Question

In the following exercise, find the inverse of each function.

$f\left(x\right)=\sqrt[4]{8x-3},x\ge \frac{3}{8}$

Short answer

The inverse of the given function is$f^{-1}\left(x\right)=\frac{x^4+3}{8}$

Step-by-step solution

Step 1. Given information  

The given function is$f\left(x\right)=\sqrt[4]{8x-3},x\ge \frac{3}{8}$

Step 2.   Interchange the variable x and y

Substituting $f\left(x\right)=y$, we get:

$y=\sqrt[4]{8x-3}$

Interchanging the variable $x$and $y$, we get:

role="math" localid="1644914797428" $x=\sqrt[4]{8y-3}$

Step 3. solve for  y

Solving for $y$, we get:

role="math" localid="1644915217218" $$\begin{gathered} x=\sqrt[4]{8y-3} \\ {x}^4=8y-3[rai\sin gexponent4bothsides] \\ 8y=x^4+3 \\ y=\frac{x^4+3}{8} \end{gathered}$$

Substituting $y=f^{-1}\left(x\right)$, we get:

role="math" localid="1644915222094" $f^{-1}\left(x\right)=\frac{x^4+3}{8}$

Step 4. Verify that the functions are inverses.    

Substituting $f^{-1}\left(x\right)$in role="math" localid="1644915076476" $f^{-1}\left(f\left(x\right)\right)=x$, we get:

$$\begin{gathered} f^{-1}\left(f\left(x\right)\right)=x \\ {f}^{-1}\left(\sqrt[4]{8x-3}\right)=x \\ \frac{{\left(\sqrt[4]{8x-3}\right)}^4-3}{8x}=x \\ \frac{8x-3+3}{8x}=x \\ \frac{8x}{8}=x \\ x=x \end{gathered}$$

This is true.

Substituting $f\left(x\right)$in $f\left(f^{-1}\left(x\right)\right)=x$, we get:

$$\begin{gathered} f\left(f^{-1}\left(x\right)\right)=x \\ f\left(\frac{x^4+3}{8}\right)=x \\ \sqrt[4]{8·\frac{x^4+3}{8}-3}=x \\ \sqrt[4]{x^4+3-3}=x \\ \sqrt[4]{x^4}=x \\ x=x \end{gathered}$$

This is true.

Thus, the obtained inverse is correct