geq 2

, so this becomes the range of

f^{-1}(x)

. The answer is D. Let's examine why the other options are incorrect. Choice A (

x \geq 7

) likely comes from incorrectly plugging in the domain value:

f(2) = 3(2) - 7 = -1

, but someone might have calculated

3 + 7 - 3 = 7

. Choice B (

x \geq -1

) confuses the range of the original function with the range of the inverse—this is actually the domain of

f^{-1}(x)

. Choice C (

x \geq \frac{7}{3}

) might result from solving

3x - 7 = 0

incorrectly. Remember: domain and range swap places when you move from a function to its inverse. Always identify both for the original function first.

Question 2

If $p(x) = \frac{x - 2}{3x + 1}$ and $q(x) = \frac{x + 2}{1 - 3x}$ , which statement about these functions is correct?

$q(x) \neq p^{-1}(x)$ because their domains do not align properly
$q(x) = p^{-1}(x)$ but their domains must be restricted for the inverse relationship
$q(x) \neq p^{-1}(x)$ because $p(q(1)) \neq 1$
$q(x) = p^{-1}(x)$ and they are inverse functions of each other (correct answer)

Explanation: When you encounter questions about inverse functions, you need to verify two key conditions: first, that composing the functions in either order yields the identity function, and second, that their domains and ranges are properly aligned. To check if

q(x) = p^{-1}(x)

, let's verify that

p(q(x)) = x

. Starting with

q(x) = \frac{x + 2}{1 - 3x}

, we substitute this into

p(x)

p(q(x)) = p\left(\frac{x + 2}{1 - 3x}\right) = \frac{\frac{x + 2}{1 - 3x} - 2}{3 \cdot \frac{x + 2}{1 - 3x} + 1}

Simplifying the numerator:

\frac{x + 2}{1 - 3x} - 2 = \frac{x + 2 - 2(1 - 3x)}{1 - 3x} = \frac{x + 2 - 2 + 6x}{1 - 3x} = \frac{7x}{1 - 3x}

Simplifying the denominator:

3 \cdot \frac{x + 2}{1 - 3x} + 1 = \frac{3(x + 2) + (1 - 3x)}{1 - 3x} = \frac{3x + 6 + 1 - 3x}{1 - 3x} = \frac{7}{1 - 3x}

Therefore:

p(q(x)) = \frac{\frac{7x}{1 - 3x}}{\frac{7}{1 - 3x}} = x

Similarly, you can verify that

q(p(x)) = x

. Since both compositions yield the identity function,

q(x) = p^{-1}(x)

. Answer choice A incorrectly suggests domain misalignment, but the domains work properly for the inverse relationship. Choice B wrongly claims domain restrictions are needed beyond what's already present. Choice C uses faulty reasoning with a specific value test that doesn't properly verify the inverse relationship. Remember: to prove two functions are inverses, show that both

f(g(x)) = x

and

g(f(x)) = x

through algebraic manipulation, not just specific value checks.

Question 3

Let $f(x) = x^2 + 6x + 5$ for $x \geq -3$ . To find $f^{-1}(x)$ , a student completes the square and gets $f(x) = (x + 3)^2 - 4$ . What is $f^{-1}(12)$ ?

$4$
$-7$
$1$ (correct answer)
$-1$

Explanation: When finding inverse functions, you're essentially solving for the input that produces a given output. The completed square form

f(x) = (x + 3)^2 - 4

makes this process much cleaner than working with the original quadratic. To find

f^{-1}(12)

, you need to solve

f(x) = 12

. Using the completed square form:

(x + 3)^2 - 4 = 12

(x + 3)^2 = 16

x + 3 = \pm 4

x = -3 \pm 4

This gives you

x = 1

x = -7

. However, the domain restriction

x \geq -3

is crucial here. Since

-7 < -3

, this solution is outside the allowed domain. Therefore,

x = 1

is the only valid solution, making

f^{-1}(12) = 1

. Answer C)

1

is correct. Answer A)

4

likely comes from solving

(x + 3)^2 = 16

and mistakenly thinking

x + 3 = 4

gives

x = 4

instead of

x = 1

. Answer B)

-7

represents the other solution to the equation that gets eliminated by the domain restriction. Answer D)

-1

might result from arithmetic errors in the solving process or confusion about the domain boundary. Study tip: Always check domain restrictions when finding inverse functions. Quadratic functions are only one-to-one (and thus invertible) when their domain is restricted, so you'll often need to eliminate one of the two solutions you get from solving the quadratic equation.

