Algebra 3 Quiz: Inverse Functions

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Inverse FunctionsQuestion 1 of 20

Which pair of functions is NOT a pair of inverse functions?

f(x)=x^2+1\ (x\ge 0),\quad g(x)=\sqrt{x-1}

f(x)=\frac{1}{x-3},\quad g(x)=\frac{1}{x}+3

f(x)=\frac{x-5}{2},\quad g(x)=2x+10

f(x)=2x^3+1,\quad g(x)=\sqrt[3]{\frac{x-1}{2}}

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Algebra 3 Quiz

Algebra 3 Quiz: Inverse Functions

Practice Inverse Functions in Algebra 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 Inverse Functions, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 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.

All questions

Question 1

Which pair of functions is NOT a pair of inverse functions?

$f(x)=x^2+1\ (x\ge 0),\quad g(x)=\sqrt{x-1}$
$f(x)=\frac{1}{x-3},\quad g(x)=\frac{1}{x}+3$
$f(x)=\frac{x-5}{2},\quad g(x)=2x+10$ (correct answer)
$f(x)=2x^3+1,\quad g(x)=\sqrt[3]{\frac{x-1}{2}}$

Explanation: Whenever you see inverse functions, think "undoing": each function should reverse the other's operation. The surest check is composition: if

f

and

g

are inverses, then

f(g(x))=x

and

g(f(x))=x

for every

x

in the appropriate domain. Among these pairs,

f(x)=\frac{x-5}{2}

and

g(x)=2x+10

do not undo each other. Composing them gives

f(g(x))=\frac{(2x+10)-5}{2}=\frac{2x+5}{2}=x+2.5,

and

g(f(x))=2\left(\frac{x-5}{2}\right)+10=x+5.

Neither composition returns

x

, so this is NOT an inverse pair. The other three pairs do work. For

f(x)=x^2+1\ (x\ge 0)

and

g(x)=\sqrt{x-1}

, the restriction

x\ge 0

is exactly why both compositions simplify to

x

. For

f(x)=\frac{1}{x-3}

and

g(x)=\frac{1}{x}+3

, substitution cancels cleanly:

f(g(x))=\frac{1}{1/x}=x

and

g(f(x))=(x-3)+3=x

. For

f(x)=2x^3+1

and

g(x)=\sqrt[3]{\frac{x-1}{2}}

, cubing and cube-rooting reverse each other, giving

x

both ways. A good strategy: when asked which pair is NOT inverse, test only the compositions—if either one fails to simplify to

x

, the pair is not inverse. Watch for sign errors when adding or subtracting constants inside the compositions.

Question 2

Let $h(x)=\frac{5x-1}{x+2}$ for $x\neq -2$ . If $h^{-1}(a)=3$ , what is the value of $a$ ?

$\frac{5}{14}$
$\frac{7}{2}$
$-\frac{7}{2}$
$\frac{14}{5}$ (correct answer)

Explanation: Whenever you see inverse notation like

h^{-1}(a)=3

, remember what the inverse means:

h^{-1}(a)

is the input that produces output

a

. So if

h^{-1}(a)=3

, then the input is

3

, which means

a=h(3)

. You do not need to solve for the inverse function—just plug

3

into

h

h(3)=\frac{5(3)-1}{3+2}=\frac{15-1}{5}=\frac{14}{5}

a=\frac{14}{5}

, and that choice is correct. The choice

\frac{5}{14}

is the reciprocal of the correct value; it could come from flipping the fraction after computing

h(3)

\frac{7}{2}

h^{-1}(3)

, not

h(3)

—it comes from solving

h(x)=3

, which swaps the roles of the input and output. The choice

-\frac{7}{2}

is the same swapping error but with a sign mistake, likely from writing the inverse formula with a reversed denominator. The correct answer

\frac{14}{5}

is positive and not related to the reciprocal or the inverse evaluated at

3

. Strategy: when you see

h^{-1}(a)=b

, immediately translate it as

h(b)=a

. If you catch yourself solving

h(x)=3

, you've likely inverted the roles of input and output—stop and plug in instead.

Question 3

A one-to-one function $f$ has domain $[-2,7)$ and range $[-5,3]$ . Which statement correctly describes $f^{-1}$ ?

Its domain is $[-2,7)$ and its range is $[-5,3]$ .
Its domain is $[-5,3]$ and its range is $[-2,7)$ . (correct answer)
Its domain is $[-5,3]$ and its range is $(-2,7]$ .
Its domain is $[-5,3]$ and its range is $[-2,7]$ .

