Find and Write an Inverse Function - Algebra 2

Find the inverse function $f^{-1}(x)$ for $f(x)=2x^3$ .
Answer: $f^{-1}(x)=\sqrt[3]{\dfrac{x}{2}}$ 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 f(x) = 2x³, let's apply this method: Start with y = 2x³, swap to get x = 2y³, then solve for y by dividing both sides by 2 to get x/2 = y³, and finally 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 properly swapping variables and solving—it undoes 'multiply by 2 then cube' with 'divide by 2 then take cube root.' Choice A shows the common error of thinking f⁻¹(x) means 1/f(x), which would be the reciprocal function, not the inverse—remember that the -1 superscript indicates inverse, not exponentiation! 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 by dividing by 2 then taking cube root), (4) The expression for y is your f⁻¹(x) = ∛(x/2). Verify your answer: compute f(f⁻¹(x)) = 2(∛(x/2))³ = 2(x/2) = x ✓
Solve $f(x)=8$ where $f(x)=2x^3$ , and express your solution as $f^{-1}(8)$ .
Answer: $f^{-1}(8)=\sqrt[3]{4}$ 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. To solve f(x) = 8 where f(x) = 2x³, we set 2x³ = 8 and solve for x: divide both sides by 2 to get x³ = 4, then take the cube root to get x = ∛4. Since f⁻¹(8) represents the input that gives output 8, we have f⁻¹(8) = ∛4. Choice D correctly identifies f⁻¹(8) = ∛4 by solving the equation 2x³ = 8 properly. We can verify: f(∛4) = 2(∛4)³ = 2(4) = 8 ✓. Choice B incorrectly gets x = 2 (which would give f(2) = 2(2³) = 16, not 8), while Choice A gets x = 4 (which would give f(4) = 2(64) = 128). 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 (divide by 2, take cube root), (4) The expression for y is your f⁻¹(x) = ∛(x/2). Inverse thinking: since f multiplies the cube by 2, f⁻¹ should divide by 2 then take the cube root—so f⁻¹(8) = ∛(8/2) = ∛4!
Write an expression for $f^{-1}(x)$ given $f(x)=3x-7$ .
Answer: $f^{-1}(x)=\dfrac{x+7}{3}$ 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.
What is $f^{-1}(x)$ if $f(x)=\dfrac{x+1}{x-1}$ with domain restriction $x\ne 1$ ?
Answer: $f^{-1}(x)=\dfrac{x+1}{x-1}$ with $x\ne 1$ 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 the rational function f(x) = (x+1)/(x-1), we use swap-and-solve: write y = (x+1)/(x-1), swap to get x = (y+1)/(y-1), then solve for y by cross-multiplying: x(y-1) = y+1, which gives xy - x = y + 1, then xy - y = x + 1, factor out y to get y(x-1) = x+1, so y = (x+1)/(x-1). Choice A correctly finds f⁻¹(x) = (x+1)/(x-1) with x ≠ 1 (since division by zero is undefined when x = 1). Remarkably, this function is its own inverse—applying it twice returns the original input! Choice B incorrectly flips the signs in the numerator and denominator, while Choice C has the wrong denominator. Inverse thinking: when a function is its own inverse (called an involution), it means f(f(x)) = x for all x in the domain—verify this by substituting f(x) back into itself!
If $f(x)=(x+2)^3$ , solve $f(x)=27$ and express the solution as $f^{-1}(27)$ .
Answer: $f^{-1}(27)=1$ 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.
If $f(x)=(x+2)^3$ , find $f^{-1}(27)$ .
Answer: $1$ 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!
For the function $p(x) = \frac{x + 5}{3x - 2}$ where $x \neq \frac{2}{3}$ , what value of $x$ satisfies $p(x) = \frac{1}{2}$ ?
Answer: $x = 12$ We solve $\frac{x + 5}{3x - 2} = \frac{1}{2}$ . Cross-multiplying: $2(x + 5) = 1(3x - 2)$ , so $2x + 10 = 3x - 2$ . Solving for $x$ : $10 + 2 = 3x - 2x$ , so $12 = x$ . We can verify: $p(12) = \frac{12 + 5}{3(12) - 2} = \frac{17}{36 - 2} = \frac{17}{34} = \frac{1}{2}$ ✓. 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.
