Find the Derivative at Each Critical Point and Determine the Local

Find the derivative at each critical point and determine the local extreme values.
- $y = \left\{ \begin{array} { l l } - x ^ { 2 } - 5 x + 10 , & x \leq 1 \\- x ^ { 2 } + 15 x - 10 , & x > 1\end{array} \right.$

A)
$\begin{array}{l|l|l|l}\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \\\hline x=\frac{5}{2} & 0 & \text { local max } & \frac{65}{4} \\x=1 & \text { undefined } & \text { local min } & 6 \\x=\frac{15}{2} & \text { undefined } & \text { local max } & \frac{185}{4}\end{array}$

B)
$\begin{array}{l|l|l|l}\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \\\hline x=\frac{5}{2} & 0 & \text { local max } & \frac{65}{4} \\x=1 & \text { undefined } & \text { local max } & 4\\x=\frac{15}{2} & \text { 0 } & \text { local min } & \frac{185}{4}\end{array}$

C)
$\begin{array}{l|l|l|l}\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \\\hline x=-\frac{5}{2} & 0 & \text { local max } & \frac{65}{4} \\\mathrm{x}=1 & \text { undefined } & \text { local min } & 4 \\\mathrm{x}=\frac{15}{2} & 0 & \text { local max } & \frac{185}{4}\end{array}$

D)
$\begin{array}{l|l|l|l}\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \\\hline x=-\frac{5}{2} & 0 & \text { local max } & \frac{65}{4} \\x=1 & \text { undefined } & \text { local min } & 6 \\x=-\frac{15}{2} & 0 & \text { local max } & \frac{185}{4}\end{array}$