Answer
\[\begin{align}
& \left( \text{a} \right)\text{ }x=1 \\
& \left( \text{b} \right)\text{Increasing on}\left( -\infty ,1 \right),\text{ decreasing on }\left( 1,\infty \right) \\
& \left( \text{c} \right)\text{Relative maximum: }\left( 1,4 \right) \\
\end{align}\]
Work Step by Step
\[\begin{align}
& \text{Let }f\left( x \right)=\left\{ \begin{matrix}
3x+1,\text{ }x\le 1 \\
5-{{x}^{2}},\text{ }x>1 \\
\end{matrix} \right. \\
& \left( \text{a} \right) \\
& \text{Differentiate } \\
& f'\left( x \right)=\left\{ \begin{matrix}
\frac{d}{dx}\left[ 3x+1 \right],\text{ }x\le 1 \\
\frac{d}{dx}\left[ 5-{{x}^{2}} \right],\text{ }x>1 \\
\end{matrix} \right. \\
& f'\left( x \right)=\left\{ \begin{matrix}
3,\text{ }x\le 1 \\
-2x,\text{ }x>1 \\
\end{matrix} \right. \\
& \text{We obtain the critical point }x=1\text{ }\left( \text{discontinuity} \right) \\
& \text{Set the intervals }\left( -\infty ,1 \right),\left( 1,\infty \right) \\
& \\
& \left( \text{b} \right)\text{Making a table of values } \\
& \begin{matrix}
\text{Interval} & \left( -\infty ,1 \right) & \left( 1,\infty \right) \\
\text{Test Value} & x=0 & x=1 \\
\text{Sign of }f'\left( x \right) & \text{ }f'\left( 0 \right)=3>0 & \text{ }f'\left( 2 \right)=-2<0 \\
\text{Conclusion} & \text{Increasing} & \text{Decreasing} \\
\end{matrix} \\
& \\
& \text{By Theorem 3}\text{.6} \\
& f'\left( x \right)\text{ changes from positive to negative at }x=1,\text{ then }f\left( x \right) \\
& \text{has a relative maximum at }\left( 1,f\left( 1 \right) \right) \\
& f\left( x \right)=\left\{ \begin{matrix}
3x+1,\text{ }x\le 1 \\
5-{{x}^{2}},\text{ }x>1 \\
\end{matrix} \right. \\
& f\left( 1 \right)=3\left( 1 \right)+1 \\
& f\left( 1 \right)=4 \\
& \text{Relative maximum: }\left( 1,4 \right) \\
\end{align}\]