Answer
\[\begin{align}
& \left( \text{a} \right)\text{ }x=0 \\
& \left( \text{b} \right)\text{Increasing on}\left( -\infty ,0 \right),\text{ decreasing on }\left( 0,\infty \right) \\
& \left( \text{c} \right)\text{Relative maximum: }\left( 0,4 \right) \\
\end{align}\]
Work Step by Step
\[\begin{align}
& \text{Let }f\left( x \right)=\left\{ \begin{matrix}
4-{{x}^{2}},\text{ }x\le 0 \\
-2x,\text{ }x>0 \\
\end{matrix} \right. \\
& \left( \text{a} \right) \\
& \text{Differentiate } \\
& f'\left( x \right)=\left\{ \begin{matrix}
\frac{d}{dx}\left[ 4-{{x}^{2}} \right],\text{ }x\le 0 \\
\frac{d}{dx}\left[ -2x \right],\text{ }x>0 \\
\end{matrix} \right. \\
& f'\left( x \right)=\left\{ \begin{matrix}
-2x,\text{ }x\le 0 \\
-2,\text{ }x>0 \\
\end{matrix} \right. \\
& f'\left( x \right)=0 \\
& -2x=0\to x=0 \\
& \text{We obtain the critical point }x=0 \\
& \text{Set the intervals }\left( -\infty ,0 \right),\left( 0,\infty \right) \\
& \\
& \left( \text{b} \right)\text{Making a table of values }\left( \text{See examples on page 180 } \right) \\
& \begin{matrix}
\text{Interval} & \left( -\infty ,0 \right) & \left( 0,\infty \right) \\
\text{Test Value} & x=-1 & x=1 \\
\text{Sign of }f'\left( x \right) & \text{ }f'\left( -1 \right)=2>0 & \text{ }f'\left( 1 \right)=-2<0 \\
\text{Conclusion} & \text{Increasing} & \text{Decreasing} \\
\end{matrix} \\
& \\
& \text{By Theorem 3}\text{.6} \\
& f'\left( x \right)\text{ changes from positive to negative at }x=0,\text{ then }f\left( x \right) \\
& \text{has a relative maximum at }\left( 0,f\left( 0 \right) \right) \\
& \text{Let }f\left( x \right)=\left\{ \begin{matrix}
4-{{x}^{2}},\text{ }x\le 0 \\
-2x,\text{ }x>0 \\
\end{matrix} \right. \\
& f\left( 0 \right)=4-{{\left( 0 \right)}^{2}} \\
& f\left( 0 \right)=4 \\
& \text{Relative maximum: }\left( 0,4 \right) \\
\end{align}\]