Explanation: When you see a question about inverse functions, remember the golden rule: the domain and range swap. For a one-to-one function, the inverse's domain is exactly the original function's range, and the inverse's range is exactly the original function's domain. The tricky part is preserving the endpoints—whether brackets or parentheses—when you swap them. Here,

f

has domain

[-2,7)

and range

[-5,3]

. So

f^{-1}

must have domain

[-5,3]

(both endpoints included, since the range of

f

is closed) and range

[-2,7)

(with

-2

included but

7

excluded, exactly as in the domain of

f

). That matches the statement "Its domain is

[-5,3]

and its range is

[-2,7)

." Now examine the other choices. The one saying "domain

[-2,7)

and range

[-5,3]

" simply repeats the original function—it forgets to swap. Another says "domain

[-5,3]

and range

(-2,7]

"—this flips the endpoint types: it incorrectly includes 7 and excludes -2, a classic sign-error when swapping intervals. The final option, "domain

[-5,3]

and range

[-2,7]

," correctly swaps the domain but wrongly changes the range to include 7, which was never in the original domain. On exam day, write down the original domain and range, then literally swap them—keeping every bracket and parenthesis exactly as it was. That single step avoids all these traps.

Question 4

Let $f(x)=\frac{1}{x-2}+3$ for $x\neq2$ . Which statement about the graph of $f^{-1}$ is true?

It has a vertical asymptote at $x=3$ and a horizontal asymptote at $y=2$ . (correct answer)
It has a vertical asymptote at $x=2$ and a horizontal asymptote at $y=3$ .
It has a vertical asymptote at $x=0$ and a horizontal asymptote at $y=3$ .
It has a vertical asymptote at $x=-3$ and a horizontal asymptote at $y=2$ .

Explanation: Whenever you see a question about the graph of an inverse function, remember that the inverse swaps the roles of the input and output. That means the domain and range of the original function become the range and domain of the inverse — so asymptotes swap too. Here,

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

has a vertical asymptote at

x=2

and a horizontal asymptote at

y=3

. For

f^{-1}

, those become a horizontal asymptote at

y=2

and a vertical asymptote at

x=3

. So the correct statement is: vertical asymptote at

x=3

and horizontal asymptote at

y=2

. The choice saying vertical asymptote at

x=2

and horizontal at

y=3

describes the original function, not its inverse — that is the common trap of forgetting to swap. The choice with vertical asymptote at

x=0

and horizontal at

y=3

may come from thinking the vertical shift creates an asymptote at zero, but the denominator

x-2

still controls the vertical asymptote, and it has already swapped for the inverse. The choice with vertical asymptote at

x=-3

and horizontal at

y=2

gets the horizontal asymptote right but uses the wrong sign for the vertical asymptote; the inverse is

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

, not

x+3

in the denominator. Study tip: when asked about inverse asymptotes, either compute the inverse quickly or remember that vertical and horizontal asymptotes swap. The original horizontal asymptote at

y=3

becomes the inverse's vertical asymptote at

x=3

Question 5

Suppose $f$ and $g$ are inverse functions. Which statement is NOT necessarily true?

The range of $g$ equals the domain of $f$ .
The graph of $g$ is the reflection of the graph of $f$ across the line $y=x$ .
For every $x$ in the domain of $f$ , $f(g(x))=x$ . (correct answer)
For every $x$ in the domain of $g$ , $f(g(x))=x$ .

Explanation: Whenever you see inverse functions, immediately think about two things: the domain/range swap and the composition cancellation properties. Inverses "undo" each other, but only within the correct domains. The false statement is the one claiming that for every

x

in the domain of

f

f(g(x))=x

. This is not necessarily true. If

x

is in the domain of

f

, then

g(x)

is only defined when

x

also lies in the domain of

g

, which equals the range of

f

. Those sets need not be the same. The guaranteed cancellation is

f(g(x))=x

for every

x

in the domain of

g

, and

g(f(x))=x

for every

x

in the domain of

f

. The other choices are true by definition: the range of

g

equals the domain of

f

, since inverse functions swap domains and ranges. The graph of

g

is indeed the reflection of the graph of

f

across the line

y=x

. And the statement that

f(g(x))=x

for every

x

in the domain of

g

is exactly the cancellation property for inverses. The trap here is mixing up the two composition directions. On the exam, write down both cancellation rules with their correct domains before answering. That makes it easy to spot which statement has the domain in the wrong place.

