Algebra 2 Quiz: Verify Functions Are Inverses

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Verify Functions Are InversesQuestion 1 of 20

A student claims that $f(x)=x^2$ (with domain restricted to $x\ge 0$ ) and $g(x)=\sqrt{x}$ are inverse functions. Verify or disprove by computing $f(g(x))$ and $g(f(x))$ using the stated domain restriction.

f(g(x))=(\sqrt{x})^2=x

but

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

, so they are not inverses.

f(g(x))=x^2

and

g(f(x))=\sqrt{x}

, so they are inverses because the outputs match the original functions.

f(g(x))=(\sqrt{x})^2=x

(for

x\ge 0

) and

g(f(x))=\sqrt{x^2}=x

(for

x\ge 0

), so they are inverses on the restricted domain.

f(g(x))=\sqrt{x^2}=|x|

and

g(f(x))=(\sqrt{x})^2=x

, so they are inverses for all real

x

without restriction.

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Algebra 2 Quiz

Algebra 2 Quiz: Verify Functions Are Inverses

Practice Verify Functions Are Inverses in Algebra 2 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 Verify Functions Are Inverses, giving you a quick way to practice the rules, question types, and explanations that matter most for Algebra 2.

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

$f(g(x))=(\sqrt{x})^2=x$ but $g(f(x))=\sqrt{x^2}=-x$ , so they are not inverses.
$f(g(x))=x^2$ and $g(f(x))=\sqrt{x}$ , so they are inverses because the outputs match the original functions.
$f(g(x))=(\sqrt{x})^2=x$ (for $x\ge 0$ ) and $g(f(x))=\sqrt{x^2}=x$ (for $x\ge 0$ ), so they are inverses on the restricted domain. (correct answer)
$f(g(x))=\sqrt{x^2}=|x|$ and $g(f(x))=(\sqrt{x})^2=x$ , so they are inverses for all real $x$ without restriction.

Explanation: This question tests your understanding that two functions are inverses if and only if their compositions both equal the identity function—meaning f(g(x)) = x AND g(f(x)) = x. To verify that f and g are inverse functions, we must check BOTH compositions: f(g(x)) should equal x (showing g undoes what f does), and g(f(x)) should equal x (showing f undoes what g does). Only when both compositions simplify to the identity function x can we conclude the functions are true inverses. One direction isn't enough—we need the bidirectional undo relationship! Here, f(g(x)) = (√x)^2 = x for x ≥ 0, and g(f(x)) = √(

x^2

) = |x| = x since x ≥ 0, both equaling x on the restricted domain. Choice A correctly determines they are inverses on the restricted domain by verifying both compositions. A distractor like choice B fails by not considering the domain restriction, incorrectly stating g(f(x)) = -x without noting x ≥ 0. The verification checklist: (1) Compute f(g(x)): substitute g(x) into f, simplify completely, (2) Check: does it equal x? If no, they're not inverses—stop. If yes, continue, (3) Compute g(f(x)): substitute f(x) into g, simplify completely, (4) Check: does it equal x? If yes, they're inverses! If no, they're not (even though first direction worked). Both must equal x for full inverse verification—this is non-negotiable! Great job considering domains—they're key for inverses like these!

Question 2

Show that $f(x)=\dfrac{x+1}{2}$ and $g(x)=2x-1$ are inverses by verifying both compositions: $f(g(x))$ and $g(f(x))$ .

$f(g(x))=\dfrac{2x-1+1}{2}=x$ and $g(f(x))=2\left(\dfrac{x+1}{2}-1\right)=x-2$ , so they are not inverses.
$f(g(x))=\dfrac{(2x-1)+1}{2}=x$ and $g(f(x))=2\left(\dfrac{x+1}{2}\right)-1=x$ , so they are inverses. (correct answer)
$f(g(x))=\dfrac{2x-1}{2}=x-\dfrac{1}{2}$ and $g(f(x))=2\left(\dfrac{x+1}{2}\right)-1=x$ , so they are not inverses.
$f(g(x))=\dfrac{x+1}{2}$ and $g(f(x))=2x-1$ , so they are inverses.

Explanation: This question tests your understanding that two functions are inverses if and only if their compositions both equal the identity function—meaning

f(g(x)) = x

AND

g(f(x)) = x

. To verify that f and g are inverse functions, we must check BOTH compositions:

f(g(x))

should equal x (showing g undoes what f does), and

g(f(x))

should equal x (showing f undoes what g does). Only when both compositions simplify to the identity function x can we conclude the functions are true inverses. One direction isn't enough—we need the bidirectional undo relationship! Let's compute:

f(g(x)) = ((2x - 1) + 1)/2 = 2x/2 = x

, and

g(f(x)) = 2*((x + 1)/2) - 1 = (x + 1) - 1 = x

, so both simplify to x. Choice A correctly verifies both compositions equal x and determines they are inverses. Choice B might miscompute

f(g(x))

by dividing incorrectly, a gentle reminder to apply operations to the entire expression. The verification checklist: (1) Compute

f(g(x))

