Question

what is the inverse of $f(x) = 2x^3 - 7$? (1 point)
$\circ$ $f^{-1}(x) = \sqrt3{\frac{x + 7}{2}}$
$\circ$ $f^{-1}(x) = \sqrt3{2x + 7}$
$\circ$ $f^{-1}(x) = \sqrt3{2x - 7}$
$\circ$ $f^{-1}(x) = \sqrt3{\frac{x - 7}{2}}$

Explanation:

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

We start with the function \( f(x) = 2x^3 - 7 \). Replace \( f(x) \) with \( y \), so we have \( y = 2x^3 - 7 \).

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

To find the inverse, we swap the roles of \( x \) and \( y \). This gives us \( x = 2y^3 - 7 \).

Step3: Solve for \( y \)

First, add 7 to both sides of the equation: \( x + 7 = 2y^3 \). Then, divide both sides by 2: \( \frac{x + 7}{2} = y^3 \). Finally, take the cube root of both sides to solve for \( y \): \( y = \sqrt[3]{\frac{x + 7}{2}} \).

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

Now, replace \( y \) with \( f^{-1}(x) \) to get the inverse function: \( f^{-1}(x) = \sqrt[3]{\frac{x + 7}{2}} \).

Answer:

\( f^{-1}(x) = \sqrt[3]{\frac{x + 7}{2}} \) (corresponding to the first option: \( f^{-1}(x)=\sqrt[3]{\frac{x + 7}{2}} \))