Question

for the function $f(x) = x^{\frac{1}{3}} - 7$, find $f^{-1}(x)$.
answer

• $f^{-1}(x) = x^3 + 7$
• $f^{-1}(x) = (x + 7)^3$
• $f^{-1}(x) = (x - 7)^3$
• $f^{-1}(x) = x^{\frac{1}{3}} + 7$

Explanation:

Step1: Let \( y = f(x) \)

Set \( y = x^{\frac{1}{3}} - 7 \).

Step2: Solve for \( x \) in terms of \( y \)

First, add 7 to both sides: \( y + 7 = x^{\frac{1}{3}} \).
Then, cube both sides to eliminate the cube root: \( (y + 7)^3 = x \).

Step3: Swap \( x \) and \( y \)

Replace \( x \) with \( y \) and \( y \) with \( x \) to get the inverse function: \( f^{-1}(x) = (x + 7)^3 \).

Answer:

\( f^{-1}(x) = (x + 7)^3 \) (the second option)