Chapter 3.4, Problem 40E

To determine

To find: $f^{-1}\left(x\right)$ and verify that f and $f^{-1}$ are inverse functions.

Answer to Problem 40E

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

Explanation of Solution

Given information:

  $f\left(x\right)={\left(x-3\right)}^3+4$

Calculation:

The $f^{-1}\left(x\right)$ can be evaluated as follows:

  $\begin{array}{c}y={\left(x-3\right)}^3+4\left[\text{Let }f\left(x\right)=y\right]\\ x={\left(y-3\right)}^3+4\left[\text{Interchange }x\text{ and }y\right]\\ {\left(y-3\right)}^3=x-4\left[\text{Solve for }y\right]\\ y-3=\sqrt[3]{x-4}\\ y=\sqrt[3]{x-4}+3\\ f^{-1}\left(x\right)=\sqrt[3]{x-4}+3\left[\text{Replace }y\text{ with }{f}^{-1}\left(x\right)\right]\end{array}$

Thus the inverse is $\sqrt[3]{x-4}+3$ .

Now evaluate $\left[f\circ f^{-1}\right]\left(x\right)$ as follows:

  $\begin{array}{c}\left[f\circ f^{-1}\right]\left(x\right)=f\left(f^{-1}\right)\\ =f\left(\sqrt[3]{x-4}+3\right)\left[\text{Put }{f}^{-1}\left(x\right)\right]\\ ={\left(\sqrt[3]{x-4}+3-3\right)}^3+4\left[\text{Put }f\left(x\right)\right]\\ ={\left(\sqrt[3]{x-4}+3-3\right)}^3+4\\ ={\left(\sqrt[3]{x-4}\right)}^3+4\\ =x-4+4\\ =x\end{array}$

Since $\left[f\circ f^{-1}\right]\left(x\right)$ is equal to x, so it proves that f and $f^{-1}$ as inverse functions.