## Precalculus (6th Edition) Blitzer

(a) Assume the above function $f\left( x \right)=\sqrt{\left( x-1 \right)}$. Put $f\left( x \right)$ with $y$: $\sqrt{\left( x-1 \right)}=y$ Interchange $x$ with $y$: $\sqrt{\left( y-1 \right)}=x$ Now solve the obtained equation for the value of $y$: Square both sides of the equation: \begin{align} & {{\left( \sqrt{\left( y-1 \right)} \right)}^{2}}={{\left( x \right)}^{2}} \\ & {{\left( y-1 \right)}^{2\times \frac{1}{2}}}={{x}^{2}} \\ & y-1={{x}^{2}} \end{align} Add $1$ to both sides of the equation: \begin{align} & y-1+1={{x}^{2}}+1 \\ & y={{x}^{2}}+1 \end{align} Put $y$ with ${{f}^{-1}}\left( x \right)$: ${{f}^{-1}}\left( x \right)={{x}^{2}}+1$ So, the inverse function is ${{f}^{-1}}\left( x \right)={{x}^{2}}+1$ , $x\ge 0$. Thus, the equation of ${{f}^{-1}}\left( x \right)$ with the function $f\left( x \right)=\sqrt{\left( x-1 \right)}$ ${{f}^{-1}}\left( x \right)={{x}^{2}}+1$ , $x\ge 0$. (b) Assume the above function $f\left( x \right)=\sqrt{\left( x-1 \right)}$. Put $f\left( x \right)$ with $y$: $\sqrt{\left( x-1 \right)}=y$ Put $y=1$ to solve for the value of $x$: $\sqrt{\left( x-1 \right)}=1$ Square both sides: \begin{align} & {{\left( \sqrt{\left( x-1 \right)} \right)}^{2}}={{\left( 1 \right)}^{2}} \\ & x-1=1 \end{align} Add $1$ to both sides of the equation: \begin{align} & x=2 \end{align} So, the point A is $\left( 2,1 \right)$. Now, put $x=5$ in the equation $\sqrt{\left( x-1 \right)}=y$ to solve for the value of $y$: \begin{align} & \sqrt{\left( 5-1 \right)}=y \\ & y=\sqrt{4} \\ & =2 \end{align} Therefore, the point $B$ is $\left( 5,2 \right)$. Put $y=0$ in the equation $\sqrt{\left( x-1 \right)}=y$ to solve the value of $x$: \begin{align} & \sqrt{\left( x-1 \right)}=y \\ & x-1=0 \\ & x=1 \end{align} Put, the point C is $\left( 1,0 \right)$. But, inverse of a function exists. Therefore, points to plot the inverse function are ${{A}_{1}}=\left( 1,2 \right)$ , ${{B}_{1}}=\left( 2,5 \right)$ , and ${{C}_{1}}=\left( 0,1 \right)$. 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 $\left[ 1,\infty \right)$ and $\left[ 0,\infty \right)$ , respectively, and domain and range of the function ${{f}^{-1}}$ are $\left[ 0,\infty \right)$ and $\left[ 1,\infty \right)$ , respectively.