: substitute g(x) into f, simplify completely, (2) Check: does it equal x? If no, they're not inverses—stop. If yes, continue, (3) Compute

g(f(x))

: substitute f(x) into g, simplify completely, (4) Check: does it equal x? If yes, they're inverses! If no, they're not (even though first direction worked). Both must equal x for full inverse verification—this is non-negotiable! Common verification error: claiming verification after only one composition. You might check

f(g(x)) = x

and declare them inverses, but without checking

g(f(x)) = x

, you haven't fully verified! While rare, it's theoretically possible for one direction to work but not the other (function pairs that are one-sided inverses). Always do both—it only takes a minute more and ensures you're correct. Complete verification = both directions = confidence!

Question 3

Are $f(x)=5-2x$ and $g(x)=\dfrac{5-x}{2}$ inverses? Verify by composition by computing $f(g(x))$ and $g(f(x))$ and checking whether both equal $x$ .

$f(g(x))=5-2\left(\dfrac{5-x}{2}\right)=x$ and $g(f(x))=\dfrac{5-(5-2x)}{2}=x$ , so they are inverses. (correct answer)
$f(g(x))=5-2\left(\dfrac{5-x}{2}\right)=x$ and $g(f(x))=\dfrac{5-(5-2x)}{2}=\dfrac{x}{2}$ , so they are not inverses.
$f(g(x))=5-\left(\dfrac{5-x}{2}\right)=\dfrac{5+x}{2}$ and $g(f(x))=x$ , so they are not inverses.
$f(g(x))=x-5$ and $g(f(x))=x+5$ , so they are inverses.

Explanation: This question tests your understanding that two functions are inverses if and only if their compositions both equal the identity function—meaning f(g(x)) = x AND g(f(x)) = x. The composition f(g(x)) means 'take the output of g and use it as input to f': substitute the entire expression for g(x) wherever you see x in f(x), then simplify. If f and g are truly inverses, this process should 'undo' everything and leave you with just x. It's like putting on shoes then taking them off—you end up back where you started (barefoot = x)! Let's compute: f(g(x)) = 5 - 2*((5 - x)/2) = 5 - (5 - x) = x, and g(f(x)) = (5 - (5 - 2x))/2 = (2x)/2 = x, so both simplify to x. Choice A correctly verifies both compositions equal x and determines they are inverses. A distractor like choice B shows one correct but halves incorrectly in the other, gently reminding us to distribute operations fully. The verification checklist: (1) Compute f(g(x)): substitute g(x) into f, simplify completely, (2) Check: does it equal x? If no, they're not inverses—stop. If yes, continue, (3) Compute g(f(x)): substitute f(x) into g, simplify completely, (4) Check: does it equal x? If yes, they're inverses! If no, they're not (even though first direction worked). Both must equal x for full inverse verification—this is non-negotiable! Common verification error: claiming verification after only one composition. You might check f(g(x)) = x and declare them inverses, but without checking g(f(x)) = x, you haven't fully verified! While rare, it's theoretically possible for one direction to work but not the other (function pairs that are one-sided inverses). Always do both—it only takes a minute more and ensures you're correct. Complete verification = both directions = confidence!

Question 4

$f(g(x))=\sqrt{x^2}=|x|$ and $g(f(x))=(\sqrt{x})^2=x$ , so they are inverses for all real $x$ without restriction.
$f(g(x))=x^2$ and $g(f(x))=\sqrt{x}$ , so they are inverses because the outputs match the original functions.
$f(g(x))=(\sqrt{x})^2=x$ but $g(f(x))=\sqrt{x^2}=-x$ , so they are not inverses.
$f(g(x))=(\sqrt{x})^2=x$ (for $x\ge 0$ ) and $g(f(x))=\sqrt{x^2}=x$ (for $x\ge 0$ ), so they are inverses on the restricted domain. (correct answer)

Explanation: This question tests your understanding that two functions are inverses if and only if their compositions both equal the identity function—meaning f(g(x)) = x AND g(f(x)) = x. To verify that f and g are inverse functions, we must check BOTH compositions: f(g(x)) should equal x (showing g undoes what f does), and g(f(x)) should equal x (showing f undoes what g does). Only when both compositions simplify to the identity function x can we conclude the functions are true inverses. One direction isn't enough—we need the bidirectional undo relationship! Here, f(g(x)) = (√x)^2 = x for x ≥ 0, and g(f(x)) = √(

x^2

Question 5

Check whether $f(x)=\dfrac{x+1}{x-2}$ and $g(x)=\dfrac{2x+1}{x-1}$ are inverse functions by verifying both compositions $f(g(x))$ and $g(f(x))$ (for values where the expressions are defined).

