Chapter 1.5, Problem 22E

To determine

To find: The $f^{-1}$ and verify that $\left(f\circ f^{-1}\right)\left(x\right)=\left(f^{-1}\circ f\right)\left(x\right)=x$ .

Answer to Problem 22E

The inverse of the given function is $f^{-1}\left(x\right)=\frac{1}{\sqrt[3]{x}}$ and it is verified that $\left(f\circ f^{-1}\right)\left(x\right)=\left(f^{-1}\circ f\right)\left(x\right)=x$ .

Explanation of Solution

Given information: The function is $f\left(x\right)=\frac{1}{x^3}$ .

Calculation:

Substitute $y$ for $f\left(x\right)$ in the equation $f\left(x\right)=\frac{1}{x^3}$ and solve the equation for $x$ in terms of $y$ .

  $\begin{array}{c}y=\frac{1}{x^3}\\ x^3=\frac{1}{y}\\ x=\sqrt[3]{\frac{1}{y}}\\ =\frac{1}{\sqrt[3]{y}}\end{array}$

Interchange $x$ and $y$ , and substitute $f^{-1}\left(x\right)$ for $y$ .

  $\begin{array}{c}y=\frac{1}{\sqrt[3]{x}}\\ f^{-1}\left(x\right)=\frac{1}{\sqrt[3]{x}}\end{array}$

Verify:

The composite function $\left(f\circ f^{-1}\right)\left(x\right)$ can be written as:

  $\left(f\circ f^{-1}\right)\left(x\right)=f\left(f^{-1}\left(x\right)\right)$

Substitute $\frac{1}{\sqrt[3]{x}}$ for $f^{-1}\left(x\right)$ to find $\left(f\circ f^{-1}\right)\left(x\right)$ for $x>0$ .

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

The composite function $\left(f^{-1}\circ f\right)\left(x\right)$ can be written as:

  $\left(f^{-1}\circ f\right)\left(x\right)=f^{-1}\left(f\left(x\right)\right)$

Substitute $\frac{1}{x^3}$ for $f\left(x\right)$ to find $\left(f^{-1}\circ f\right)\left(x\right)$ for $x>0$ .

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

It is verified that $\left(f\circ f^{-1}\right)\left(x\right)=\left(f^{-1}\circ f\right)\left(x\right)=x$ .

Therefore, the inverse of the given function is $f^{-1}\left(x\right)=\frac{1}{\sqrt[3]{x}}$ and it is verified that $\left(f\circ f^{-1}\right)\left(x\right)=\left(f^{-1}\circ f\right)\left(x\right)=x$ .