Question

For the function f(x) = (25 – 10), find f-'(x). of-1(x) = (V)+10 3 of-'(x) = V(x + 10) of-1(x) = x3 + 10 f-1(x) = x3 + 10

Answer 1

Answer::

$f^{-1}(x)=\sqrt[5]{x^3+10}$

Explanation:

Given f(x) defined below:

$f(x)=(x^5-10)^{\frac{1}{3}}$

To solve for the inverse, follow the steps below:

Step 1: Rewrite the equation using y.

$y=(x^5-10)^{\frac{1}{3}}$

Step 2: Next, swap x and y:

$x=(y^5-10)^{\frac{1}{3}}$

Step 3: Solve for y.

$\begin{gathered} x^3=y^5-10 \\ y^5=x^3+10 \\ y=\sqrt[5]{x^3+10} \end{gathered}$

Step 4: Replace y with the inverse function:

$f^{-1}(x)=\sqrt[5]{x^3+10}$