Find the Derivative at Each Critical Point and Determine the Local

Multiple Choice

Find the derivative at each critical point and determine the local extreme values.
- $y = \left\{ \begin{array} { l l } - \frac { 1 } { 4 } x ^ { 2 } - \frac { 1 } { 2 } x + \frac { 15 } { 4 } , & x \leq 1 \\x ^ { 3 } - 6 x ^ { 2 } + 8 x , & x > 1\end{array} \right.$

A)
$\begin{array}{l|l|l|l}\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \\\hline x=-1 & 0 & \text { local min } & 4 \\x=3.15 & 0 & \text { local max } & -3.08\end{array}$

B)
$\begin{array}{l|l|l|l}\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \\\hline x=-1 & \text { undefined } & \text { local min } & 4 \\x=3.15 & 0 & \text { local max } & -3.08\end{array}$

C)
$\begin{array}{l|l|l|l}\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \\\hline x=-1 & 0 & \text { local max } & 4 \\x=3.15 & 0 & \text { local min } & -3.08\end{array}$

D)
$\begin{array}{l|l|l|l}\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \\\hline x=1 & 0 & \text { local max } & 4 \\x=3.15 & \text { undefined } & \text { local max } & -3.08\end{array}$