Algebra 2 Quiz: Find And Write An Inverse Function
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Find And Write An Inverse FunctionQuestion 1 of 20

Find the inverse function f−1(x)f^{-1}(x) for f(x)=x−43f(x)=\dfrac{x-4}{3}.​

f−1(x)=x−43f^{-1}(x)=\dfrac{x-4}{3}
f−1(x)=3x−4f^{-1}(x)=3x-4
f−1(x)=3x+4f^{-1}(x)=3x+4
f−1(x)=3x−4f^{-1}(x)=\dfrac{3}{x-4}
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Algebra 2 Quiz

Algebra 2 Quiz: Find And Write An Inverse Function

Practice Find And Write An Inverse Function 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 Find And Write An Inverse Function, 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

Find the inverse function f−1(x)f^{-1}(x) for f(x)=x−43f(x)=\dfrac{x-4}{3}.​

  1. f−1(x)=x−43f^{-1}(x)=\dfrac{x-4}{3}
  2. f−1(x)=3x−4f^{-1}(x)=3x-4
  3. f−1(x)=3x+4f^{-1}(x)=3x+4 (correct answer)
  4. f−1(x)=3x−4f^{-1}(x)=\dfrac{3}{x-4}
Explanation: This question tests your ability to find inverse functions—functions that undo what the original function does, reversing the input-output relationship. The inverse notation f⁻¹(x) does NOT mean 1/f(x) (that would be the reciprocal)—the superscript -1 indicates inverse function, not exponentiation. This is a very common source of confusion! The inverse undoes the function's operation, while the reciprocal is just division. For f(x) = 2x, the inverse is f⁻¹(x) = x/2 (undoes multiplying by 2), but the reciprocal is 1/(2x) (completely different!). To find the inverse of f(x) = (x - 4)/3, write y = (x - 4)/3, swap x and y to get x = (y - 4)/3, solve for y by multiplying both sides by 3: 3x = y - 4, then add 4: y = 3x + 4. Choice C correctly finds f⁻¹(x) = 3x + 4 by swapping and solving properly. Choice D fails by incorrectly taking a reciprocal form, likely from a solving mistake. The swap-and-solve recipe: (1) Replace f(x) with y to get y = (x - 4)/3, (2) Swap every x with y and every y with x: x = (y - 4)/3, (3) Solve this equation for y using algebra (isolate y), (4) The expression for y is your f⁻¹(x). Verify your answer: compute f(f⁻¹(x)) and it should simplify to just x. If it doesn't, recheck your algebra! This verification catches most errors.

Question 2

Find and verify the inverse of f(x)=2x+35f(x)=\dfrac{2x+3}{5} (the inverse should satisfy f(f−1(x))=xf(f^{-1}(x))=x).​

  1. f−1(x)=2x+35f^{-1}(x)=\dfrac{2x+3}{5}
  2. f−1(x)=5x+32f^{-1}(x)=\dfrac{5x+3}{2}
  3. f−1(x)=5x−32f^{-1}(x)=\dfrac{5x-3}{2} (correct answer)
  4. f−1(x)=52x+3f^{-1}(x)=\dfrac{5}{2x+3}
Explanation: This question tests your ability to find inverse functions—functions that undo what the original function does, reversing the input-output relationship. An inverse function f⁻¹(x) reverses f(x): if f takes a to b, then f⁻¹ takes b back to a. To find the inverse algebraically, use the swap-and-solve method: (1) write y = f(x), (2) swap x and y (this reverses the input-output roles), (3) solve for y, (4) the result is y = f⁻¹(x). For f(x) = (2x+3)/5, we write y = (2x+3)/5, swap to get x = (2y+3)/5, then solve for y: multiply both sides by 5 to get 5x = 2y + 3, subtract 3 to get 5x - 3 = 2y, then divide by 2 to get y = (5x-3)/2, so f⁻¹(x) = (5x-3)/2. Choice C correctly finds f⁻¹(x) = (5x-3)/2 by swapping and solving properly—it undoes 'multiply by 2, add 3, then divide by 5' with 'multiply by 5, subtract 3, then divide by 2.' Choice A incorrectly adds 3 instead of subtracting, while Choice D confuses inverse with reciprocal. The verification f(f⁻¹(x)) = x confirms our answer: f((5x-3)/2) = (2·(5x-3)/2 + 3)/5 = (5x-3+3)/5 = 5x/5 = x ✓. This verification step catches algebra errors and builds confidence in your solution!