Question 4

Consider the function $h(x) = \sqrt{x + 4} - 2$ where $x \geq -4$ . If $h^{-1}(x)$ represents the inverse function, which statement correctly verifies that $h^{-1}(x) = x^2 + 4x$ for $x \geq 0$ ?

$h(h^{-1}(3)) = h(21) = \sqrt{25} - 2 = 3$ and $h^{-1}(h(5)) = h^{-1}(1) = 5$ (correct answer)
$h(h^{-1}(3)) = h(21) = \sqrt{21} - 2 = 3$ and $h^{-1}(h(5)) = h^{-1}(1) = 5$
$h(h^{-1}(3)) = h(9) = \sqrt{13} - 2 = 3$ and $h^{-1}(h(5)) = h^{-1}(1) = 5$
$h(h^{-1}(3)) = h(21) = \sqrt{25} - 2 = 3$ and $h^{-1}(h(0)) = h^{-1}(-2) = 0$

Explanation: To verify inverse functions, we need both

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

and

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

. First,

h^{-1}(3) = 3^2 + 4(3) = 9 + 12 = 21

, so

h(h^{-1}(3)) = h(21) = \sqrt{21 + 4} - 2 = \sqrt{25} - 2 = 5 - 2 = 3

✓. Next,

h(5) = \sqrt{5 + 4} - 2 = \sqrt{9} - 2 = 3 - 2 = 1

, so

h^{-1}(h(5)) = h^{-1}(1) = 1^2 + 4(1) = 5

✓. Choice B incorrectly computes

\sqrt{21 + 4}

\sqrt{21}

. Choice C uses

h^{-1}(3) = 9

instead of 21. Choice D uses an invalid test point since

h^{-1}(-2)

is undefined for

x \geq 0

Question 5

Consider the piecewise function $f(x) = \begin{cases} 2x + 1 & \text{if } x < 0 \\ x^2 + 1 & \text{if } x \geq 0 \end{cases}$ . For what values of $x$ does $f$ have an inverse function when restricted to the domain $x \geq -2$ ?

The function has an inverse for $x \geq -2$ without any additional restrictions
The function has an inverse only when further restricted to $x \geq 0$ (correct answer)
The function has an inverse only when further restricted to $x \geq 1$
The function cannot have an inverse for any restriction of the given domain

Explanation: For a function to have an inverse, it must be one-to-one (injective). On

[-2, 0)

f(x) = 2x + 1

gives values from

f(-2) = -3

f(0^-) = 1

. On

[0, \infty)

f(x) = x^2 + 1

gives values from

f(0) = 1

\infty

. The issue is that

f(x) = x^2 + 1

for

x \geq 0

is not one-to-one since

f(a) = f(-a)

for any

a > 0

. However, when we restrict to

x \geq 0

, the quadratic piece

x^2 + 1

is one-to-one (increasing). Also, we need to check if there's overlap in range values. The linear piece on

[-2, 0)

has range

[-3, 1)

and the quadratic piece on

[0, \infty)

has range

[1, \infty)

. At

x = 0

, both pieces would give

f(0) = 1

, but since

x = 0

belongs to the quadratic piece, there's no issue. The function is one-to-one on

[0, \infty)

. Choice A ignores that the quadratic piece needs restriction. Choice C over-restricts. Choice D incorrectly concludes no inverse exists.

Question 6

Let $f(x) = \frac{2x + 3}{x - 1}$ where $x \neq 1$ . If $g(x)$ is the inverse function of $f(x)$ , what is $g(5)$ ?

$\frac{8}{3}$ (correct answer)
$\frac{13}{3}$
$\frac{5}{2}$
$2$

Explanation: To find g(5), we need to solve f(x) = 5. Setting

\frac{2x + 3}{x - 1} = 5

, we get

2x + 3 = 5(x - 1) = 5x - 5

. This gives us

2x + 3 = 5x - 5

, so

8 = 3x

, and

x = \frac{8}{3}

. Therefore

g(5) = \frac{8}{3}

. Choice B results from incorrectly solving

2x + 3 = 5x + 5

. Choice C comes from setting up

\frac{2x + 3}{x - 1} = 5

but solving

2x + 3 = 5x - 1

. Choice D results from the error

2x + 3 = 5(x - 1)

becoming

2x + 3 = 5x - 3

Question 7

Let $f(x) = x^2 - 4x + 5$ for $x \geq 2$ . If $f^{-1}$ exists on this restricted domain, what is $f^{-1}(1) + f^{-1}(5)$ ?

