Find the derivative at each critical point and determine the local extreme values.
-$y = x \left( 16 - x ^ { 2 } \right)$
A)
$\begin{array}{l|l|l|r}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline x=-2.31 & 0 & \text { local max } & 24.63 \
x=2.31 & 0 & \text { local min } & -49.27
\end{array}$

B)
$\begin{array}{l|l|l|r}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline \mathrm{x}=2.31 & 0 & \text { local max } & -49.27 \
\mathrm{x}=-2.31 & 0 & \text { local min } & 24.63
\end{array}$

C)
$\begin{array}{l|l|l|r}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline x=-2.31 & 0 & \text { local max } & -49.27 \
x=2.31 & 0 & \text { local min } & 24.63
\end{array}$

D)
$\begin{array}{l|l|l|r}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline \mathrm{x}=2.31 & 0 & \text { local max } & 24.63 \
\mathrm{x}=-2.31 & 0 & \text { local min } & -24.63
\end{array}$

Question

Find the derivative at each critical point and determine the local extreme values.
-$y = x \left( 16 - x ^ { 2 } \right)$
A)
$\begin{array}{l|l|l|r}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline x=-2.31 & 0 & \text { local max } & 24.63 \
x=2.31 & 0 & \text { local min } & -49.27
\end{array}$

B)
$\begin{array}{l|l|l|r}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline \mathrm{x}=2.31 & 0 & \text { local max } & -49.27 \
\mathrm{x}=-2.31 & 0 & \text { local min } & 24.63
\end{array}$

C)
$\begin{array}{l|l|l|r}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline x=-2.31 & 0 & \text { local max } & -49.27 \
x=2.31 & 0 & \text { local min } & 24.63
\end{array}$

D)
$\begin{array}{l|l|l|r}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline \mathrm{x}=2.31 & 0 & \text { local max } & 24.63 \
\mathrm{x}=-2.31 & 0 & \text { local min } & -24.63
\end{array}$

Quizplus · Accepted Answer

First, find the derivative of $y = x(16 - x^2)$, which is $y' = 16 - 3x^2$. Setting $y' = 0$ gives $x = \pm \sqrt{\frac{16}{3}}$, or approximately $x = \pm 2.31$. Evaluating the second derivative, $y'' = -6x$, at these critical points shows that $y''(-2.31) < 0$ (indicating a local maximum) and $y''(2.31) > 0$ (indicating a local minimum). The value of the function at $x = 2.31$ is $24.63$ and at $x = -2.31$ is $-24.63$, not as listed in options A, B, or C.

Find the Derivative at Each Critical Point and Determine the Local