$f(g(x))=\dfrac{\frac{2x+1}{x-1}+1}{\frac{2x+1}{x-1}-2}=\dfrac{\frac{3x}{x-1}}{\frac{3}{x-1}}=x$ and $g(f(x))=\dfrac{2\frac{x+1}{x-2}+1}{\frac{x+1}{x-2}-1}=\dfrac{\frac{3x}{x-2}}{\frac{3}{x-2}}=x$ , so they are inverses. (correct answer)
$f(g(x))=\dfrac{\frac{2x+1}{x-1}+1}{\frac{2x+1}{x-1}-2}=\dfrac{3x}{3}=1$ and $g(f(x))=x$ , so they are not inverses.
$f(g(x))=\dfrac{\frac{2x+1}{x-1}+1}{\frac{2x+1}{x-1}-2}=\dfrac{\frac{3x}{x-1}}{\frac{1}{x-1}}=3x\ne x$ , so they are not inverses.
$f(g(x))=x$ but $g(f(x))=\dfrac{\frac{2x+2}{x-2}+1}{\frac{x+1}{x-2}-1}=\dfrac{\frac{2x+3}{x-2}}{\frac{3}{x-2}}=\dfrac{2x+3}{3}\ne x$ , so they are not inverses.

Explanation: This question tests your understanding that two functions are inverses if and only if their compositions both equal the identity function—meaning f(g(x)) = x AND g(f(x)) = x. The composition f(g(x)) means 'take the output of g and use it as input to f': substitute the entire expression for g(x) wherever you see x in f(x), then simplify. If f and g are truly inverses, this process should 'undo' everything and leave you with just x. It's like putting on shoes then taking them off—you end up back where you started (barefoot = x)! For f(g(x)) = f((2x+1)/(x-1)) = ((2x+1)/(x-1) + 1)/((2x+1)/(x-1) - 2) = ((2x+1+x-1)/(x-1))/((2x+1-2x+2)/(x-1)) = (3x/(x-1))/(3/(x-1)) = 3x/3 = x ✓, and g(f(x)) = g((x+1)/(x-2)) = (2(x+1)/(x-2) + 1)/((x+1)/(x-2) - 1) = ((2x+2+x-2)/(x-2))/((x+1-x+2)/(x-2)) = (3x/(x-2))/(3/(x-2)) = 3x/3 = x ✓. Choice A correctly shows both compositions equal x and verifies they are inverses. Choice B makes algebraic errors in g(f(x)), Choice C incorrectly simplifies f(g(x)) to 1, and Choice D incorrectly simplifies the denominator in f(g(x)). Common verification error: claiming verification after only one composition. You might check f(g(x)) = x and declare them inverses, but without checking g(f(x)) = x, you haven't fully verified! While rare, it's theoretically possible for one direction to work but not the other (function pairs that are one-sided inverses). Always do both—it only takes a minute more and ensures you're correct. Complete verification = both directions = confidence!

Question 6

Demonstrate that $f(x)=7-2x$ and $g(x)=\dfrac{7-x}{2}$ are inverse functions by showing both $f(g(x))=x$ and $g(f(x))=x$ .

$f(g(x))=7-2\left(\dfrac{7-x}{2}\right)=x$ and $g(f(x))=\dfrac{7-(7-2x)}{2}=x$ , so they are inverses. (correct answer)
$f(g(x))=7-2\left(\dfrac{7-x}{2}\right)=x$ and $g(f(x))=\dfrac{7-(7-2x)}{2}=\dfrac{x}{2}$ , so they are inverses.
$f(g(x))=7-\left(\dfrac{7-x}{2}\right)=\dfrac{7+x}{2}$ and $g(f(x))=\dfrac{7-(7-2x)}{2}=x$ , so they are not inverses.
$f(g(x))=7-2\left(\dfrac{7-x}{2}\right)=x$ , so they are inverses (no need to check $g(f(x))$ ).

Explanation: This question tests your understanding that two functions are inverses if and only if their compositions both equal the identity function—meaning f(g(x)) = x AND g(f(x)) = x. The composition f(g(x)) means 'take the output of g and use it as input to f': substitute the entire expression for g(x) wherever you see x in f(x), then simplify. If f and g are truly inverses, this process should 'undo' everything and leave you with just x. It's like putting on shoes then taking them off—you end up back where you started (barefoot = x)! Computing f(g(x)) = 7 - 2*( (7 - x)/2 ) = 7 - (7 - x) = x, and g(f(x)) = [7 - (7 - 2x)] / 2 = (2x)/2 = x, both equaling x. Choice A correctly verifies both compositions and shows the inverse relationship. Choice C corrects the error of incomplete verification, emphasizing that both must be checked. Common verification error: claiming verification after only one composition. You might check f(g(x)) = x and declare them inverses, but without checking g(f(x)) = x, you haven't fully verified! While rare, it's theoretically possible for one direction to work but not the other (function pairs that are one-sided inverses). Always do both—it only takes a minute more and ensures you're correct. Complete verification = both directions = confidence! You've got this—linear functions like these are perfect for practice!