Determine the inverse function for $f(x)=3x-7$ .
Answer: $f^{-1}(x)=\dfrac{x+7}{3}$ 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!
Find the inverse function $f^{-1}(x)$ for $f(x)=2x^3$ .
Answer: $f^{-1}(x)=\sqrt[3]{\dfrac{x}{2}}$ 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.
What is $f^{-1}(x)$ if $f(x)=x^3+5$ ?
Answer: $f^{-1}(x)=\sqrt[3]{x-5}$ 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, let's apply this method: Start with y = x³ + 5, swap variables to get x = y³ + 5, then solve for y by subtracting 5 from both sides to get x - 5 = y³, and 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 properly swapping and solving—it undoes 'cube then add 5' with 'subtract 5 then take cube root.' Choice C shows the common error of thinking f⁻¹(x) means 1/f(x), which would be the reciprocal function, not the inverse—remember that the -1 superscript indicates inverse, not exponentiation! The swap-and-solve recipe: (1) Replace f(x) with y to get y = x³ + 5, (2) Swap every x with y and every y with x: x = y³ + 5, (3) Solve this equation for y using algebra (subtract 5, then take cube root), (4) The expression for y is your f⁻¹(x) = ∛(x-5). Inverse thinking: ask yourself 'what operations does f do, and in what order?' then reverse the order and undo each operation—if f(x) = x³ + 5 does 'cube, then add 5,' the inverse should do 'subtract 5, then take cube root': ∛(x-5)!
Determine the inverse function for $f(x)=3x-7$ .
Answer: $f^{-1}(x)=\dfrac{x+7}{3}$ 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!
Write an expression for $f^{-1}(x)$ given $f(x)=x^3+5$ .
Answer: $f^{-1}(x)=\sqrt[3]{x-5}$ 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 ✓.
What function undoes $f(x)=\dfrac{x-4}{3}$ ? (That is, find $f^{-1}(x)$ .)
Answer: $f^{-1}(x)=3x+4$ 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 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!
If $g(x) = 2x^3 - 1$ and $g^{-1}(x)$ represents the inverse function of $g(x)$ , which equation could be used to find $g^{-1}(15)$ ?
Answer: $2x^3 - 1 = 15$ To find $g^{-1}(15)$ , we need to solve the equation $g(x) = 15$ . Since $g(x) = 2x^3 - 1$ , we set up the equation $2x^3 - 1 = 15$ . The solution to this equation gives us the value of $x$ such that $g(x) = 15$ , which is exactly $g^{-1}(15)$ . Choice B incorrectly uses $+1$ instead of $-1$ . Choice C represents a step in finding the inverse function formula but isn't the equation to solve for $g^{-1}(15)$ . Choice D confuses the process by incorporating part of the inverse formula incorrectly.
The function $h(x) = \frac{4x - 3}{x + 2}$ has domain $x \neq -2$ . If $h^{-1}(x)$ exists, what is the expression for $h^{-1}(x)$ ?
Answer: $h^{-1}(x) = \frac{2x + 3}{4 - x}$ where $x \neq 4$ To find the inverse, we start with $y = \frac{4x - 3}{x + 2}$ and solve for $x$ . Cross-multiplying: $y(x + 2) = 4x - 3$ , so $yx + 2y = 4x - 3$ . Collecting $x$ terms: $yx - 4x = -3 - 2y$ , so $x(y - 4) = -3 - 2y$ . Therefore, $x = \frac{-3 - 2y}{y - 4} = \frac{-(3 + 2y)}{y - 4} = \frac{3 + 2y}{4 - y}$ . Switching $x$ and $y$ : $h^{-1}(x) = \frac{3 + 2x}{4 - x} = \frac{2x + 3}{4 - x}$ . The domain restriction $x \neq 4$ comes from the denominator. Choice B has the wrong sign in the denominator. Choice C incorrectly swaps numerator and denominator relationships. Choice D has the correct form but wrong coefficient arrangement.
Find the inverse function $f^{-1}(x)$ for $f(x)=\dfrac{4}{x+1}$ with domain restriction $x\ne -1$ .