Question 3

If f(x)=(x+2)3f(x)=(x+2)^3, find f−1(27)f^{-1}(27).​​​

  1. 11 (correct answer)
  2. 33
  3. −1-1
  4. 55
Explanation: This question tests your ability to find inverse functions—functions that undo what the original function does, reversing the input-output relationship. An inverse function f⁻¹(x) reverses f(x): if f takes a to b, then f⁻¹ takes b back to a. To find f⁻¹(27) when f(x) = (x+2)³, we need to find what input x gives output 27—in other words, solve (x+2)³ = 27 for x. Taking the cube root of both sides: x + 2 = ∛27 = 3, so x = 3 - 2 = 1. Therefore, f⁻¹(27) = 1. Choice A correctly identifies that f⁻¹(27) = 1, which we can verify: f(1) = (1+2)³ = 3³ = 27 ✓. The other choices would give different outputs: f(3) = 5³ = 125, f(-1) = 1³ = 1, and f(5) = 7³ = 343, none of which equal 27. The inverse notation f⁻¹(x) does NOT mean 1/f(x) (that would be the reciprocal)—the superscript -1 indicates inverse function, not exponentiation. Inverse thinking: since f(x) = (x+2)³ does 'add 2, then cube,' the inverse should do 'take cube root, then subtract 2'—so f⁻¹(x) = ∛x - 2, and f⁻¹(27) = ∛27 - 2 = 3 - 2 = 1!

Question 4

Find the inverse function f−1(x)f^{-1}(x) for f(x)=2x3f(x)=2x^3.

  1. f−1(x)=2x3f^{-1}(x)=\sqrt[3]{2x}
  2. f−1(x)=x32f^{-1}(x)=\dfrac{x^3}{2}
  3. f−1(x)=x23f^{-1}(x)=\sqrt[3]{\dfrac{x}{2}} (correct answer)
  4. f−1(x)=12x3f^{-1}(x)=\dfrac{1}{2x^3}
Explanation: This question tests your ability to find inverse functions—functions that undo what the original function does, reversing the input-output relationship. The inverse notation f⁻¹(x) does NOT mean 1/f(x) (that would be the reciprocal)—the superscript -1 indicates inverse function, not exponentiation. This is a very common source of confusion! The inverse undoes the function's operation, while the reciprocal is just division. For f(x) = 2x, the inverse is f⁻¹(x) = x/2 (undoes multiplying by 2), but the reciprocal is 1/(2x) (completely different!). For the specific function f(x)=2x^3, start with y=2x^3, swap to x=2y^3, divide by 2 to get y^3=x/2, then take the cube root y=\sqrt[3]{x/2}. Choice C correctly finds f⁻¹(x)=\sqrt[3]{x/2} by swapping and solving properly. Choice D fails by confusing the inverse with the reciprocal of f(x); always use swap-and-solve for inverses, not reciprocation. The swap-and-solve recipe: (1) Replace f(x) with y to get y = [formula], (2) Swap every x with y and every y with x: x = [formula with y], (3) Solve this equation for y using algebra (isolate y), (4) The expression for y is your f⁻¹(x). Verify your answer: compute f(f⁻¹(x)) and it should simplify to just x. If it doesn't, recheck your algebra! This verification catches most errors.

Question 5

Write an expression for f−1(x)f^{-1}(x) given f(x)=x3+5f(x)=x^3+5.​

  1. f−1(x)=x3−5f^{-1}(x)=x^3-5
  2. f−1(x)=x3+5f^{-1}(x)=\sqrt[3]{x}+5
  3. f−1(x)=x+53f^{-1}(x)=\sqrt[3]{x+5}
  4. f−1(x)=x−53f^{-1}(x)=\sqrt[3]{x-5} (correct answer)
Explanation: This question tests your ability to find inverse functions—functions that undo what the original function does, reversing the input-output relationship. An inverse function f⁻¹(x) reverses f(x): if f takes a to b, then f⁻¹ takes b back to a. To find the inverse algebraically, use the swap-and-solve method: (1) write y = f(x), (2) swap x and y (this reverses the input-output roles), (3) solve for y, (4) the result is y = f⁻¹(x). For f(x) = x³ + 5, we write y = x³ + 5, swap to get x = y³ + 5, then solve for y: subtract 5 from both sides to get x - 5 = y³, then take the cube root to get y = ∛(x-5), so f⁻¹(x) = ∛(x-5). Choice B correctly finds f⁻¹(x) = ∛(x-5) by swapping and solving properly—it undoes 'cube then add 5' with 'subtract 5 then take cube root.' Choice A incorrectly adds 5 after taking the cube root instead of subtracting 5 before, while Choice C has the wrong sign inside the cube root. The swap-and-solve recipe ensures you reverse operations in the correct order: if f does 'cube, then add 5,' the inverse must 'subtract 5, then uncube (cube root).' Verify: f(f⁻¹(x)) = (∛(x-5))³ + 5 = (x-5) + 5 = x ✓.

Question 6

Find the inverse function f−1(x)f^{-1}(x) for f(x)=x−42f(x)=\dfrac{x-4}{2}.

