## Precalculus (6th Edition) Blitzer

a)The value of $h\left( 5 \right)$ is $8$. b)The value of $h\left( 0 \right)$ is $3$. c)The value of $h\left( 3 \right)$ is $6$.
(a) As $5\ne 3$. So, the first part of the piecewise function will be applicable for $x=5$. \begin{align} & h\left( x \right)=\frac{{{x}^{2}}-9}{x-3} \\ & h\left( 5 \right)=\frac{{{5}^{2}}-9}{5-3} \\ & h\left( 5 \right)=\frac{16}{2} \\ & h\left( 5 \right)=8 \end{align} Hence, the value of $h\left( 5 \right)$ is $8$. (b) As $0\ne 3$. So, the first part of the piecewise function will be applicable for $x=0$. \begin{align} & h\left( x \right)=\frac{{{x}^{2}}-9}{x-3} \\ & h\left( 0 \right)=\frac{{{0}^{2}}-9}{0-3} \\ & h\left( 0 \right)=\frac{-9}{-3} \\ & h\left( 0 \right)=3 \end{align} Hence, the value of $h\left( 0 \right)$ is $3$. (c) As $3=3$. So, the second part of the piecewise function will be applicable for $x=3$. \begin{align} & h\left( x \right)=6 \\ & h\left( 3 \right)=6 \end{align} Hence, the value of $h\left( 3 \right)$ is $6$.