Algebra 2 Quiz: Find And Write An Inverse Function
20 questions · exam conditions
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Find And Write An Inverse FunctionQuestion 1 of 20

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

f−1(x)=xx−2f^{-1}(x)=\dfrac{x}{x-2}, where x≠2x\ne 2
f−1(x)=2x1−xf^{-1}(x)=\dfrac{2x}{1-x}, where x≠1x\ne 1
f−1(x)=x+2xf^{-1}(x)=\dfrac{x+2}{x}, where x≠0x\ne 0
f−1(x)=1−x2xf^{-1}(x)=\dfrac{1-x}{2x}, where x≠0x\ne 0
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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

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 2

Write an expression for f−1(x)f^{-1}(x) given f(x)=3x−7f(x)=3x-7.

  1. f−1(x)=x+73f^{-1}(x)=\dfrac{x+7}{3} (correct answer)
  2. f−1(x)=3x+7f^{-1}(x)=3x+7
  3. f−1(x)=x−73f^{-1}(x)=\dfrac{x-7}{3}
  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!). For f(x)=3x-7, set y=3x-7, swap to x=3y-7, add 7 to get x+7=3y, divide by 3: y=(x+7)/3. Choice A correctly finds f⁻¹(x)=(x+7)/3 by swapping and solving properly. Choice D confuses inverse with reciprocal; remember, reciprocal would be 1/(3x-7), but inverse requires swap-and-solve. 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 3

If r(x)=2x+73r(x) = \sqrt[3]{2x + 7} and r−1(x)r^{-1}(x) is the inverse function, which expression represents r−1(x)r^{-1}(x)?

  1. r−1(x)=2x3−71r^{-1}(x) = \frac{2x^3 - 7}{1} for all real xx
  2. r−1(x)=x3+72r^{-1}(x) = \frac{x^3 + 7}{2} for all real xx
  3. r−1(x)=(x−7)32r^{-1}(x) = \frac{(x - 7)^3}{2} for all real xx
  4. r−1(x)=x3−72r^{-1}(x) = \frac{x^3 - 7}{2} for all real xx (correct answer)
Explanation: When you encounter inverse function problems, remember that finding the inverse means "undoing" the original function's operations in reverse order. You're essentially solving for the input when given the output. To find r−1(x)r^{-1}(x) when r(x)=2x+73r(x) = \sqrt[3]{2x + 7}, start by setting y=2x+73y = \sqrt[3]{2x + 7}. Now solve for xx in terms of yy. First, cube both sides: y3=2x+7y^3 = 2x + 7. Then isolate xx: subtract 7 from both sides to get y3−7=2xy^3 - 7 = 2x, then divide by 2 to get x=y3−72x = \frac{y^3 - 7}{2}. Finally, swap variables to express the inverse function: r−1(x)=x3−72r^{-1}(x) = \frac{x^3 - 7}{2}. Choice A incorrectly writes 2x3−72x^3 - 7 in the numerator instead of x3−7x^3 - 7, suggesting the student multiplied x3x^3 by 2 instead of dividing the entire expression by 2. Choice B gives x3+72\frac{x^3 + 7}{2}, which represents adding 7 instead of subtracting it—this comes from incorrectly handling the sign when moving 7 to the other side of the equation. Choice C shows (x−7)32\frac{(x - 7)^3}{2}, which suggests the student subtracted 7 before cubing instead of cubing first then subtracting 7. The correct answer is D: r−1(x)=x3−72r^{-1}(x) = \frac{x^3 - 7}{2}. Study tip: Always verify your inverse by checking that r(r−1(x))=xr(r^{-1}(x)) = x. This catch-all method will reveal any algebraic errors in your work.

Question 4

Solve f(x)=27f(x)=27 where f(x)=(x+2)3f(x)=(x+2)^3, and express your solution as f−1(27)f^{-1}(27).​​​

  1. f−1(27)=3f^{-1}(27)=3
  2. f−1(27)=1f^{-1}(27)=1 (correct answer)
  3. f−1(27)=253f^{-1}(27)=\sqrt[3]{25}
  4. f−1(27)=5f^{-1}(27)=5
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(27) when f(x) = (x + 2)³, we need to find the x-value that makes f(x) = 27. Set (x + 2)³ = 27, take the cube root of both sides to get x + 2 = 3, then subtract 2 to get x = 1. Therefore, f⁻¹(27) = 1. Choice B correctly identifies f⁻¹(27) = 1 by solving (x + 2)³ = 27 to find x = 1. Choice A incorrectly gives 3, which is the cube root of 27 but doesn't account for the +2 in the function; Choice C gives an unnecessarily complex expression; Choice D gives 5, which seems to come from nowhere. The swap-and-solve recipe: (1) Replace f(x) with y to get y = (x + 2)³, (2) Swap every x with y and every y with x: x = (y + 2)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. When asked for f⁻¹(27), you're finding the input that gives output 27—it's like asking 'what number, when you add 2 and cube it, gives 27?' The answer is 1, because (1 + 2)³ = 3³ = 27.

