Answer
a) ${{f}^{-1}}\left( x \right)=\sqrt[3]{x}+1$.
b) they are inverses
Work Step by Step
(a)
Consider the function:
$f\left( x \right)={{\left( x-1 \right)}^{3}}$
Step 1: Replace the function $f\left( x \right)$ with y in the equation of $f\left( x \right)$.
$y={{\left( x-1 \right)}^{3}}$
Step 2: Interchange the variables x and y.
$x={{\left( y-1 \right)}^{3}}$
Step 3: Solve the equation for y.
The variable y has to be isolated. As $\sqrt[3]{{{y}^{3}}}=y$ , take the cube root on both the sides of the equation. So, the equation becomes,
${{\left( x \right)}^{{1}/{3}\;}}={{\left( {{\left( y-1 \right)}^{3}} \right)}^{{1}/{3}\;}}$
Simplify the powers. So,
$\sqrt[3]{x}=y-1$
Add $1$ to both the sides of the equation and simplify.
$\begin{align}
& \sqrt[3]{x}+1=y-1+1 \\
& \sqrt[3]{x}+1=y \\
\end{align}$
Because the obtained equation defines y as a function of x, then the inverse function exists for the function f.
Step 4: Replace y by ${{f}^{-1}}\left( x \right)$ in the equation obtained in step 3.
Thus, the inverse function${{f}^{-1}}\left( x \right)$ is obtained as,
${{f}^{-1}}\left( x \right)=\sqrt[3]{x}+1$
Therefore, the inverse function is ${{f}^{-1}}\left( x \right)=\sqrt[3]{x}+1$.
(b)
Consider the function:
$f\left( x \right)={{\left( x-1 \right)}^{3}}$.
The inverse of the function is obtained as ${{f}^{-1}}\left( x \right)=\sqrt[3]{x}+1$ in part (a).
The inverse of an equation can be verified by showing that $f\left( {{f}^{-1}}\left( x \right) \right)=x\text{ and }{{f}^{-1}}\left( f\left( x \right) \right)=x\text{ }$. First calculate $f\left( {{f}^{-1}}\left( x \right) \right)$.
Replace ${{f}^{-1}}\left( x \right)=\sqrt[3]{x}+1$ and evaluate the function f.
$\begin{align}
& f\left( {{f}^{-1}}\left( x \right) \right)=f\left( \sqrt[3]{x}+1 \right) \\
& ={{\left( \sqrt[3]{x}+1-1 \right)}^{3}} \\
& ={{\left( \sqrt[3]{x} \right)}^{3}} \\
& =x
\end{align}$
Now, calculate ${{f}^{-1}}\left( f\left( x \right) \right)$. Replace $f\left( x \right)={{\left( x-1 \right)}^{3}}$ and evaluate the inverse function ${{f}^{-1}}$.
$\begin{align}
& {{f}^{-1}}\left( f\left( x \right) \right)={{f}^{-1}}\left( {{\left( x-1 \right)}^{3}} \right) \\
& =\sqrt[3]{{{\left( x-1 \right)}^{3}}}+1 \\
& =x-1+1 \\
& =x
\end{align}$
Thus, it verifies that $f\left( {{f}^{-1}}\left( x \right) \right)=x\text{ and }{{f}^{-1}}\left( f\left( x \right) \right)=x$. Therefore, the obtained inverse function ${{f}^{-1}}\left( x \right)=\sqrt[3]{x}+1$ is correct.
Hence, the functions $f$ and ${{f}^{-1}}$ are inverses of each other.
