Question

cubic & cube root functions as inverses quick check
the function $f(x) = x^3 + 4$ is a translation of the parent cubic function. what is the inverse? (1 point)
$\circ$ $f^{-1}(x) = \sqrt3{x} - 4$
$\circ$ $f^{-1}(x) = \sqrt3{x - 4}$
$\circ$ $f^{-1}(x) = \sqrt3{x + 4}$
$\circ$ $f^{-1}(x) = \sqrt3{x} + 4$

Explanation:

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

We start with the function \( f(x) = x^3 + 4 \). Replace \( f(x) \) with \( y \), so we have \( y = x^3 + 4 \).

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

To find the inverse, we swap the roles of \( x \) and \( y \). This gives us \( x = y^3 + 4 \).

Step3: Solve for \( y \)

Subtract 4 from both sides of the equation: \( x - 4 = y^3 \). Then, take the cube root of both sides to solve for \( y \). The cube root of \( y^3 \) is \( y \), and the cube root of \( x - 4 \) is \( \sqrt[3]{x - 4} \). So, \( y = \sqrt[3]{x - 4} \).

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

Now, replace \( y \) with the inverse function notation \( f^{-1}(x) \). We get \( f^{-1}(x) = \sqrt[3]{x - 4} \).

Answer:

\( f^{-1}(x) = \sqrt[3]{x - 4} \) (the second option: \( f^{-1}(x)=\sqrt[3]{x - 4} \))