Question

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

Explanation:

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

Given \( f(x)=x^{\frac{1}{3}} - 7 \), so \( y=x^{\frac{1}{3}} - 7 \).

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

Add 7 to both sides: \( y + 7=x^{\frac{1}{3}} \).
Cube both sides: \( (y + 7)^{3}=x \).

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

The inverse function \( f^{-1}(x)=(x + 7)^{3} \).

Answer:

\( f^{-1}(x)=(x + 7)^{3} \) (the option with this expression, e.g., if it's the fourth option as per the image, identify it by its position or label; assuming the option is \( f^{-1}(x)=(x + 7)^{3} \) among the choices, that's the correct one)