Question 5

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 6

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 f(x) = 3x - 7, we write y = 3x - 7, swap to get x = 3y - 7, then solve for y: add 7 to both sides to get x + 7 = 3y, then divide by 3 to get y = (x+7)/3, so f⁻¹(x) = (x+7)/3. Choice C correctly finds f⁻¹(x) = (x+7)/3 by swapping and solving properly—it undoes 'multiply by 3 then subtract 7' with 'add 7 then divide by 3.' Choice B incorrectly subtracts 7 instead of adding 7, while Choice D confuses inverse with reciprocal (3/(x-7) is NOT the inverse!). The swap-and-solve recipe ensures you reverse operations in the correct order: if f does 'multiply by 3, then subtract 7,' the inverse must 'add 7, then divide by 3.' Inverse thinking helps predict the answer before computing!

Question 7

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 8

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 9

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 10

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 11

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 12

Find the inverse function f−1(x)f^{-1}(x) for f(x)=4x+1f(x)=\dfrac{4}{x+1} with domain restriction x≠−1x \ne -1.

  1. f−1(x)=14x+1f^{-1}(x)=\dfrac{1}{4x+1}
  2. f−1(x)=4x+1f^{-1}(x)=\dfrac{4}{x+1} with x≠−1x \ne -1
  3. f−1(x)=4x−1f^{-1}(x)=\dfrac{4}{x}-1 with x≠0x \ne 0 (correct answer)
  4. f−1(x)=x+14f^{-1}(x)=\dfrac{x+1}{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)f^{-1}(x) reverses f(x)f(x): if ff takes a to b, then f−1f^{-1} takes b back to a. To find the inverse algebraically, use the swap-and-solve method: (1) write y=f(x)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)y = f^{-1}(x). For f(x)=4x+1f(x) = \dfrac{4}{x+1}, we write y=4x+1y = \dfrac{4}{x+1}, swap to get x=4y+1x = \dfrac{4}{y+1}, then solve for y: multiply both sides by (y+1)(y+1) to get x(y+1)=4x(y+1) = 4, which gives xy+x=4xy + x = 4, then xy=4−xxy = 4 - x, so y=4−xx=4x−1y = \dfrac{4 - x}{x} = \dfrac{4}{x} - 1. Therefore f−1(x)=4x−1f^{-1}(x) = \dfrac{4}{x} - 1. Choice B correctly finds f−1(x)=4x−1f^{-1}(x) = \dfrac{4}{x} - 1 with x≠0x \ne 0 by swapping and solving properly. Choice A incorrectly gets 4/(x+1)4/(x+1) which would make it its own inverse (not true here), while Choice D confuses inverse with reciprocal. The domain restriction x≠0x \ne 0 comes from division by x in the inverse formula. Verify: f(f−1(x))=f(4x−1)=4(4x−1)+1=44x=xf(f^{-1}(x)) = f(\dfrac{4}{x} - 1) = \dfrac{4}{(\dfrac{4}{x} - 1) + 1} = \dfrac{4}{\dfrac{4}{x}} = x ✓.

Question 13

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} (correct answer)
  3. f−1(x)=3x−7f^{-1}(x)=\dfrac{3}{x-7}
  4. f−1(x)=x−73f^{-1}(x)=\dfrac{x-7}{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 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 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)=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 15

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 16

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)=2x+4f^{-1}(x)=2x+4 (correct answer)
  2. f−1(x)=2x−4f^{-1}(x)=2x-4
  3. f−1(x)=2x−4f^{-1}(x)=\dfrac{2}{x-4}
  4. f−1(x)=x−42f^{-1}(x)=\dfrac{x-4}{2}
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!

Question 17

Find and verify the inverse of f(x)=2x+35f(x)=\dfrac{2x+3}{5} by choosing the correct f−1(x)f^{-1}(x) (so that 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, let's apply swap-and-solve: Start with y = (2x+3)/5, swap to get x = (2y+3)/5, multiply both sides by 5 to get 5x = 2y + 3, subtract 3 from both sides to get 5x - 3 = 2y, and 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 properly swapping and solving—it undoes 'multiply by 2, add 3, then divide by 5' with 'multiply by 5, subtract 3, then divide by 2.' We can verify: f(f⁻¹(x)) = f((5x-3)/2) = (2·(5x-3)/2 + 3)/5 = ((5x-3) + 3)/5 = 5x/5 = x ✓. 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 (multiply by 5, subtract 3, divide by 2), (4) The expression for y is your f⁻¹(x). Inverse thinking: the function does three operations in order (×2, +3, ÷5), so the inverse reverses these (×5, -3, ÷2)!

Question 18

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 19

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!

Question 20

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!