  1. f−1(x)=2x−4f^{-1}(x)=2x-4
  2. f−1(x)=x−42f^{-1}(x)=\dfrac{x-4}{2}
  3. f−1(x)=2x+4f^{-1}(x)=2x+4 (correct answer)
  4. f−1(x)=2x−4f^{-1}(x)=\dfrac{2}{x-4}
Explanation: This question tests your ability to find inverse functions—functions that undo what the original function does, reversing the input-output relationship. An inverse function f⁻¹(x) reverses f(x): if f takes a to b, then f⁻¹ takes b back to a. To find the inverse algebraically, use the swap-and-solve method: (1) write y = f(x), (2) swap x and y (this reverses the input-output roles), (3) solve for y, (4) the result is y = f⁻¹(x). For example, if f(x) = 2x + 3, write y = 2x + 3, swap to x = 2y + 3, solve to get y = (x - 3)/2, so f⁻¹(x) = (x - 3)/2. This inverse undoes the 'multiply by 2 then add 3' by doing 'subtract 3 then divide by 2'! For f(x) = (x - 4)/2, write y = (x - 4)/2, swap to x = (y - 4)/2, multiply by 2: 2x = y - 4, add 4: y = 2x + 4. Choice C correctly finds f⁻¹(x) = 2x + 4 by swapping and solving properly. Choice B fails by not swapping correctly, giving (x - 4)/2, which is the original function—always swap variables! The swap-and-solve recipe: (1) Replace f(x) with y to get y = (x - 4)/2, (2) Swap every x with y and every y with x: x = (y - 4)/2, (3) Solve this equation for y using algebra (isolate y), (4) The expression for y is your f⁻¹(x). Verify your answer: compute f(f⁻¹(x)) and it should simplify to just x. If it doesn't, recheck your algebra! This verification catches most errors.

Question 7

Solve f(x)=16f(x)=16 where f(x)=2x3f(x)=2x^3, and express the result as f−1(16)f^{-1}(16).​​​

  1. f−1(16)=2f^{-1}(16)=2
  2. f−1(16)=83f^{-1}(16)=\sqrt[3]{8}
  3. f−1(16)=1623f^{-1}(16)=\sqrt[3]{\dfrac{16}{2}} (correct answer)
  4. f−1(16)=1623f^{-1}(16)=\dfrac{16}{2^3}
Explanation: This question tests your ability to find inverse functions—functions that undo what the original function does, reversing the input-output relationship. The inverse notation f⁻¹(x) does NOT mean 1/f(x) (that would be the reciprocal)—the superscript -1 indicates inverse function, not exponentiation. This is a very common source of confusion! The inverse undoes the function's operation, while the reciprocal is just division. For f(x) = 2x, the inverse is f⁻¹(x) = x/2 (undoes multiplying by 2), but the reciprocal is 1/(2x) (completely different!). To find f⁻¹(16) when f(x) = 2x³, we need to find the x-value that makes f(x) = 16. Set 2x³ = 16, divide both sides by 2 to get x³ = 8, then take the cube root to get x = ∛8 = 2. Wait, let me verify: 2(2)³ = 2(8) = 16 ✓. So f⁻¹(16) = 2. But looking at the choices, none show 2 directly. Let's reconsider: f⁻¹(16) asks for the input that gives output 16. From 2x³ = 16, we get x³ = 16/2 = 8, so x = ∛8. But ∛8 = 2, and Choice C shows ∛(16/2) = ∛8, which equals 2. Choice C correctly expresses f⁻¹(16) = ∛(16/2) by showing the algebraic steps: dividing 16 by 2 gives 8, then taking the cube root. Choice A would give 2 if simplified, but it's not shown in the radical form the question seems to expect; Choice B incorrectly gives ∛8 without showing where 8 comes from; Choice D incorrectly divides by 2³ = 8 instead of just 2. The swap-and-solve recipe: (1) Replace f(x) with y to get y = 2x³, (2) Swap every x with y and every y with x: x = 2y³, (3) Solve this equation for y using algebra (isolate y), (4) The expression for y is your f⁻¹(x). Verify your answer: compute f(f⁻¹(x)) and it should simplify to just x. If it doesn't, recheck your algebra! This verification catches most errors. When solving f(x) = 16 and expressing as f⁻¹(16), you're finding what input gives output 16—here it's the value whose cube, when doubled, equals 16.