Question 7

A student claims $f(x)=x^2$ (with domain restricted to $x\ge 0$ ) and $g(x)=\sqrt{x}$ are inverses. Verify the claim by showing whether both $f(g(x))$ and $g(f(x))$ equal $x$ on the appropriate domains.

$f(g(x))=(\sqrt{x})^2=x$ for $x\ge0$ , and $g(f(x))=\sqrt{x^2}=x$ for $x\ge0$ , so they are inverses (with $f$ restricted to $x\ge0$ ). (correct answer)
$f(g(x))=\sqrt{x^2}=|x|$ and $g(f(x))=(\sqrt{x})^2=x$ , so they are not inverses.
$f(g(x))=x$ but $g(f(x))=|x|$ , so they are not inverses even when restricting $x\ge0$ .
Since $f(g(x))=x$ , that alone proves they are inverses.

x^2

) = |x| = x for x ≥ 0, so both equal x. Choice A correctly verifies both compositions equal x with the domain restriction and determines they are inverses. Choice B fails to account for the domain, incorrectly concluding g(f(x)) = |x| makes them non-inverses without noting |x| = x for x ≥ 0. Common verification error: claiming verification after only one composition. You might check f(g(x)) = x and declare them inverses, but without checking g(f(x)) = x, you haven't fully verified! While rare, it's theoretically possible for one direction to work but not the other (function pairs that are one-sided inverses). Always do both—it only takes a minute more and ensures you're correct. Complete verification = both directions = confidence!

Question 8

A student claims $f(x)=x^2$ and $g(x)=\sqrt{x}$ are inverses. Verify or disprove by composition, assuming $f$ has domain $x \ge 0$ . Compute $f(g(x))$ and $g(f(x))$ and decide if both equal $x$ (for allowed inputs).

$f(g(x)) = \sqrt{x^2} = |x|$ and $g(f(x)) = (\sqrt{x})^2 = x$ , so they are not inverses even when $x \ge 0$ .
$f(g(x)) = x^2$ and $g(f(x)) = x^2$ , so they are inverses.
$f(g(x)) = (\sqrt{x})^2 = x$ for $x \ge 0$ and $g(f(x)) = \sqrt{x^2} = x$ for $x \ge 0$ , so they are inverses on $x \ge 0$ . (correct answer)
$f(g(x)) = x$ but $g(f(x)) = \sqrt{x^2} = -x$ , so they are not inverses.

Explanation: This question tests your understanding that two functions are inverses if and only if their compositions both equal the identity function—meaning

f(g(x)) = x

AND

g(f(x)) = x

. The composition

f(g(x))

means 'take the output of g and use it as input to f': substitute the entire expression for g(x) wherever you see x in f(x), then simplify. If f and g are truly inverses, this process should 'undo' everything and leave you with just x. It's like putting on shoes then taking them off—you end up back where you started (barefoot = x)! Let's compute:

f(g(x)) = (\sqrt{x})^2 = x

for

x \ge 0

, and

g(f(x)) = \sqrt{x^2} = |x| = x

for

x \ge 0

, so both simplify to x on the restricted domain. Choice A correctly verifies both compositions equal x and determines they are inverses on

x \ge 0

. A distractor like choice B swaps the compositions and overlooks the domain restriction, gently correcting that

\sqrt{x^2} = x

when

x \ge 0

. The verification checklist: (1) Compute

f(g(x)

): substitute g(x) into f, simplify completely, (2) Check: does it equal x? If no, they're not inverses—stop. If yes, continue, (3) Compute

g(f(x)

): substitute f(x) into g, simplify completely, (4) Check: does it equal x? If yes, they're inverses! If no, they're not (even though first direction worked). Both must equal x for full inverse verification—this is non-negotiable! Common verification error: claiming verification after only one composition. You might check

f(g(x)) = x

and declare them inverses, but without checking

g(f(x)) = x

Question 9

Demonstrate that $f(x)=7-2x$ and $g(x)=\dfrac{7-x}{2}$ are inverse functions by showing both $f(g(x))=x$ and $g(f(x))=x$ .

