Answer
\[\begin{align}
& \left( \text{a} \right)\text{ }x=-3 \\
& \left( \text{b} \right)\text{Increasing on}\left( -3,\infty \right),\text{ decreasing on }\left( -\infty ,3 \right) \\
& \left( \text{c} \right)\text{Relative minimum: }\left( -3,-1 \right) \\
\end{align}\]
Work Step by Step
\[\begin{align}
& f\left( x \right)=\left| x+3 \right|-1 \\
& \left( \text{a} \right) \\
& \text{Differentiating} \\
& f'\left( x \right)=-\frac{x+3}{\left| x+3 \right|} \\
& \text{Using the definition of the absolute value}\text{, we can write} \\
& \text{the derivative of the function as} \\
& f'\left( x \right)=\left\{ \begin{matrix}
-1,\text{ }x<-3 \\
1,\text{ }x>-3 \\
\end{matrix} \right. \\
& \text{The derivative is not defined at }x=-3,\text{ we obtain the critical } \\
& \text{point }x=-3 \\
& \text{Set the intervals }\left( -\infty ,-3 \right),\left( -3,\infty \right) \\
& \\
& \left( \text{b} \right)\text{Making a table of values }\left( \text{See examples on page 180 } \right) \\
& \begin{matrix}
\text{Interval} & \left( -\infty ,-3 \right) & \left( -3,\infty \right) \\
\text{Test Value} & x=-4 & x=0 \\
\text{Sign of }f'\left( x \right) & \text{ }f'\left( -4 \right)=-1<0 & \text{ }f'\left( 0 \right)=1>0 \\
\text{Conclusion} & \text{Decreasing} & \text{Increasing} \\
\end{matrix} \\
& \\
& \left( \text{c} \right)\text{By Theorem 3}\text{.6} \\
& f'\left( x \right)\text{ changes from negative to positive at }x=-3,\text{ then }f\left( x \right) \\
& \text{has a relative minimum at }\left( -3,f\left( -3 \right) \right) \\
& f\left( -3 \right)=\left| -3+3 \right|-1 \\
& f\left( -3 \right)=-1 \\
& \text{Relative minimum: }\left( -3,-1 \right) \\
\end{align}\]