Question 6

Let $f(x)=\frac{2x+3}{x-4}$ for $x\neq 4$ . If $f^{-1}$ denotes the inverse function, which expression and domain correctly define $f^{-1}$ ?

$\frac{4x+3}{x-2},\ x\neq 2$ (correct answer)
$\frac{4x-3}{x-2},\ x\neq 2$
$\frac{2x+3}{x-4},\ x\neq 4$
$\frac{2x-3}{x+4},\ x\neq -4$

Explanation: When you see an inverse-function question, your task is to swap the roles of input and output: solve the equation for

x

in terms of

y

, then replace

y

with

x

. Start with

y=\frac{2x+3}{x-4}

. Multiply through to get

y(x-4)=2x+3

, so

yx-4y=2x+3

. Gather the

x

-terms:

yx-2x=4y+3

. Factor out

x

x(y-2)=4y+3

, so

x=\frac{4y+3}{y-2}

. Rename

y

x

, giving

\frac{4x+3}{x-2}

, and the denominator tells you

x\neq 2

. The version with

\frac{4x-3}{x-2}

comes from moving the

3

with the wrong sign; the constant term must be

4y+3

, not

4y-3

. The expression

\frac{2x+3}{x-4}

is the original function itself — it appears if you stop after one algebraic step or confuse a function with its inverse. Finally,

\frac{2x-3}{x+4}

has both sign errors and the wrong domain; it may result from incorrectly solving

y(x-4)=2x-3

or misplacing terms. Also notice the inverse's domain excludes

2

, not

4

-4

. On this exam, always finish by checking the domain from the inverse's denominator, not the original's. A quick way to verify: plug a simple value into

f

, then into your proposed inverse — you should get back to your starting number.

Question 7

The graph of $g$ is the reflection of the graph of $f(x)=\sqrt{x-2}+3,\ x\ge2$ across the line $y=x$ . Which equation and domain define $g$ ?

$g(x)=(x-3)^2+2,\ x\ge 3$ (correct answer)
$g(x)=(x-3)^2+2,\ x\ge 2$
$g(x)=(x+3)^2-2,\ x\ge -3$
$g(x)=(x-3)^2-2,\ x\ge 3$

Explanation: Whenever you see a reflection across the line

y=x

, think inverse function. The graph of

g

is the inverse of

f

, so you need to swap the roles of

x

and

y

— and remember that the domain of

g

is the range of

f

. Start with

y=\sqrt{x-2}+3

. Solve for

x

: subtract 3, square both sides, then add 2:

x=(y-3)^2+2.

Swap

x

and

y

to get

g(x)=(x-3)^2+2.

Since

f

is defined for

x\ge 2

and its output starts at 3 and increases, the range of

f

y\ge 3

. Therefore

g

has domain

x\ge 3

. The choice

(x-3)^2+2,\ x\ge 2

is a trap: it uses

f

's domain instead of

g

's. The choice

(x+3)^2-2,\ x\ge -3

comes from mis-handling signs when solving — subtracting 3 instead of adding and subtracting 2 instead of adding. The choice

(x-3)^2-2,\ x\ge 3

has the right domain but the wrong constant: you must add 2, not subtract it, after solving for

x

. On reflection questions, compute the inverse algebraically, then set its domain equal to the original function's range. That single check eliminates most wrong answers.

Question 8

Let $g(x)=\ln(2x-3)+4$ for $x>\frac{3}{2}$ . Which expression defines $g^{-1}(x)$ ?

$\frac{1}{2}e^{x-4}-\frac{3}{2},\ x\in\mathbb{R}$
$\frac{1}{2}e^{x+4}-\frac{3}{2},\ x\in\mathbb{R}$
$2e^{x-4}-3,\ x\in\mathbb{R}$
$\frac{1}{2}e^{x-4}+\frac{3}{2},\ x\in\mathbb{R}$ (correct answer)

Explanation: When you see a question about an inverse function, remember that you are swapping the roles of input and output: start with

y=\ln(2x-3)+4

, then solve for

x

in terms of

y

. That equation becomes

y-4=\ln(2x-3)