$f(g(x))=7-2\left(\dfrac{7-x}{2}\right)=x$ , so they are inverses (no need to check $g(f(x))$ ).
$f(g(x))=7-2\left(\dfrac{7-x}{2}\right)=x$ and $g(f(x))=\dfrac{7-(7-2x)}{2}=\dfrac{x}{2}$ , so they are inverses.
$f(g(x))=7-\left(\dfrac{7-x}{2}\right)=\dfrac{7+x}{2}$ and $g(f(x))=\dfrac{7-(7-2x)}{2}=x$ , so they are not inverses.
$f(g(x))=7-2\left(\dfrac{7-x}{2}\right)=x$ and $g(f(x))=\dfrac{7-(7-2x)}{2}=x$ , so they are inverses. (correct answer)

Explanation: This question tests your understanding that two functions are inverses if and only if their compositions both equal the identity function—meaning f(g(x)) = x AND g(f(x)) = x. The composition f(g(x)) means 'take the output of g and use it as input to f': substitute the entire expression for g(x) wherever you see x in f(x), then simplify. If f and g are truly inverses, this process should 'undo' everything and leave you with just x. It's like putting on shoes then taking them off—you end up back where you started (barefoot = x)! Computing f(g(x)) = 7 - 2*( (7 - x)/2 ) = 7 - (7 - x) = x, and g(f(x)) = [7 - (7 - 2x)] / 2 = (2x)/2 = x, both equaling x. Choice A correctly verifies both compositions and shows the inverse relationship. Choice C corrects the error of incomplete verification, emphasizing that both must be checked. Common verification error: claiming verification after only one composition. You might check f(g(x)) = x and declare them inverses, but without checking g(f(x)) = x, you haven't fully verified! While rare, it's theoretically possible for one direction to work but not the other (function pairs that are one-sided inverses). Always do both—it only takes a minute more and ensures you're correct. Complete verification = both directions = confidence! You've got this—linear functions like these are perfect for practice!

Question 10

Verify the inverse relationship between the temperature conversion formulas $C(F)=\dfrac{5}{9}(F-32)$ and $F(C)=\dfrac{9}{5}C+32$ by showing that $C(F(C))=C$ and $F(C(F))=F$ .

$C(F(C))=\dfrac{5}{9}\left(\dfrac{9}{5}C+32\right)=C+\dfrac{160}{9}\ne C$ , so they are not inverses.
$C(F(C))=C$ ; therefore the formulas are inverses without needing to check $F(C(F))$ .
$C(F(C))=\dfrac{5}{9}\left(\dfrac{9}{5}C\right)+32=C+32\ne C$ and $F(C(F))=F$ , so they are not inverses.
$C(F(C))=\dfrac{5}{9}\left(\left(\dfrac{9}{5}C+32\right)-32\right)=\dfrac{5}{9}\cdot\dfrac{9}{5}C=C$ and $F(C(F))=\dfrac{9}{5}\left(\dfrac{5}{9}(F-32)\right)+32=F-32+32=F$ , so they are inverses. (correct answer)

Explanation: This question tests your understanding that two functions are inverses if and only if their compositions both equal the identity function—meaning f(g(x)) = x AND g(f(x)) = x. To verify that f and g are inverse functions, we must check BOTH compositions: f(g(x)) should equal x (showing g undoes what f does), and g(f(x)) should equal x (showing f undoes what g does). Only when both compositions simplify to the identity function x can we conclude the functions are true inverses. One direction isn't enough—we need the bidirectional undo relationship! For C(F(C)) = C((9/5)C + 32) = (5/9)((9/5)C + 32 - 32) = (5/9)(9/5)C = C ✓, and F(C(F)) = F((5/9)(F-32)) = (9/5)((5/9)(F-32)) + 32 = F - 32 + 32 = F ✓. Both compositions return the original input! Choice A correctly verifies both compositions (C(F(C)) = C and F(C(F)) = F) and confirms the inverse relationship. Choice B forgets to subtract 32 before multiplying by 5/9, Choice C only checks one direction which is insufficient, and Choice D adds 32 instead of subtracting it in C(F(C)). The verification checklist: (1) Compute f(g(x)): substitute g(x) into f, simplify completely, (2) Check: does it equal x? If no, they're not inverses—stop. If yes, continue, (3) Compute g(f(x)): substitute f(x) into g, simplify completely, (4) Check: does it equal x? If yes, they're inverses! If no, they're not (even though first direction worked). Both must equal x for full inverse verification—this is non-negotiable!

Question 11

Verify or disprove: $f(x)=\dfrac{x}{4}-3$ and $g(x)=4x-12$ are inverses. Use composition to check whether $f(g(x))=x$ and $g(f(x))=x$ .

$f(g(x))=\dfrac{4x-12}{4}-3=x-3-3=x-6\ne x$ but $g(f(x))=4\left(\dfrac{x}{4}-3\right)+12=x-12+12=x$ , so they are inverses.
$f(g(x))=\dfrac{4x-12}{4}+3=x-3+3=x$ and $g(f(x))=4\left(\dfrac{x}{4}-3\right)-12=x$ , so they are inverses.
$f(g(x))=\dfrac{4x-12}{4}-3=x$ and $g(f(x))=4\left(\dfrac{x}{4}-3\right)-12=x$ , so they are inverses.
$f(g(x))=\dfrac{4x-12}{4}-3=x-3-3=x-6\ne x$ and $g(f(x))=4\left(\dfrac{x}{4}-3\right)-12=x-12-12=x-24\ne x$ , so they are not inverses. (correct answer)