Answer: $f^{-1}(x)=\dfrac{4}{x}-1$ with $x\ne 0$ 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) = 4/(x+1), we write y = 4/(x+1), swap to get x = 4/(y+1), then solve for y: multiply both sides by (y+1) to get x(y+1) = 4, which gives xy + x = 4, then xy = 4 - x, so y = (4-x)/x = 4/x - 1. Therefore f⁻¹(x) = 4/x - 1. Choice B correctly finds f⁻¹(x) = 4/x - 1 with x ≠ 0 by swapping and solving properly. Choice A incorrectly gets 4/(x+1) which would make it its own inverse (not true here), while Choice D confuses inverse with reciprocal. The domain restriction x ≠ 0 comes from division by x in the inverse formula. Verify: f(f⁻¹(x)) = f(4/x - 1) = 4/((4/x - 1) + 1) = 4/(4/x) = x ✓.
Write an expression for $f^{-1}(x)$ given $f(x)=3x-7$ .
Answer: $f^{-1}(x)=\dfrac{x+7}{3}$ 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 the linear function f(x) = 3x - 7, let's apply swap-and-solve: Start with y = 3x - 7, swap variables to get x = 3y - 7, then solve for y by adding 7 to both sides to get x + 7 = 3y, and finally divide by 3 to get y = (x + 7)/3, so f⁻¹(x) = (x + 7)/3. Choice B correctly finds f⁻¹(x) = (x + 7)/3 by properly swapping and solving—it undoes 'multiply by 3 then subtract 7' with 'add 7 then divide by 3.' Choice C shows a sign error (using x - 7 instead of x + 7), while Choice D incorrectly treats this as a reciprocal function rather than finding the true inverse. 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 (add 7, then divide by 3), (4) The expression for y is your f⁻¹(x) = (x + 7)/3. 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!
Given $f(x) = (x - 1)^3 + 4$ , if $f(a) = 12$ , then what is the value of $a$ ?
Answer: $a = 3$ To find $a$ , we solve $f(a) = 12$ . Setting up the equation: $(a - 1)^3 + 4 = 12$ . Subtracting 4: $(a - 1)^3 = 8$ . Taking the cube root: $a - 1 = \sqrt[3]{8} = 2$ . Therefore, $a = 2 + 1 = 3$ . We can verify: $f(3) = (3-1)^3 + 4 = 2^3 + 4 = 8 + 4 = 12$ ✓. Choice B gives $a = 2$ , which would yield $f(2) = 1^3 + 4 = 5$ . Choice C gives $a = 1$ , yielding $f(1) = 0^3 + 4 = 4$ . Choice D gives $a = 4$ , yielding $f(4) = 3^3 + 4 = 31$ .
Solve $f(x)=16$ where $f(x)=2x^3$ , and express your answer as $f^{-1}(16)$ .
Answer: $f^{-1}(16)=\sqrt[3]{\frac{16}{2}}$ 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) = 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!
If $r(x) = \sqrt[3]{2x + 7}$ and $r^{-1}(x)$ is the inverse function, which expression represents $r^{-1}(x)$ ?
Answer: $r^{-1}(x) = \frac{x^3 - 7}{2}$ for all real $x$ 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 $r^{-1}(x)$ when $r(x) = \sqrt[3]{2x + 7}$ , start by setting $y = \sqrt[3]{2x + 7}$ . Now solve for $x$ in terms of $y$ . First, cube both sides: $y^3 = 2x + 7$ . Then isolate $x$ : subtract 7 from both sides to get $y^3 - 7 = 2x$ , then divide by 2 to get $x = \frac{y^3 - 7}{2}$ . Finally, swap variables to express the inverse function: $r^{-1}(x) = \frac{x^3 - 7}{2}$ . Choice A incorrectly writes $2x^3 - 7$ in the numerator instead of $x^3 - 7$ , suggesting the student multiplied $x^3$ by 2 instead of dividing the entire expression by 2. Choice B gives $\frac{x^3 + 7}{2}$ , 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 $\frac{(x - 7)^3}{2}$ , which suggests the student subtracted 7 before cubing instead of cubing first then subtracting 7. The correct answer is D: $r^{-1}(x) = \frac{x^3 - 7}{2}$ . Study tip: Always verify your inverse by checking that $r(r^{-1}(x)) = x$ . This catch-all method will reveal any algebraic errors in your work.

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