$4$
$6$ (correct answer)
$8$
$2$

Explanation: First, let's complete the square:

f(x) = (x - 2)^2 + 1

. For

x \geq 2

, this function is one-to-one with range

y \geq 1

. To find

f^{-1}(1)

, solve

f(x) = 1

(x - 2)^2 + 1 = 1

, so

(x - 2)^2 = 0

, giving

x = 2

. To find

f^{-1}(5)

, solve

f(x) = 5

(x - 2)^2 + 1 = 5

, so

(x - 2)^2 = 4

, giving

x - 2 = \pm 2

. Since

x \geq 2

, we take

x - 2 = 2

, so

x = 4

. Therefore

f^{-1}(1) + f^{-1}(5) = 2 + 4 = 6

. Choice A uses only the positive solution from the first equation. Choice C would result from taking both solutions

x = 0

and

x = 4

for the second equation, ignoring the domain restriction. Choice D represents just

f^{-1}(1)

Question 8

The function $f(x) = \frac{ax + b}{cx + d}$ has inverse $f^{-1}(x) = \frac{dx - b}{-cx + a}$ . If $f(2) = 3$ and $f^{-1}(3) = 2$ , which relationship must hold?

$ad - bc = 1$ and the given conditions are automatically satisfied by the inverse relationship
$ad - bc \neq 0$ and the given conditions provide independent constraints on the coefficients
$ad - bc = -1$ and $f(2) = 3$ implies $f^{-1}(3) = 2$ by the definition of inverse functions (correct answer)
$ad + bc = 0$ and both conditions must be verified separately to ensure the inverse formula

Explanation: For a rational function to have an inverse, the determinant

ad - bc \neq 0

. The given inverse formula is correct when

ad - bc = -1

(or we could multiply by -1 to get the standard form). Since f and

f^{-1}

are inverses, if

f(2) = 3

, then automatically

f^{-1}(3) = 2

by definition - these aren't independent conditions. Choice A has the wrong determinant value. Choice B incorrectly suggests the conditions are independent. Choice D gives an impossible determinant condition (this would make the function non-invertible).

Question 9

Functions $u(x) = \frac{2x - 1}{3}$ and $v(x) = \frac{3x + 1}{2}$ are tested to see if they are inverses. What is the value of $u(v(x)) + v(u(x))$ ?

$\frac{4x + 1}{3}$
$x$
$2x$ (correct answer)
$\frac{6x + 2}{6}$

Explanation: When you encounter questions about function composition, you're testing whether two functions "undo" each other. If functions are true inverses, then

u(v(x)) = x

and

v(u(x)) = x

, so their sum would equal

2x

. Let's calculate each composition step by step. For

u(v(x))

, substitute

v(x) = \frac{3x + 1}{2}

into

u(x) = \frac{2x - 1}{3}

u(v(x)) = u\left(\frac{3x + 1}{2}\right) = \frac{2 \cdot \frac{3x + 1}{2} - 1}{3} = \frac{3x + 1 - 1}{3} = \frac{3x}{3} = x

For

v(u(x))

, substitute

u(x) = \frac{2x - 1}{3}

into

v(x) = \frac{3x + 1}{2}

v(u(x)) = v\left(\frac{2x - 1}{3}\right) = \frac{3 \cdot \frac{2x - 1}{3} + 1}{2} = \frac{2x - 1 + 1}{2} = \frac{2x}{2} = x

Therefore,

u(v(x)) + v(u(x)) = x + x = 2x

, confirming these functions are indeed inverses. Choice A,

\frac{4x + 1}{3}

, results from incorrectly combining terms during composition. Choice B,

x

, comes from mistakenly thinking the sum equals just one composition instead of both. Choice D,

\frac{6x + 2}{6}

, appears to stem from algebraic errors in the composition process, though it does simplify to

x + \frac{1}{3}

. Remember: when checking if functions are inverses, compute both compositions carefully. If each equals

x

, they're inverses, and their sum will always be

2x

Question 10

Given that $g(x) = \frac{ax + b}{cx + d}$ where $ad - bc \neq 0$ , and $g(1) = 2$ , $g(2) = 3$ , if $g^{-1}(2) = 1$ , what additional relationship must hold?