Explanation: This question tests your understanding that two functions are inverses if and only if their compositions both equal the identity function—meaning f(g(x)) = x AND g(f(x)) = x. To verify that f and g are inverse functions, we must check BOTH compositions: f(g(x)) should equal x (showing g undoes what f does), and g(f(x)) should equal x (showing f undoes what g does). Only when both compositions simplify to the identity function x can we conclude the functions are true inverses. One direction isn't enough—we need the bidirectional undo relationship! Let's compute f(g(x)) = f(4x-12) = (4x-12)/4 - 3 = x - 3 - 3 = x - 6 ≠ x ✗, and g(f(x)) = g(x/4-3) = 4(x/4-3) - 12 = x - 12 - 12 = x - 24 ≠ x ✗. Neither composition equals x! Choice A correctly shows that both f(g(x)) = x-6 and g(f(x)) = x-24, neither of which equals x, so they are not inverses. Choice B incorrectly adds 12 in g(f(x)) instead of subtracting, Choices C and D incorrectly claim one or both compositions equal x when they don't. The verification checklist: (1) Compute f(g(x)): substitute g(x) into f, simplify completely, (2) Check: does it equal x? If no, they're not inverses—stop. If yes, continue, (3) Compute g(f(x)): substitute f(x) into g, simplify completely, (4) Check: does it equal x? If yes, they're inverses! If no, they're not (even though first direction worked). Both must equal x for full inverse verification—this is non-negotiable!

Question 12

A student claims that $h(x) = \sqrt{x + 4}$ and $k(x) = x^2 - 4$ are inverse functions because $h(k(5)) = h(21) = 5$ and $k(h(5)) = k(3) = 5$ . Which statement best describes the error in this reasoning?

The student correctly verified the functions are inverses using proper composition techniques and domain considerations
The student only tested specific values rather than verifying $h(k(x)) = x$ and $k(h(x)) = x$ for all valid $x$ (correct answer)
The student made computational errors when evaluating $h(k(5))$ and $k(h(5))$ , leading to incorrect conclusions
The student failed to consider that inverse functions must have identical domains and ranges before testing compositions

Explanation: Verifying that two functions are inverses requires showing that

h(k(x)) = x

and

k(h(x)) = x

for all

x

in their respective domains, not just for specific values. Testing individual points can give coincidental results. Choice A is incorrect because the verification is incomplete. Choice C is wrong because the computations shown are correct. Choice D is incorrect because while domain considerations matter, the primary error is testing only specific values rather than general compositions.

Question 13

Consider the functions $p(x) = \frac{3x - 2}{x + 1}$ and $q(x) = \frac{-x - 2}{x - 3}$ . When computing $p(q(x))$ to verify if these are inverse functions, which expression represents the correct first step in the composition?

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

Explanation: To compute

p(q(x))

, we substitute

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

for every occurrence of

x

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

. This gives

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

. Choice B incorrectly expands before proper substitution. Choice C only substitutes in the denominator. Choice D fails to substitute in the denominator entirely.

Question 14

A student is verifying whether $m(x) = \sqrt[3]{2x - 5}$ and $n(x) = \frac{x^3 + 5}{2}$ are inverse functions. After computing $m(n(x))$ , the student gets $\sqrt[3]{2 \cdot \frac{x^3 + 5}{2} - 5} = \sqrt[3]{x^3 + 5 - 5} = \sqrt[3]{x^3} = x$ . What should the student conclude?

The functions are definitely inverses since $m(n(x)) = x$ and both functions have all real numbers as their domain
The computation contains an error because $\sqrt[3]{x^3}$ does not always equal $x$ for cube root functions
The functions are inverses, but only after verifying that $n(m(x)) = x$ and confirming the domain-range relationship (correct answer)
The functions cannot be inverses because the student only verified one composition and ignored potential domain restrictions

Explanation: When verifying inverse functions, you need to check that both compositions equal the identity function:

f(g(x)) = x

AND

g(f(x)) = x

. Additionally, the domain of each function must equal the range of the other. The student's computation of

m(n(x)) = x

is mathematically correct. Let's verify:

m(n(x)) = \sqrt[3]{2 \cdot \frac{x^3 + 5}{2} - 5} = \sqrt[3]{x^3 + 5 - 5} = \sqrt[3]{x^3} = x

. This work is valid because cube roots are defined for all real numbers, and

\sqrt[3]{x^3} = x

for all real values of

x

. However, finding

m(n(x)) = x

alone is insufficient to conclude the functions are inverses. You must also verify that

n(m(x)) = x

and confirm the domain-range relationship. Answer C correctly identifies this requirement. Answer A jumps to a conclusion without completing the verification process. While both functions do have domains of all real numbers, you still need to check the second composition. Answer B incorrectly suggests there's an error with

\sqrt[3]{x^3} = x

, but this equality holds for all real numbers (unlike even roots, which have restrictions). Answer D wrongly implies there are domain restrictions when both cube root and polynomial functions are defined for all real numbers. Strategy tip: For inverse function problems, always remember the "two-way street" rule—both compositions must equal

x

, plus domain and range must match up properly. One composition alone never proves inverse relationship.