. Exponentiating both sides gives

e^{y-4}=2x-3

. Add

3

to get

e^{y-4}+3=2x

, and divide by

2

x=\frac12 e^{y-4}+\frac32

. So the inverse is

\frac12 e^{x-4}+\frac32

, and since the original function has range all real numbers, its inverse has domain

x\in\mathbb{R}

. The choice

\frac12 e^{x-4}-\frac32

uses the right exponential setup but signs incorrectly: it replaces the needed "plus

\frac32

" with "minus

\frac32

." The choice

\frac12 e^{x+4}-\frac32

makes two mistakes: it uses

x+4

instead of

x-4

, and it again has the wrong constant term. The choice

2e^{x-4}-3

shows what happens if you multiply by

2

and subtract

3

instead of dividing by

2

and adding

\frac32

— that reflects forgetting to undo the original coefficient of

2

and the constant shift correctly. A useful check: compose the inverse with the original function. If both produce

x

, your algebra is right. Also watch the constant carefully — when solving

e^{y-4}=2x-3

, you must add

3

before dividing by

2

, not subtract

3

Question 9

The function $f(x)=x^2-6x+10$ is not one-to-one over its natural domain. Which of the following domain restrictions would NOT produce a one-to-one function?

$x\ge 5$
$x\ge 2$ (correct answer)
$x\le 3$
$x\le 0$

Explanation: Whenever you see a quadratic function and a question about being one-to-one, think about symmetry. Write

f(x)=x^2-6x+10=(x-3)^2+1,

so the parabola's vertex is at

x=3

. The function decreases on

(-\infty,3]

and increases on

[3,\infty)

. A restricted domain is one-to-one only if it stays entirely on one side of that turning point. The restriction

x\ge 2

does NOT work, because its domain includes values on both sides of

x=3

: from

2

3

the graph is decreasing, and from

3

onward it is increasing. This creates repeated outputs; for example,

f(2)=2

and

f(4)=2

, so two different

x

-values give the same

y

-value. The other restrictions do keep the domain on one side:

x\ge 5

lies completely to the right of

3

;

x\le 3

is the left side plus the vertex; and

x\le 0

is also entirely left of

3

. Each passes the horizontal line test on that restricted domain. The trap here is assuming that any restriction starting below the vertex will work. It won't if it also includes values above the vertex. Study tip: complete the square to find the axis

x=h

, then make sure the restricted domain is a subset of

(-\infty,h]

[h,\infty)

. If it straddles

h

, check symmetric pairs of inputs.

Question 10

A student says that the inverse of $f(x)=\frac{1}{x-1}$ is $f^{-1}(x)=\frac{1}{x}-1$ . Which response is correct?

Yes; reflecting across $y=x$ gives $\frac{1}{x}-1$ .
No; the inverse is $\frac{1}{x}+1$ , not $\frac{1}{x}-1$ . (correct answer)
No; $f$ is not one-to-one, so it does not have an inverse.
Yes; the reciprocal of $f$ is $x-1$ , so that is its inverse.

Explanation: Whenever you're asked for an inverse function, the key move is to swap input and output — exchange

x

and

y

, then solve for

y

. Start with

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

. Swap:

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

. Multiply:

x(y-1)=1

, so

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

, and therefore

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

. The correct inverse is

\frac{1}{x}+1

, not

\frac{1}{x}-1

. The student's answer likely came from a sign error after the swap. The choice saying "reflecting across

y=x

" describes the right method, but carrying out that reflection gives

\frac{1}{x}+1

, not

\frac{1}{x}-1

. So that version is also wrong — the algebra mistake happens during the reflection step. The choice claiming

f

is not one-to-one is mistaken: if

\frac{1}{a-1}=\frac{1}{b-1}

, then

a-1=b-1

, so

a=b

. The function passes the horizontal line test and does have an inverse. Finally, the choice saying "the reciprocal of

f

x-1

" trades on a common vocabulary trap: the reciprocal of the function value is

x-1

, but a reciprocal is not an inverse function. Those two meanings of "inverse" are different except in special cases like

f(x)=\frac{1}{x}

. Study takeaway: whenever a question asks for an inverse function, swap

x

and

y

, solve for

y

, and test with composition. If you see the word "reciprocal," remember that the inverse function is found by solving, not by flipping the expression.

Question 11

If $f(x)=\sqrt[3]{2x-5}+1$ , which of the following is $f^{-1}(x)$ ?

