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!

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!

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.

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!

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 $f^{-1}(x) = \dfrac{x}{2}$ (undoes multiplying by 2), but the reciprocal is $1/(2x)$ (completely different!). For $f(x)=\dfrac{x-4}{3}$, set $y=\dfrac{x-4}{3}$, swap $x=\dfrac{y-4}{3}$, multiply by 3: $3x=y-4$, add 4: $y=3x+4$. Choice C correctly finds $f^{-1}(x)=3x+4$ 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 = $ [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.

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.

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.'

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!

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.

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.