Find the derivative at each critical point and determine the local extreme values.
-$y = \left\{ \begin{array} { l l } 8 - x , & x

Question

Find the derivative at each critical point and determine the local extreme values.
-$y = \left\{ \begin{array} { l l } 8 - x , & x < 0 \ 8 + 7 x - x ^ { 2 } , & x \geq 0 \end{array} \right.$

A)
$\begin{array}{l|l|l|l}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline x=0 & \text { undefined } & \text { local min } & 8 \
x=\frac{7}{2} & 0 & \text { local max } & \frac{81}{4}
\end{array}$

B)
$\begin{array}{l|l|l|l}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline x=8 & \text { undefined } & \text { local min } & 8 \
x=0 & 0 & \text { local max } & \frac{81}{4}
\end{array}$

C)
$\begin{array}{l|l|l|l}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline x=0 & \text { undefined } & \text { local min } & 8 \
x=\frac{9}{2} & 0 & \text { local max } & \frac{113}{4}
\end{array}$

D)
$\begin{array}{l|l|l|l}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline x=0 & \text { undefined } & \text { local min } & -8 \
x=\frac{7}{2} & 0 & \text { local max } & -\frac{17}{4}
\end{array}$

Quizplus · Accepted Answer

For $x < 0$, the derivative is constant (-1) and there are no critical points. For $x \geq 0$, the derivative of $8 + 7x - x^2$ is $7 - 2x$, which equals 0 when $x = \frac{7}{2}$. At $x = 0$, the function changes from linear to quadratic, making it a critical point. Evaluating the function at $x = \frac{7}{2}$ gives a local maximum value of $\frac{81}{4}$. At $x = 0$, the function value is 8, indicating a local minimum due to the change in the function's structure.

Find the Derivative at Each Critical Point and Determine the Local