Question

Find the inverse of the function ,$f\left(x\right)=\sqrt[5]{3x-2}$

Short answer

The inverse of, $f\left(x\right)=\sqrt[5]{3x-2}$is

$f^{-1}\left(x\right)=\frac{x^5+2}{3}$

Step-by-step solution

Given information

Given the function,$f\left(x\right)=\sqrt[5]{3x-2}$

Replace f(x) with yand then x,y

Replacing, we get

$$\begin{gathered} f\left(x\right)=\sqrt[5]{3x-2} \\ y=\sqrt[5]{3x-2} \\ x=\sqrt[5]{3y-2} \\ {x}^5=3y-2 \end{gathered}$$

Solve further and calculate the inverse

Further solving, we get

$$\begin{gathered} x^5=3y-2 \\ {x}^5+2=3y \\ y=\frac{x^5+2}{3} \\ {f}^{-1}\left(x\right)=\frac{x^5+2}{3} \end{gathered}$$

Verify f(f-1(x))=x

Substituting, we get

$$\begin{gathered} f\left(\frac{x^5+2}{3}\right)=\sqrt[5]{3\left(\frac{x^5+2}{3}\right)-2} \\ f\left(\frac{x^5+2}{3}\right)=\sqrt[5]{x^5+2-2} \\ f\left(\frac{x^5+2}{3}\right)=\sqrt[5]{x^5} \\ f\left(\frac{x^5+2}{3}\right)=x \end{gathered}$$

Verify f-1(f(x))=x

Substituting, we get

$$\begin{gathered} f^{-1}\left(\sqrt[5]{3x-2}\right)=\sqrt[5]{3\left(\frac{x^5+2}{3}\right)-2} \\ {f}^{-1}\left(\sqrt[5]{3x-2}\right)=\sqrt[5]{x^5+2-2} \\ {f}^{-1}\left(\sqrt[5]{3x-2}\right)=\sqrt[5]{x^5} \\ {f}^{-1}\left(\sqrt[5]{3x-2}\right)=x \end{gathered}$$

So these are inverses