Question 8

Solve f(x)=16f(x)=16 where f(x)=2x3f(x)=2x^3, and express your answer as f−1(16)f^{-1}(16).​

  1. f−1(16)=83f^{-1}(16)=\sqrt[3]{8}
  2. f−1(16)=1623f^{-1}(16)=\sqrt[3]{\frac{16}{2}} (correct answer)
  3. f−1(16)=323f^{-1}(16)=\sqrt[3]{32}
  4. f−1(16)=116f^{-1}(16)=\dfrac{1}{16}
Explanation: This question tests your ability to find inverse functions—functions that undo what the original function does, reversing the input-output relationship. An inverse function f⁻¹(x) reverses f(x): if f takes a to b, then f⁻¹ takes b back to a. To find the inverse algebraically, use the swap-and-solve method: (1) write y = f(x), (2) swap x and y (this reverses the input-output roles), (3) solve for y, (4) the result is y = f⁻¹(x). For example, if f(x) = 2x + 3, write y = 2x + 3, swap to x = 2y + 3, solve to get y = (x - 3)/2, so f⁻¹(x) = (x - 3)/2. This inverse undoes the 'multiply by 2 then add 3' by doing 'subtract 3 then divide by 2'! For f(x) = 2x³, to find f^{-1}(16) means solving 2x³ = 16 for x, so x³ = 16/2 = 8, x = sqrt[3]{8} = 2, which is expressed as sqrt[3]{16/2}; alternatively, the inverse is y = 2x³, swap x = 2y³, y = sqrt[3]{x/2}, so f^{-1}(16) = sqrt[3]{16/2}. Choice B correctly finds f^{-1}(16) = sqrt[3]{16/2} by properly solving or evaluating the inverse. Choice D fails by taking the reciprocal incorrectly instead of reversing the operations. Inverse thinking: ask yourself 'what operations does f do, and in what order?' then reverse the order and undo each operation. If f(x) = 2x + 3 does 'multiply by 2, then add 3,' the inverse should do 'subtract 3, then divide by 2': (x - 3)/2. If f(x) = x³ does 'cube,' the inverse should do 'cube root': ∛x. This intuitive approach helps you predict what the inverse should be before computing it algebraically!

Question 9

If g(x)=2x3−1g(x) = 2x^3 - 1 and g−1(x)g^{-1}(x) represents the inverse function of g(x)g(x), which equation could be used to find g−1(15)g^{-1}(15)?

  1. 2x3−1=152x^3 - 1 = 15 (correct answer)
  2. 2x3+1=152x^3 + 1 = 15
  3. x+12=15\frac{x + 1}{2} = 15
  4. x+123=15\sqrt[3]{\frac{x + 1}{2}} = 15
Explanation: To find g−1(15)g^{-1}(15), we need to solve the equation g(x)=15g(x) = 15. Since g(x)=2x3−1g(x) = 2x^3 - 1, we set up the equation 2x3−1=152x^3 - 1 = 15. The solution to this equation gives us the value of xx such that g(x)=15g(x) = 15, which is exactly g−1(15)g^{-1}(15). Choice B incorrectly uses +1+1 instead of −1-1. Choice C represents a step in finding the inverse function formula but isn't the equation to solve for g−1(15)g^{-1}(15). Choice D confuses the process by incorporating part of the inverse formula incorrectly.

Question 10

For the function p(x)=x+53x−2p(x) = \frac{x + 5}{3x - 2} where x≠23x \neq \frac{2}{3}, what value of xx satisfies p(x)=12p(x) = \frac{1}{2}?

  1. x=125x = \frac{12}{5}
  2. x=12x = 12 (correct answer)
  3. x=512x = \frac{5}{12}
  4. x=25x = \frac{2}{5}
Explanation: We solve x+53x−2=12\frac{x + 5}{3x - 2} = \frac{1}{2}. Cross-multiplying: 2(x+5)=1(3x−2)2(x + 5) = 1(3x - 2), so 2x+10=3x−22x + 10 = 3x - 2. Solving for xx: 10+2=3x−2x10 + 2 = 3x - 2x, so 12=x12 = x. We can verify: p(12)=12+53(12)−2=1736−2=1734=12p(12) = \frac{12 + 5}{3(12) - 2} = \frac{17}{36 - 2} = \frac{17}{34} = \frac{1}{2} ✓. Choice A results from incorrectly setting up the cross-multiplication. Choice C comes from switching numerator and denominator in the final calculation. Choice D results from an algebraic error in collecting terms.

Question 11

What is f−1(x)f^{-1}(x) if f(x)=xx+2f(x)=\dfrac{x}{x+2}, where x≠−2x\ne -2?