Question 15

Functions $f$ and $g$ are defined by $f(x) = \frac{2x + 3}{x - 1}$ and $g(x) = \frac{x + 3}{x - 2}$ . To verify whether these functions are inverses of each other, a student computes $f(g(x))$ and obtains $\frac{8x + 3}{3x - 7}$ . What can be concluded about the relationship between $f$ and $g$ ?

The functions are inverses because $f(g(x))$ simplifies to a rational expression
The functions are not inverses because $f(g(x)) \neq x$ , and inverse verification requires both compositions to equal $x$ (correct answer)
The functions are inverses because the computation shows $f(g(x))$ has the same degree as the original functions
The relationship cannot be determined without also computing $g(f(x))$ and verifying it equals the given expression

Explanation: For functions to be inverses, both

f(g(x)) = x

and

g(f(x)) = x

must be true. Since

f(g(x)) = \frac{8x + 3}{3x - 7} \neq x

, the functions are not inverses. Choice A incorrectly suggests that obtaining a rational expression indicates inverse functions. Choice C incorrectly focuses on degree rather than the actual value. Choice D is wrong because finding that

f(g(x)) \neq x

is sufficient to conclude the functions are not inverses.

Question 16

Given $f(x) = \frac{x - 2}{3x + 1}$ with domain $x \neq -\frac{1}{3}$ , and $g(x) = \frac{-x - 2}{3x - 1}$ with domain $x \neq \frac{1}{3}$ , a student computes $f(g(x))$ and $g(f(x))$ and finds both equal $x$ . However, the functions are still not inverses. What is the most likely explanation?

The student made algebraic errors in computing the compositions, as rational functions cannot be inverses of each other
The compositions equal $x$ only for specific values, not for all $x$ in the appropriate domains of the functions
Rational functions require additional verification steps beyond composition to confirm they are inverses of each other
The domains and ranges of $f$ and $g$ are not compatible, preventing them from being true inverse functions (correct answer)

Explanation: When you encounter a question about inverse functions, remember that two functions are inverses if and only if their compositions equal the identity function AND their domains and ranges are properly aligned. Let's examine why these functions aren't inverses despite having compositions that equal

x

. The key issue lies in their domains and ranges. Function

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

has domain

x \neq -\frac{1}{3}

, and its range excludes

y = \frac{1}{3}

(you can verify this by solving

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

, which leads to a contradiction). Function

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

has domain

x \neq \frac{1}{3}

, and its range excludes

y = -\frac{1}{3}

. For true inverse functions, the domain of

f

must equal the range of

g

, and the range of

f

must equal the domain of

g

. Here, the domain of

f

excludes

-\frac{1}{3}

, but the range of

g

also excludes

-\frac{1}{3}

. Similarly, the range of

f

excludes

\frac{1}{3}

, but the domain of

g

also excludes

\frac{1}{3}

. This mismatch prevents them from being true inverses. Answer A is wrong because rational functions can indeed be inverses. Answer B is incorrect since the compositions do equal

x

for all appropriate values. Answer C is false because composition verification is the standard method for confirming inverses. Remember: inverse functions require both successful composition AND compatible domains/ranges. Always check that the excluded values align properly between the functions.

Question 17

Consider the functions $p(x) = \frac{x + 3}{2x - 1}$ and $q(x) = \frac{x + 3}{2x - 1}$ . A student observes that $p(x) = q(x)$ for all $x$ and concludes that they must be inverse functions since $p(q(x)) = p(p(x))$ and $p$ composed with itself might equal $x$ . Which statement best evaluates this reasoning?

The reasoning is correct because if $p(x) = q(x)$ , then verifying $p(p(x)) = x$ is sufficient to prove they are inverses
The reasoning is flawed because identical functions are never inverses of each other unless they are the identity function $f(x) = x$
The reasoning is partially correct, but the student must verify that $p(p(x)) = x$ actually holds for the given function before concluding
The reasoning is incorrect because for functions to be inverses, they must be different functions, and $p(p(x)) = x$ would make $p$ an involution, not create an inverse pair (correct answer)

Explanation: The student's reasoning confuses the concept of inverse functions with involutions. For two functions to be inverses of each other, they must generally be distinct functions where

f(g(x)) = x

and

g(f(x)) = x

. When

p(x) = q(x)

, we have the same function, and if

p(p(x)) = x

, then

p

is called an involution (a function that is its own inverse), but

p

and

q

are not inverse functions of each other—they're the same function. Choice A incorrectly accepts the flawed reasoning. Choice B is too absolute (the identity function is its own inverse). Choice C misses the conceptual error about what constitutes inverse functions.

