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 $f^{-1}$(27), solve $(x+2)^3$=27, take cube root x+2=3, then x=1. Choice B correctly finds $f^{-1}$(27)=1 by solving properly. Choice A fails by forgetting to subtract the 2 after taking the cube root; always reverse all operations in the function. Inverse thinking: ask yourself 'what operations does f do, and in what order?' then reverse the order and undo each operation. If f(x) = $(x+2)^3$ does 'add 2, then cube,' the inverse should do 'cube root, then subtract 2' to get back to x.

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 $f(x)=x^3$+5, set $y=x^3$+5, swap $x=y^3$+5, subtract 5: $x-5=y^3$, take 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 confuses inverse with reciprocal; remember to isolate the variable correctly after swapping. Inverse thinking: ask yourself 'what operations does f do, and in what order?' then reverse the order and undo each operation. If f(x) = $x^3$ + 5 does 'cube, then add 5,' the inverse should do 'subtract 5, then cube root': \sqrt[3]{x-5}. 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 $f^{-1}(16)$, solve $2x^3=16$, divide by 2: $x^3=8$, take cube root $x=2$, or equivalently $\sqrt[3]{16/2}=\sqrt[3]{8}=2$. Choice B correctly finds $f^{-1}(16)=\sqrt[3]{16/2}=2$ by solving properly. Choice C reverses the fraction inside the root, a common solving mistake; remember to divide by 2 before rooting. 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 'cube, then multiply by 2,' the inverse should do 'divide by 2, then cube root': $\sqrt[3]{x/2}$. 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!). For f(x)=(2x+3)/5, set y=(2x+3)/5, swap x=(2y+3)/5, multiply by 5: 5x=2y+3, subtract 3: 5x-3=2y, divide by 2: y=(5x-3)/2. Choice B correctly finds f⁻¹(x)=(5x-3)/2 by swapping and solving properly. Choice D confuses inverse with reciprocal; use swap-and-solve and verify by composing $f(f^{-1}$(x))=x. 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!). 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.

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 $f(x)=\dfrac{x+1}{x-1}$, set $y=\dfrac{x+1}{x-1}$, swap to $x=\dfrac{y+1}{y-1}$, then solve: $x(y-1)=y+1$, $xy-x=y+1$, $xy-y=x+1$, $y(x-1)=x+1$, $y=\dfrac{x+1}{x-1}$. Choice B correctly finds $f^{-1}(x)=\dfrac{x+1}{x-1}$ (same as $f$) by swapping and solving properly. Choice A might result from incorrectly swapping signs during algebra; double-check by verifying $f(f^{-1}(x))=x$ to catch such errors. Inverse thinking: ask yourself 'what operations does $f$ do, and in what order?' then reverse the order and undo each operation. If $f(x) = 2x + 3$ does 'multiply by 2, then add 3,' the inverse should do 'subtract 3, then divide by 2': $(x - 3)/2$. 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 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!

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 $f(x) = \sqrt[3]{x - 8}$, write $y = \sqrt[3]{x - 8}$, swap x and y to get $x = \sqrt[3]{y - 8}$, solve for y by cubing both sides: $x^3 = y - 8$, then add 8: $y = x^3 + 8$. Choice B correctly finds $f^{-1}(x) = x^3 + 8$ by swapping and solving properly. Choice D fails by incorrectly incorporating a reciprocal and mishandling the cube root. The swap-and-solve recipe: (1) Replace f(x) with y to get $y = \sqrt[3]{x - 8}$, (2) Swap every x with y and every y with x: $x = \sqrt[3]{y - 8}$, (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.