## Precalculus (6th Edition) Blitzer

(a) Suppose the above function $f\left( x \right)=\sqrt[3]{x-1}$. Take $f\left( x \right)$ with $y$: $\sqrt[3]{x-1}=y$ Interchange $x$ with $y$: $\sqrt[3]{y-1}=x$ Solve the obtained equation for the value of $y$: Take the cube of both sides of the equation: \begin{align} & {{\left( \sqrt[3]{y-1} \right)}^{3}}={{\left( x \right)}^{3}} \\ & {{\left( y-1 \right)}^{3\times \frac{1}{3}}}={{\left( x \right)}^{3}} \\ & y-1={{\left( x \right)}^{3}} \end{align} Add $1$ to both sides of the equation: \begin{align} & y-1+1={{x}^{3}}+1 \\ & y={{x}^{3}}+1 \end{align} Take $y$ with ${{f}^{-1}}\left( x \right)$: ${{f}^{-1}}\left( x \right)={{x}^{3}}+1$ So, the inverse function is ${{f}^{-1}}\left( x \right)={{x}^{3}}+1$. Thus, the equation of ${{f}^{-1}}\left( x \right)$ with the function $f\left( x \right)=\sqrt{x}+2$ is ${{f}^{-1}}\left( x \right)={{x}^{3}}+1$. (b) Suppose the above equation $f\left( x \right)=\sqrt[3]{x-1}$. Put $f\left( x \right)$ with $y$: $\sqrt[3]{x-1}=y$ Put $y=1$ to solve for the value of $x$: $\sqrt[3]{x-1}=1$ Take the cube of both sides: \begin{align} & {{\left( \sqrt[3]{x-1} \right)}^{3}}=1 \\ & x-1=1 \\ & x=2 \end{align} So, the point A is $\left( 2,1 \right)$. Now, put $x=1$ in the equation $\sqrt[3]{x-1}=y$ to solve for the value of $y$: \begin{align} & \sqrt[3]{1-1}=y \\ & y=0 \end{align} So, the point $B$ is $\left( 1,0 \right)$. Put $y=2$ in the equation $\sqrt[3]{x-1}=y$ to solve for the value of $x$: \begin{align} & \sqrt[3]{x-1}=2 \\ & x-1=8 \\ & x=9 \end{align} So, the point C is $\left( 9,2 \right)$. Since inverse of a function exists. Therefore, points to the plot inverse function are ${{A}_{1}}=\left( 1,2 \right)$ , ${{B}_{1}}=\left( 0,1 \right)$ and ${{C}_{1}}=\left( 2,9 \right)$. Plot all the points on the same graph for both the function and its inverse as shown above: (c) The domain and range of the function $f$ are $\left( \infty ,-\infty \right)$ and the domain and range of the function ${{f}^{-1}}$ are $\left( \infty ,-\infty \right)$.