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

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

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

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

Question

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

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

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

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

Quizplus · Accepted Answer

The derivatives of the given piecewise function are -2x - 5 for $x \leq 1$ and -2x + 15 for $x > 1$. Setting these derivatives to zero gives critical points at $x = -\frac{5}{2}$ and $x = \frac{15}{2}$, respectively. At $x = 1$, the derivative is undefined because the piecewise functions have different slopes on either side of $x = 1$. Evaluating the function at these critical points gives the local extremum values.

Find the Derivative at Each Critical Point and Determine the Local