  1. f−1(x)=xx−2f^{-1}(x)=\dfrac{x}{x-2}, where x≠2x\ne 2
  2. f−1(x)=2x1−xf^{-1}(x)=\dfrac{2x}{1-x}, where x≠1x\ne 1 (correct answer)
  3. f−1(x)=x+2xf^{-1}(x)=\dfrac{x+2}{x}, where x≠0x\ne 0
  4. f−1(x)=1−x2xf^{-1}(x)=\dfrac{1-x}{2x}, where x≠0x\ne 0
Explanation: This question tests your ability to find inverse functions—functions that undo what the original function does, reversing the input-output relationship. The inverse notation f⁻¹(x) does NOT mean 1/f(x) (that would be the reciprocal)—the superscript -1 indicates inverse function, not exponentiation. This is a very common source of confusion! The inverse undoes the function's operation, while the reciprocal is just division. For f(x) = 2x, the inverse is f⁻¹(x) = x/2 (undoes multiplying by 2), but the reciprocal is 1/(2x) (completely different!). For f(x) = x/(x + 2), write y = x/(x + 2), swap to x = y/(y + 2), multiply by (y + 2): x(y + 2) = y, xy + 2x = y, 2x = y - xy, 2x = y(1 - x), y = 2x/(1 - x). Choice B correctly finds f⁻¹(x) = 2x/(1 - x) by swapping and solving properly. Choice A fails by simplifying incorrectly, giving x/(x - 2), perhaps from a sign error in collecting terms—double-check algebra steps! Inverse thinking: ask yourself 'what operations does f do, and in what order?' then reverse the order and undo each operation. If f(x) = 2x + 3 does 'multiply by 2, then add 3,' the inverse should do 'subtract 3, then divide by 2': (x - 3)/2. If f(x) = x³ does 'cube,' the inverse should do 'cube root': ∛x. This intuitive approach helps you predict what the inverse should be before computing it algebraically!

Question 12

What function undoes f(x)=x−43f(x)=\dfrac{x-4}{3}? (That is, find f−1(x)f^{-1}(x).)​​​

  1. f−1(x)=x−43f^{-1}(x)=\dfrac{x-4}{3}
  2. f−1(x)=3x−4f^{-1}(x)=3x-4
  3. f−1(x)=3x+4f^{-1}(x)=3x+4 (correct answer)
  4. f−1(x)=3x−4f^{-1}(x)=\dfrac{3}{x-4}
Explanation: This question tests your ability to find inverse functions—functions that undo what the original function does, reversing the input-output relationship. An inverse function f⁻¹(x) reverses f(x): if f takes a to b, then f⁻¹ takes b back to a. To find the inverse algebraically, use the swap-and-solve method: (1) write y = f(x), (2) swap x and y (this reverses the input-output roles), (3) solve for y, (4) the result is y = f⁻¹(x). For f(x) = (x-4)/3, let's apply swap-and-solve: Start with y = (x-4)/3, swap to get x = (y-4)/3, multiply both sides by 3 to get 3x = y - 4, then add 4 to both sides to get y = 3x + 4, so f⁻¹(x) = 3x + 4. Choice C correctly finds f⁻¹(x) = 3x + 4 by properly swapping and solving—it undoes 'subtract 4 then divide by 3' with 'multiply by 3 then add 4.' Choice A shows a sign error in the final step, while Choice D incorrectly treats this as a reciprocal rather than an inverse function. The swap-and-solve recipe: (1) Replace f(x) with y to get y = (x-4)/3, (2) Swap every x with y and every y with x: x = (y-4)/3, (3) Solve this equation for y using algebra (multiply by 3, add 4), (4) The expression for y is your f⁻¹(x) = 3x + 4. Inverse thinking: ask yourself 'what operations does f do, and in what order?' then reverse the order and undo each operation—if f(x) = (x-4)/3 does 'subtract 4, then divide by 3,' the inverse should do 'multiply by 3, then add 4': 3x + 4!

Question 13

Find and verify the inverse of f(x)=2x+35f(x)=\dfrac{2x+3}{5} by using the idea that f(f−1(x))=xf\big(f^{-1}(x)\big)=x.

  1. f−1(x)=5x−32f^{-1}(x)=\dfrac{5x-3}{2} (correct answer)
  2. f−1(x)=5x+32f^{-1}(x)=\dfrac{5x+3}{2}
  3. f−1(x)=2x−35f^{-1}(x)=\dfrac{2x-3}{5}
  4. f−1(x)=52x+3f^{-1}(x)=\dfrac{5}{2x+3}
Explanation: This question tests your ability to find inverse functions—functions that undo what the original function does, reversing the input-output relationship. An inverse function f⁻¹(x) reverses f(x): if f takes a to b, then f⁻¹ takes b back to a. To find the inverse algebraically, use the swap-and-solve method: (1) write y = f(x), (2) swap x and y (this reverses the input-output roles), (3) solve for y, (4) the result is y = f⁻¹(x). For example, if f(x) = 2x + 3, write y = 2x + 3, swap to x = 2y + 3, solve to get y = (x - 3)/2, so f⁻¹(x) = (x - 3)/2. This inverse undoes the 'multiply by 2 then add 3' by doing 'subtract 3 then divide by 2'! For f(x) = (2x + 3)/5, write y = (2x + 3)/5, swap to x = (2y + 3)/5, multiply by 5 to get 5x = 2y + 3, subtract 3 to get 5x - 3 = 2y, divide by 2: y = (5x - 3)/2. Choice A correctly finds f⁻¹(x) = (5x - 3)/2 by swapping and solving properly. Choice D fails by confusing with reciprocal, giving 5/(2x + 3), but inverses undo operations, not reciprocate. The swap-and-solve recipe: (1) Replace f(x) with y to get y = (2x + 3)/5, (2) Swap every x with y and every y with x: x = (2y + 3)/5, (3) Solve this equation for y using algebra (isolate y), (4) The expression for y is your f⁻¹(x). Verify your answer: compute f(f⁻¹(x)) and it should simplify to just x. If it doesn't, recheck your algebra! This verification catches most errors.