$g^{-1}(3) = 2$ by the symmetry property of inverse functions applied to the given conditions
$g^{-1}(1) = 0$ because the inverse relationship requires this complementary mapping for consistency
$g(0) = \frac{b}{d}$ and this value determines the remaining inverse relationship through the formula
No additional relationship is required since $g^{-1}(2) = 1$ is automatically satisfied when $g(1) = 2$ (correct answer)

Explanation: Since g and

g^{-1}

are inverse functions, if

g(1) = 2

, then automatically

g^{-1}(2) = 1

by the definition of inverse functions. This is not an additional constraint but rather a consequence of the given information. The condition

g^{-1}(2) = 1

doesn't provide new information beyond

g(1) = 2

. Choice A incorrectly suggests we can deduce

g^{-1}(3) = 2

, but while

g(2) = 3

does imply

g^{-1}(3) = 2

, the question asks what MUST hold given

g^{-1}(2) = 1

. Choice B proposes an unrelated constraint. Choice C gives a true but irrelevant formula for

g(0)

Question 11

Function $f(x) = \frac{x}{x - 3}$ where $x \neq 3$ has an inverse. If we compute $(f^{-1} \circ f)(5)$ , which step correctly identifies a potential error?

The composition should equal 5, but we must verify $f(5)$ is in the domain of $f^{-1}$
The composition should equal 5, but we must verify 5 is in the domain of $f^{-1} \circ f$ (correct answer)
The composition equals $\frac{5}{2}$ , and we should check this matches $f^{-1}(f(5))$
The composition should equal $f^{-1}(5)$ , then we apply $f$ to get the final result

Explanation: For inverse functions,

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

for all

x

in the domain of

f

. Since

5 \neq 3

, it's in the domain of

f

, so

(f^{-1} \circ f)(5) = 5

. However, we should verify that 5 is indeed in the domain where this composition is defined. The domain of

f^{-1} \circ f

is the set of

x

values where

f(x)

is defined AND

f(x)

is in the domain of

f^{-1}

. Choice A focuses on the wrong condition. Choice C gives an incorrect value and reasoning. Choice D misunderstands how function composition works.

Question 12

Consider $f(x) = \frac{1}{x + 2}$ for $x \neq -2$ . Which expression represents $f^{-1}(x)$ and what is its domain?

$f^{-1}(x) = \frac{1}{x} - 2$ with domain $x \neq 0$ (correct answer)
$f^{-1}(x) = \frac{1 - 2x}{x}$ with domain $x \neq 0$
$f^{-1}(x) = \frac{1}{x} - 2$ with domain $x > 0$
$f^{-1}(x) = \frac{2x - 1}{x}$ with domain $x \neq 0$

Explanation: To find the inverse, let

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

. Solving for x:

y(x + 2) = 1

, so

yx + 2y = 1

, giving

yx = 1 - 2y

, and

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

. Therefore

f^{-1}(x) = \frac{1}{x} - 2

. The domain excludes

x = 0

since we can't divide by zero. We can verify:

f(f^{-1}(x)) = f\left(\frac{1}{x} - 2\right) = \frac{1}{\frac{1}{x} - 2 + 2} = \frac{1}{\frac{1}{x}} = x

. Choice B shows the intermediate step before simplification. Choice C incorrectly restricts to positive values only. Choice D has algebraic errors in the numerator.

Question 13

If $f(x) = \sqrt{x + 3} - 1$ where $x \geq -3$ , then $f^{-1}(2)$ equals which value?