$\frac{(x+1)^3+5}{2},\ x\in\mathbb{R}$
$\frac{(x+1)^3-5}{2},\ x\in\mathbb{R}$
$\frac{(x-1)^3-5}{2},\ x\in\mathbb{R}$
$\frac{(x-1)^3+5}{2},\ x\in\mathbb{R}$ (correct answer)

Explanation: Whenever you need an inverse, solve for the original input in terms of the original output. Start with

y=\sqrt[3]{2x-5}+1.

Undo the operations in reverse: subtract 1 first, then cube both sides, then add 5, and divide by 2.

y-1=\sqrt[3]{2x-5}

\Rightarrow (y-1)^3=2x-5

\Rightarrow x=\frac{(y-1)^3+5}{2}.

Now swap

x

and

y

to get the inverse:

f^{-1}(x)=\frac{(x-1)^3+5}{2},\quad x\in\mathbb{R}.

This matches the choice with

(x-1)^3+5

over 2. The wrong choices encode common sign errors. The choice with

(x+1)^3+5

adds101 instead of subtracting it when reversing the outer

+1

. The choice with

(x+1)^3-5

makes that same sign error andalso subtracts 5 instead of adding it, as if the constant inside the cube root had the opposite sign. The choice with

(x-1)^3-5

correctly subtracts the 1, but then subtracts 5 when isolating

x

, forgetting that

2x-5

must be undone by adding 5, not subtracting it. Study tip: after you find an inverse, always check by composing

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

and simplifying. If it doesn't equal

x

, you'll instantly see whether you used the wrong sign on a constant or reversed the order of operations.

Question 12

The function $f(x)=x^2-4x+7$ is restricted to $x\ge2$ . Which of the following is $f^{-1}(x)$ ?

$2+\sqrt{x+3},\ x\ge -3$
$2-\sqrt{x-3},\ x\ge3$
$2+\sqrt{x-3},\ x\ge3$ (correct answer)
$-2+\sqrt{x-3},\ x\ge3$

Explanation: When you see a restricted quadratic, you're really being asked: can I undo this function, and which half of the parabola is allowed? Since a full parabola fails the horizontal line test, the restriction

x\ge 2

selects the right branch, making an inverse possible. Rewrite

f(x)=x^2-4x+7

by completing the square:

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

. Because

x\ge2

, the output is at least

3

, so the range is

[3,\infty)

. To invert, set

y=(x-2)^2+3

, subtract

3

, and take the square root:

\sqrt{y-3}=x-2

. Since

x\ge2

, the square root must be the positive one, so

x=2+\sqrt{y-3}

. Swap variables to get

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

, with domain

x\ge3

because the original range becomes the inverse's domain. The choice

2+\sqrt{x+3},\ x\ge-3

uses the wrong constant inside the root; it comes from confusing the vertex shift with the range. The choice

2-\sqrt{x-3},\ x\ge3

is the negative square-root branch, which would be the inverse of the left side of the parabola, not the restricted

x\ge2

side. Finally,

-2+\sqrt{x-3},\ x\ge3

misplaces the vertex: subtracting

2

instead of adding it reflects a sign error when solving for

x

. On inverse-function questions, always complete the square, identify the original range, and choose the root sign that matches the restricted domain. That single habit prevents most of these traps.

Question 13

The function $f$ is defined by $f(x)=x^2-6x+5$ on the domain $(-\infty,3]$ . Which expression gives $f^{-1}(x)$ ?

$3-\sqrt{x+4}$ (correct answer)
$3+\sqrt{x+4}$
$-3+\sqrt{x+4}$
$-3-\sqrt{x+4}$

Explanation: Complete the square:

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

. On

(-\infty,3]

x-3\le0

, so solving

y=(x-3)^2-4

gives

x-3=-\sqrt{y+4}

, hence

f^{-1}(x)=3-\sqrt{x+4}

. The plus-root option is the inverse for the domain

[3,\infty)

; the options with

-3

have the horizontal shift wrong.

Question 14

Let $f(x)=\frac{2x+1}{x-3}$ . Which statement about $f^{-1}$ is true?

$f^{-1}(x)=\frac{3x-1}{x-2}$ , domain $x\ne2$
$f^{-1}(x)=\frac{3x+1}{x+2}$ , domain $x\ne-2$
$f^{-1}(x)=\frac{3x-1}{x+2}$ , domain $x\ne-2$
$f^{-1}(x)=\frac{3x+1}{x-2}$ , domain $x\ne2$ (correct answer)

Explanation: Set

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

, so

yx-3y=2x+1

. Then

x(y-2)=3y+1

, so

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

. Because

y=2

is not in the range, the domain is

x\ne2

. The options with

3x-1

have a sign error in the numerator; the options with denominator

x+2

have the wrong denominator sign.