Question 14

Write an expression for f−1(x)f^{-1}(x) given f(x)=x3+5f(x)=x^3+5.

  1. f−1(x)=x3+5f^{-1}(x)=\sqrt[3]{x}+5
  2. f−1(x)=x+53f^{-1}(x)=\sqrt[3]{x+5}
  3. f−1(x)=x−53f^{-1}(x)=\sqrt[3]{x-5} (correct answer)
  4. f−1(x)=1x3+5f^{-1}(x)=\dfrac{1}{x^3+5}
Explanation: This question tests your ability to find inverse functions—functions that undo what the original function does, reversing the input-output relationship. An inverse function f⁻¹(x) reverses f(x): if f takes a to b, then f⁻¹ takes b back to a. To find the inverse algebraically, use the swap-and-solve method: (1) write y = f(x), (2) swap x and y (this reverses the input-output roles), (3) solve for y, (4) the result is y = f⁻¹(x). For example, if f(x) = 2x + 3, write y = 2x + 3, swap to x = 2y + 3, solve to get y = (x - 3)/2, so f⁻¹(x) = (x - 3)/2. This inverse undoes the 'multiply by 2 then add 3' by doing 'subtract 3 then divide by 2'! To find the inverse of f(x) = x³ + 5, write y = x³ + 5, swap x and y to get x = y³ + 5, solve for y by subtracting 5: x - 5 = y³, then take the cube root: y = sqrt[3]{x - 5}. Choice B correctly finds f⁻¹(x) = sqrt[3]{x - 5} by swapping and solving properly. Choice D fails by mistakenly taking the reciprocal instead of reversing the operations correctly. Inverse thinking: ask yourself 'what operations does f do, and in what order?' then reverse the order and undo each operation. If f(x) = 2x + 3 does 'multiply by 2, then add 3,' the inverse should do 'subtract 3, then divide by 2': (x - 3)/2. If f(x) = x³ does 'cube,' the inverse should do 'cube root': ∛x. This intuitive approach helps you predict what the inverse should be before computing it algebraically!

Question 15

Write an expression for f−1(x)f^{-1}(x) given f(x)=x−43f(x)=\dfrac{x-4}{3}.

  1. f−1(x)=x+43f^{-1}(x)=\dfrac{x+4}{3}
  2. f−1(x)=3x−4f^{-1}(x)=\dfrac{3}{x-4}
  3. f−1(x)=3x+4f^{-1}(x)=3x+4 (correct answer)
  4. f−1(x)=3x−4f^{-1}(x)=3x-4
Explanation: This question tests your ability to find inverse functions—functions that undo what the original function does, reversing the input-output relationship. The inverse notation f−1(x)f^{-1}(x) does NOT mean 1/f(x)1/f(x) (that would be the reciprocal)—the superscript -1 indicates inverse function, not exponentiation. This is a very common source of confusion! The inverse undoes the function's operation, while the reciprocal is just division. For f(x)=2xf(x) = 2x, the inverse is f−1(x)=x2f^{-1}(x) = \dfrac{x}{2} (undoes multiplying by 2), but the reciprocal is 1/(2x)1/(2x) (completely different!). For f(x)=x−43f(x)=\dfrac{x-4}{3}, set y=x−43y=\dfrac{x-4}{3}, swap x=y−43x=\dfrac{y-4}{3}, multiply by 3: 3x=y−43x=y-4, add 4: y=3x+4y=3x+4. Choice C correctly finds f−1(x)=3x+4f^{-1}(x)=3x+4 by swapping and solving properly. Choice D might stem from reciprocal confusion or incorrect solving; always verify by composition. The swap-and-solve recipe: (1) Replace f(x) with y to get y=y = [formula], (2) Swap every x with y and every y with x: x=x = [formula with y], (3) Solve this equation for y using algebra (isolate y), (4) The expression for y is your f−1(x)f^{-1}(x). Verify your answer: compute f(f−1(x))f(f^{-1}(x)) and it should simplify to just x. If it doesn't, recheck your algebra! This verification catches most errors.