$12$
$9$
$\sqrt{6}$
$6$ (correct answer)

Explanation: When you encounter inverse function problems, remember that

f^{-1}(2)

asks "what input value gives an output of 2?" Rather than finding the inverse function formula, you can solve this directly by setting

f(x) = 2

. Starting with

f(x) = \sqrt{x + 3} - 1 = 2

, add 1 to both sides:

\sqrt{x + 3} = 3

. Square both sides to eliminate the square root:

x + 3 = 9

. Solving for

x

x = 6

. Let's verify:

f(6) = \sqrt{6 + 3} - 1 = \sqrt{9} - 1 = 3 - 1 = 2

✓ Now examining the wrong answers: Choice A (

12

) likely comes from incorrectly setting up

\sqrt{x + 3} = 2

instead of

\sqrt{x + 3} = 3

, leading to

x + 3 = 4

, so

x = 1

. But this gives

f(1) = 1

, not 2. Some students might then add more steps incorrectly. Choice B (

9

) represents stopping at

x + 3 = 9

without subtracting 3. Choice C (

\sqrt{6}

) might result from confusing the inverse process or not fully simplifying

\sqrt{9}

. Study tip: For inverse function evaluation problems, don't waste time finding the complete inverse function. Instead, set

f(x)

equal to your target output and solve for

x

. Always verify your answer by plugging it back into the original function.

Question 14

Let $p(x) = 2x + 1$ and $q(x) = x^2$ for $x \geq 0$ . If $r(x) = p(q(x))$ , what is $r^{-1}(19)$ ?

$3$ (correct answer)
$9$
$\sqrt{9}$
$\pm 3$

Explanation: First find

r(x) = p(q(x)) = p(x^2) = 2x^2 + 1

. To find

r^{-1}(19)

, solve

r(x) = 19

2x^2 + 1 = 19

, so

2x^2 = 18

, giving

x^2 = 9

, and

x = \pm 3

. However, since

q(x) = x^2

is restricted to

x \geq 0

, the composition

r(x)

inherits this domain restriction, so

r^{-1}(19) = 3

. Choice B gives the value of

x^2

rather than

x

. Choice C is equivalent to choice A but unnecessarily complex. Choice D ignores the domain restriction that makes

r(x)

one-to-one.

Question 15

If $f(x) = \frac{2x + 3}{x - 1}$ where $x \neq 1$ , and $g(x)$ is the inverse function of $f(x)$ , what is $g(5)$ ?

$\frac{8}{3}$ (correct answer)
$\frac{13}{4}$
$\frac{7}{3}$
$2$

Explanation: To find g(5), we need to find the value of x such that f(x) = 5. Setting up the equation:

\frac{2x + 3}{x - 1} = 5

. Cross-multiplying:

2x + 3 = 5(x - 1) = 5x - 5

. Solving:

2x + 3 = 5x - 5

, so

8 = 3x

, giving

x = \frac{8}{3}

. Therefore,

g(5) = \frac{8}{3}

. Choice B results from incorrectly setting up

2x + 3 = 5x + 5

. Choice C comes from the error

2x + 3 = 5x - 1

. Choice D results from solving

2x + 3 = 4(x - 1)

instead of

5(x - 1)

Question 16

A function $g(x) = ax + b$ has an inverse $g^{-1}(x) = \frac{x - 4}{3}$ . If $g(g^{-1}(7)) + g^{-1}(g(-2)) = k$ , what is the value of $k$ ?

$3$
$9$
$5$ (correct answer)
$1$

Explanation: When you encounter inverse function problems, remember that inverse functions "undo" each other. If

g

and

g^{-1}

are inverses, then

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

and

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

for all values in their domains. Let's find the original function

g(x)

first. Since

g^{-1}(x) = \frac{x - 4}{3}

, we can find

g(x)

by swapping variables and solving. Let

y = \frac{x - 4}{3}

, so

3y = x - 4

, which gives us

x = 3y + 4

. Therefore,

g(x) = 3x + 4

. Now we can evaluate

g(g^{-1}(7)) + g^{-1}(g(-2))

. Using the inverse function property,

g(g^{-1}(7)) = 7

. Similarly,

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

. Therefore,

k = 7 + (-2) = 5

. Looking at the wrong answers: Choice A (

3

) might result from incorrectly calculating

g^{-1}(7) = \frac{7-4}{3} = 1

and then adding

1 + 2 = 3

. Choice B (

9

) could come from adding

g^{-1}(7) = 1

and

g(-2) = 3(-2) + 4 = -2

incorrectly, getting

1 + 8 = 9

. Choice D (

1

) might result from finding only

g^{-1}(7) = 1

and stopping there. Remember: when you see

f(f^{-1}(a))

f^{-1}(f(a))

, these always simplify to just

a

. This inverse function property is your shortcut to avoid unnecessary calculations.