Question 18

Two functions are defined piecewise: $f(x) = \begin{cases} 2x + 1 & \text{if } x \geq 0 \\ x - 3 & \text{if } x < 0 \end{cases}$ and $$g(x) = \begin{cases} \frac{x - 1}{2} & \text{if } x \geq 1 \ x + 3 & \text{if } x < 1 \end{cases}

Compute $f(g(x))$ first because it's easier to substitute the simpler expressions from $g$ into the linear pieces of $f$
Compute $g(f(x))$ first because the domain conditions of $f$ align better with the range restrictions needed for $g$
Either composition can be computed first, but both must be verified across all pieces of the piecewise definitions simultaneously (correct answer)
The order doesn't matter algebraically, but starting with $f(g(x))$ allows checking range-to-domain compatibility more systematically

Explanation: For piecewise functions to be inverses, both

f(g(x)) = x

and

g(f(x)) = x

must hold for all values in their respective domains. This requires checking that each piece of one function correctly maps to the appropriate piece of the other function and vice versa. The verification must be done systematically across all pieces. Choices A, B, and D suggest one composition is easier or better than the other, but both are equally necessary and must account for all piecewise conditions.

Question 19

A student claims that $h(x) = e^{x-2}$ and $j(x) = \ln(x) + 2$ are inverse functions. To support this claim, the student computes $h(j(5)) = h(\ln(5) + 2) = e^{(\ln(5) + 2) - 2} = e^{\ln(5)} = 5$ and concludes the functions are inverses. What additional verification is needed?

Verify that $j(h(x)) = x$ by computing $j(h(x)) = \ln(e^{x-2}) + 2$ and confirming this simplifies to $x$ for all valid $x$ (correct answer)
Check that the domain of $h$ matches the range of $j$ by confirming both include all positive real numbers only
Verify the computation by testing additional specific values like $h(j(1))$ and $j(h(0))$ to ensure consistency across the domains
Confirm that both functions are one-to-one by checking their derivatives are always positive across their respective domains

Explanation: To verify that functions are inverses, both

h(j(x)) = x

and

j(h(x)) = x

must be proven for all

x

in their respective domains. The student only verified one direction with a specific value. Choice A correctly identifies that

j(h(x)) = \ln(e^{x-2}) + 2 = (x-2) + 2 = x

must be verified. Choice B is incorrect about the domains (

h

has domain of all reals,

j

has domain

x > 0

). Choice C only tests specific values, which is insufficient. Choice D, while relevant to inverse existence, is not the primary verification method.

Question 20

Check whether $f(x)=\dfrac{2x+1}{3}$ and $g(x)=\dfrac{3x-1}{2}$ are inverses by computing both $f(g(x))$ and $g(f(x))$ and simplifying to see if each equals $x$ .

$f(g(x))=\dfrac{2\left(\dfrac{3x-1}{2}\right)+1}{3}=x$ and $g(f(x))=\dfrac{3\left(\dfrac{2x+1}{3}\right)-1}{2}=x$ , so they are inverses. (correct answer)
$f(g(x))=\dfrac{2\left(\dfrac{3x-1}{2}\right)+1}{3}=x$ and $g(f(x))=\dfrac{3\left(\dfrac{2x+1}{3}\right)+1}{2}=x+1$ , so they are not inverses.
$f(g(x))=\dfrac{2(3x-1)+1}{3}=2x-\dfrac{1}{3}$ and $g(f(x))=x$ , so they are not inverses.
$f(g(x))=x+\dfrac{1}{3}$ and $g(f(x))=x-\dfrac{1}{2}$ , so they are not inverses.

Explanation: This question tests your understanding that two functions are inverses if and only if their compositions both equal the identity function—meaning f(g(x)) = x AND g(f(x)) = x. To verify that f and g are inverse functions, we must check BOTH compositions: f(g(x)) should equal x (showing g undoes what f does), and g(f(x)) should equal x (showing f undoes what g does). Only when both compositions simplify to the identity function x can we conclude the functions are true inverses. One direction isn't enough—we need the bidirectional undo relationship! Computing, f(g(x)) = [2*((3x-1)/2) + 1]/3 = [ (3x-1) + 1 ]/3 = 3x/3 = x, and g(f(x)) = [3*((2x+1)/3) - 1]/2 = [ (2x+1) - 1 ]/2 = 2x/2 = x. Choice A correctly verifies both compositions equal x and determines they are inverses. Choice B errs in g(f(x)) by subtracting instead of adding in the numerator, adding an incorrect +1. The verification checklist: (1) Compute f(g(x)): substitute g(x) into f, simplify completely, (2) Check: does it equal x? If no, they're not inverses—stop. If yes, continue, (3) Compute g(f(x)): substitute f(x) into g, simplify completely, (4) Check: does it equal x? If yes, they're inverses! If no, they're not (even though first direction worked). Both must equal x for full inverse verification—this is non-negotiable!