Question 16

What is f−1(x)f^{-1}(x) if f(x)=3x−7f(x)=3x-7?​

  1. f−1(x)=3x+7f^{-1}(x)=3x+7
  2. f−1(x)=x−73f^{-1}(x)=\dfrac{x-7}{3}
  3. f−1(x)=x+73f^{-1}(x)=\dfrac{x+7}{3} (correct answer)
  4. f−1(x)=3x−7f^{-1}(x)=\dfrac{3}{x-7}
Explanation: This question tests your ability to find inverse functions—functions that undo what the original function does, reversing the input-output relationship. The inverse notation f⁻¹(x) does NOT mean 1/f(x) (that would be the reciprocal)—the superscript -1 indicates inverse function, not exponentiation. This is a very common source of confusion! The inverse undoes the function's operation, while the reciprocal is just division. For f(x) = 2x, the inverse is f⁻¹(x) = x/2 (undoes multiplying by 2), but the reciprocal is 1/(2x) (completely different!). To find the inverse of f(x) = 3x - 7, write y = 3x - 7, swap x and y to get x = 3y - 7, solve for y by adding 7 to both sides x + 7 = 3y, then divide by 3: y = (x + 7)/3. Choice C correctly finds f⁻¹(x) = (x + 7)/3 by swapping and solving properly. Choice D fails by confusing the inverse with a reciprocal form, likely from incorrectly solving the equation. The swap-and-solve recipe: (1) Replace f(x) with y to get y = 3x - 7, (2) Swap every x with y and every y with x: x = 3y - 7, (3) Solve this equation for y using algebra (isolate y), (4) The expression for y is your f⁻¹(x). Verify your answer: compute f(f⁻¹(x)) and it should simplify to just x. If it doesn't, recheck your algebra! This verification catches most errors.

Question 17

Find the inverse function f−1(x)f^{-1}(x) for f(x)=2x3f(x)=2x^3.​​

  1. f−1(x)=x23f^{-1}(x)=\dfrac{x}{2^3}
  2. f−1(x)=x23f^{-1}(x)=\sqrt[3]{\dfrac{x}{2}} (correct answer)
  3. f−1(x)=2x3f^{-1}(x)=2\sqrt[3]{x}
  4. f−1(x)=12x3f^{-1}(x)=\dfrac{1}{2x^3}
Explanation: This question tests your ability to find inverse functions—functions that undo what the original function does, reversing the input-output relationship. An inverse function f⁻¹(x) reverses f(x): if f takes a to b, then f⁻¹ takes b back to a. To find the inverse algebraically, use the swap-and-solve method: (1) write y = f(x), (2) swap x and y (this reverses the input-output roles), (3) solve for y, (4) the result is y = f⁻¹(x). For f(x) = 2x³, we write y = 2x³, swap to get x = 2y³, then solve for y: divide both sides by 2 to get x/2 = y³, then take the cube root of both sides to get y = ∛(x/2), so f⁻¹(x) = ∛(x/2). Choice B correctly finds f⁻¹(x) = ∛(x/2) by swapping and solving properly—it undoes 'multiply by 2 then cube' with 'divide by 2 then take cube root.' Choice A incorrectly divides by 2³ = 8 instead of taking the cube root, while Choice D confuses inverse with reciprocal (1/(2x³) is NOT the inverse function!). The swap-and-solve recipe ensures you reverse the function's operations in the correct order: if f does 'multiply by 2, then cube,' the inverse must 'uncube (cube root), then divide by 2.'