Question 15

Suppose $g(x)=f^{-1}(x)=\frac{x-2}{x+3}$ . Which expression defines $f(x)$ ?

$f(x)=\frac{3x+2}{1-x}$ (correct answer)
$f(x)=\frac{3x-2}{1-x}$
$f(x)=\frac{2x+3}{x-1}$
$f(x)=\frac{3x+2}{x-1}$

Explanation: Since

g=f^{-1}

f

is the inverse of

g

. Solving

y=\frac{x-2}{x+3}

for

x

gives

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

, so

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

. The options with

3x-2

or denominator

x-1

come from sign errors when moving terms.

Question 16

For $f(x)=\frac{ax+3}{x-2}$ , it is true that $f^{-1}(x)=f(x)$ for all $x$ in the domains. What is $a$ ?

1
2 (correct answer)
3
4

Explanation: Find the inverse: solving

y=\frac{ax+3}{x-2}

gives

x=\frac{2y+3}{y-a}

, so

f^{-1}(x)=\frac{2x+3}{x-a}

. Setting this equal to

\frac{ax+3}{x-2}

and comparing coefficients gives

a=2

. The other numeric choices do not make the two rational expressions identical.

Question 17

The graph of $f(x)=4+\sqrt{x-6}$ is reflected across the line $y=x$ , producing the graph of $f^{-1}$ . If the point $(5,k)$ lies on $f^{-1}$ , what is $k$ ?

5
7 (correct answer)
9
12

Explanation: A point

(5,k)

on the reflected graph means

f^{-1}(5)=k

, so

f(k)=5

. Solve

4+\sqrt{k-6}=5

\sqrt{k-6}=1

, so

k=7

. The values 5, 9, and 12 come from treating

(5,k)

as a point on

f

or from solving the radical incorrectly.

Question 18

A student finds $f^{-1}(x)=4+\sqrt{x+11}$ for $f(x)=x^2-8x+5$ with domain $x\le4$ . Which statement best describes the student's error?

The student solved for $y$ instead of $x$ .
The student should have added 11 outside the square root instead of subtracting it.
The student used the branch on the right side of the vertex, which corresponds to $x\ge4$ , not $x\le4$ . (correct answer)
The student reflected the graph over the x-axis instead of over the line $y=x$ .

Explanation: The function is

(x-4)^2-11

. Since the domain is

x\le4

x-4

is nonpositive, so the inverse is

4-\sqrt{x+11}

. The student's plus-root form would be the inverse on

x\ge4

. The other error descriptions do not account for the plus sign.

Question 19

Let $f(x)=\sqrt{2x-7}+3$ . Which statement about $f^{-1}$ is true?

$f^{-1}(x)=\frac{(x-3)^2-7}{2}$ , domain $x\ge3$ , range $y\le\frac72$
$f^{-1}(x)=\frac{(x+3)^2-7}{2}$ , domain $x\ge\frac72$ , range $y\ge3$
$f^{-1}(x)=\frac{(x-3)^2+7}{2}$ , domain $x\ge\frac72$ , range $y\ge3$
$f^{-1}(x)=\frac{(x-3)^2+7}{2}$ , domain $x\ge3$ , range $y\ge\frac72$ (correct answer)

Explanation: The original domain is

[\frac72,\infty)

and range is

[3,\infty)

, so the inverse has domain

[3,\infty)

and range

[\frac72,\infty)

. Solving

y=\sqrt{2x-7}+3

gives

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

. The sign-swapped options use

(x+3)^2

-7

, and the domain/range-swapped option reverses the original domain and range.

Question 20

Let $f(x)=2^{x-1}+3$ . Which expression gives $f^{-1}(x)$ with its correct domain?

$1+\log_2(x+3)$ , domain $x>-3$
$-1+\log_2(x-3)$ , domain $x>3$
$1+\log_2(x-3)$ , domain $x>3$ (correct answer)
$\log_2(x-3)-3$ , domain $x>3$

Explanation: Solve

y=2^{x-1}+3

2^{x-1}=y-3

, so

x=1+\log_2(y-3)

. The inverse is

1+\log_2(x-3)

, defined for

x>3

. Options with

x+3

forget to subtract 3; options with

-1

-3

apply the wrong shifts.