## Precalculus (6th Edition) Blitzer

a) ${{f}^{-1}}\left( x \right)=\sqrt[3]{x+1}$ b) they are inverses
(a) Consider the function, $f\left( x \right)={{x}^{3}}-1$ Step-1-Replace $f\left( x \right)$ with $y$ in $f\left( x \right)$ \begin{align} & f\left( x \right)={{x}^{3}}-1 \\ & y={{x}^{3}}-1 \end{align} Step-2-Interchange $x$ and $y$ \begin{align} & y={{x}^{3}}-1 \\ & x={{y}^{3}}-1 \\ \end{align} Step-3-Solve for $y$. Add $1$ to each side \begin{align} & x={{y}^{3}}-1 \\ & x+1={{y}^{3}}-1+1 \\ & x+1={{y}^{3}} \end{align} Take the cube root of each side \begin{align} & x+1={{y}^{3}} \\ & \sqrt[3]{x+1}=\sqrt[3]{{{y}^{3}}} \\ & \sqrt[3]{x+1}=y \end{align} Step-4-Replace $y$ in step 3 by ${{f}^{-1}}\left( x \right)$ \begin{align} & y=\sqrt[3]{x+1} \\ & {{f}^{-1}}\left( x \right)=\sqrt[3]{x+1} \end{align} Therefore, the required inverse of the function is ${{f}^{-1}}\left( x \right)=\sqrt[3]{x+1}$ (b) Consider the functions: $f\left( x \right)={{x}^{3}}-1$ and ${{f}^{-1}}\left( x \right)=\sqrt[3]{x+1}$ Replace $x$ with ${{f}^{-1}}\left( x \right)$ \begin{align} & f\left( {{f}^{-1}}\left( x \right) \right)={{\left( {{f}^{-1}}\left( x \right) \right)}^{3}}-1 \\ & ={{\left( \sqrt[3]{x+1} \right)}^{3}}-1 \end{align} Simplify \begin{align} & f\left( g\left( x \right) \right)={{\left( \sqrt[3]{x+1} \right)}^{3}}-1 \\ & =\left( x+1 \right)-1 \\ & =x \end{align} Now, find ${{f}^{-1}}\left( f\left( x \right) \right)$ The equation for ${{f}^{-1}}\left( x \right)$ is given as: ${{f}^{-1}}\left( x \right)=\sqrt[3]{x+1}$ Replace $x$ with $f\left( x \right)$ \begin{align} & {{f}^{-1}}\left( f\left( x \right) \right)=\sqrt[3]{f\left( x \right)+1} \\ & =\sqrt[3]{\left( {{x}^{3}}-1 \right)+1} \end{align} Simplify \begin{align} & {{f}^{-1}}\left( f\left( x \right) \right)=\sqrt[3]{\left( {{x}^{3}}-1 \right)+1} \\ & =\sqrt[3]{{{x}^{3}}} \\ & =x \end{align} Thus, $f\left( {{f}^{-1}}\left( x \right) \right)=x$ and ${{f}^{-1}}\left( f\left( x \right) \right)=x$ It can be easily observed that ${{f}^{-1}}$ can be expressed in $f$ if we add 1 and then take the cube root over the whole expression. Hence, the functions $f$ and ${{f}^{-1}}$ are inverses of each other.