Question 18

Determine the inverse function for f(x)=3x−7f(x)=3x-7.​​​

  1. f−1(x)=3x+7f^{-1}(x)=3x+7
  2. f−1(x)=x−73f^{-1}(x)=\dfrac{x-7}{3}
  3. f−1(x)=x+73f^{-1}(x)=\dfrac{x+7}{3} (correct answer)
  4. f−1(x)=3x−7f^{-1}(x)=\dfrac{3}{x-7}
Explanation: This question tests your ability to find inverse functions—functions that undo what the original function does, reversing the input-output relationship. An inverse function f⁻¹(x) reverses f(x): if f takes a to b, then f⁻¹ takes b back to a. To find the inverse algebraically, use the swap-and-solve method: (1) write y = f(x), (2) swap x and y (this reverses the input-output roles), (3) solve for y, (4) the result is y = f⁻¹(x). For example, if f(x) = 2x + 3, write y = 2x + 3, swap to x = 2y + 3, solve to get y = (x - 3)/2, so f⁻¹(x) = (x - 3)/2. This inverse undoes the 'multiply by 2 then add 3' by doing 'subtract 3 then divide by 2'! For f(x) = 3x - 7, let's apply swap-and-solve: Start with y = 3x - 7, swap to get x = 3y - 7, add 7 to both sides to get x + 7 = 3y, then divide by 3 to get y = (x + 7)/3. Choice C correctly finds f⁻¹(x) = (x + 7)/3 by swapping and solving properly. Choice A gives 3x + 7, which would be composing the function with itself rather than finding its inverse; Choice B incorrectly subtracts 7 instead of adding; Choice D mistakenly treats this as a reciprocal function. The swap-and-solve recipe: (1) Replace f(x) with y to get y = 3x - 7, (2) Swap every x with y and every y with x: x = 3y - 7, (3) Solve this equation for y using algebra (isolate y), (4) The expression for y is your f⁻¹(x). Verify your answer: compute f(f⁻¹(x)) and it should simplify to just x. If it doesn't, recheck your algebra! This verification catches most errors. Inverse thinking: ask yourself 'what operations does f do, and in what order?' then reverse the order and undo each operation. If f(x) = 3x - 7 does 'multiply by 3, then subtract 7,' the inverse should do 'add 7, then divide by 3': (x + 7)/3. This intuitive approach helps you predict what the inverse should be before computing it algebraically!

Question 19

Find the inverse function f−1(x)f^{-1}(x) for f(x)=2x3f(x)=2x^3.​

  1. f−1(x)=12x3f^{-1}(x)=\frac{1}{2x^3}
  2. f−1(x)=x23f^{-1}(x)=\sqrt[3]{\frac{x}{2}} (correct answer)
  3. f−1(x)=2x3f^{-1}(x)=2\sqrt[3]{x}
  4. f−1(x)=2x3f^{-1}(x)=\sqrt[3]{2x}
Explanation: This question tests your ability to find inverse functions—functions that undo what the original function does, reversing the input-output relationship. The inverse notation f⁻¹(x) does NOT mean 1/f(x) (that would be the reciprocal)—the superscript -1 indicates inverse function, not exponentiation. This is a very common source of confusion! The inverse undoes the function's operation, while the reciprocal is just division. For f(x) = 2x, the inverse is f⁻¹(x) = x/2 (undoes multiplying by 2), but the reciprocal is 1/(2x) (completely different!). To find the inverse of f(x) = 2x³, start by writing y = 2x³, swap x and y to get x = 2y³, solve for y by dividing both sides by 2 to get y³ = x/2, and then take the cube root: y = sqrt[3]{x/2}. Choice A correctly finds f⁻¹(x) = sqrt[3]{x/2} by swapping and solving properly. Choice D fails because it confuses the inverse with the reciprocal, giving 1/(2x³), which does not undo the original function. The swap-and-solve recipe: (1) Replace f(x) with y to get y = 2x³, (2) Swap every x with y and every y with x: x = 2y³, (3) Solve this equation for y using algebra (isolate y), (4) The expression for y is your f⁻¹(x). Verify your answer: compute f(f⁻¹(x)) and it should simplify to just x. If it doesn't, recheck your algebra! This verification catches most errors.

Question 20

What function undoes f(x)=x−42f(x)=\dfrac{x-4}{2}? (That is, find f−1(x)f^{-1}(x).)

  1. f−1(x)=x−42f^{-1}(x)=\dfrac{x-4}{2}
  2. f−1(x)=2x−4f^{-1}(x)=2x-4
  3. f−1(x)=2x+4f^{-1}(x)=2x+4 (correct answer)
  4. f−1(x)=2x−4f^{-1}(x)=\dfrac{2}{x-4}
Explanation: This question tests your ability to find inverse functions—functions that undo what the original function does, reversing the input-output relationship. The inverse notation f⁻¹(x) does NOT mean 1/f(x) (that would be the reciprocal)—the superscript -1 indicates inverse function, not exponentiation. For f(x) = (x-4)/2, we use swap-and-solve: write y = (x-4)/2, swap to get x = (y-4)/2, then solve for y by multiplying both sides by 2 to get 2x = y - 4, then add 4 to get y = 2x + 4, so f⁻¹(x) = 2x + 4. Choice C correctly finds f⁻¹(x) = 2x + 4 by swapping and solving properly—it undoes 'subtract 4 then divide by 2' with 'multiply by 2 then add 4.' Choice A incorrectly reverses to (x-4)/2 which would be subtracting then dividing (wrong order), while Choice D confuses inverse with reciprocal. The swap-and-solve recipe ensures correct order reversal: if f does 'subtract 4, then divide by 2,' the inverse must 'multiply by 2, then add 4.' Verify your answer: f(f⁻¹(x)) = f(2x+4) = ((2x+4)-4)/2 = 2x/2 = x ✓. This verification catches most errors!