Question

what is the inverse of $f(x) = \sqrt3{x + 4}$? (1 point)
\\( \circ \\ f^{-1}(x) = x^3 - 4 \\)
\\( \circ \\ f^{-1}(x) = x^3 + 4 \\)
\\( \circ \\ f^{-1}(x) = (x - 4)^3 \\)
\\( \circ \\ f^{-1}(x) = (x + 4)^3 \\)

Explanation:

Step1: Replace \( f(x) \) with \( y \)

We start by writing the function as \( y = \sqrt[3]{x + 4} \).

Step2: Swap \( x \) and \( y \)

To find the inverse, we swap the roles of \( x \) and \( y \), so we get \( x = \sqrt[3]{y + 4} \).

Step3: Solve for \( y \)

First, we cube both sides of the equation to eliminate the cube root. Cubing the left side gives \( x^3 \), and cubing the right side gives \( y + 4 \) (since \( (\sqrt[3]{a})^3=a \)). So we have \( x^3 = y + 4 \). Then, we subtract 4 from both sides to solve for \( y \): \( y = x^3 - 4 \).

Step4: Replace \( y \) with \( f^{-1}(x) \)

We replace \( y \) with the inverse function notation, so \( f^{-1}(x)=x^3 - 4 \).

Answer:

\( f^{-1}(x)=x^3 - 4 \) (the first option: \( f^{-1}(x)=x^3 - 4 \))