Find the derivative at each critical point and determine the local extreme values.
-$y=x^{2} \sqrt{9-x}$
A)
$\begin{array}{l|l|l|l}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline x=0 & 0 & \text { min } & 0 \
x=9 & \text { undefined } & \text { min } & 0 \
x=\frac{36}{5} & 0 & \text { local max } & \frac{3888}{125} \sqrt{5}
\end{array}$

B)
$\begin{array}{l|l|l|l}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline x=9 & \text { undefined } & \text { min } & 0 \
x=\frac{36}{5} & 0 & \text { local max } & \frac{3888}{125} \sqrt{5}
\end{array}$

C)
$\begin{array}{l|l|l|l}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline x=0 & 0 & \min & 0 \
x=9 & 0 & \min & 0 \
x=\frac{36}{5} & 0 & \text { local max } & \frac{3888}{125} \sqrt{5}
\end{array}$

D)
$\begin{array}{l|l|l|l}
\text { Critical Pt.| } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline \mathrm{x}=\frac{36}{5} & 0 & \text { local max } & \frac{3888}{125} \sqrt{5}
\end{array}$

Question

Find the derivative at each critical point and determine the local extreme values.
-$y=x^{2} \sqrt{9-x}$
A)
$\begin{array}{l|l|l|l}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline x=0 & 0 & \text { min } & 0 \
x=9 & \text { undefined } & \text { min } & 0 \
x=\frac{36}{5} & 0 & \text { local max } & \frac{3888}{125} \sqrt{5}
\end{array}$

B)
$\begin{array}{l|l|l|l}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline x=9 & \text { undefined } & \text { min } & 0 \
x=\frac{36}{5} & 0 & \text { local max } & \frac{3888}{125} \sqrt{5}
\end{array}$

C)
$\begin{array}{l|l|l|l}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline x=0 & 0 & \min & 0 \
x=9 & 0 & \min & 0 \
x=\frac{36}{5} & 0 & \text { local max } & \frac{3888}{125} \sqrt{5}
\end{array}$

D)
$\begin{array}{l|l|l|l}
\text { Critical Pt.| } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline \mathrm{x}=\frac{36}{5} & 0 & \text { local max } & \frac{3888}{125} \sqrt{5}
\end{array}$

Quizplus · Accepted Answer

The derivative of $y=x^{2} \sqrt{9-x}$ yields critical points at $x=0$, $x=9$ (where the derivative is undefined due to the square root of a negative number becoming imaginary), and $x=\frac{36}{5}$. At $x=0$ and $x=9$, the function has minimum values of 0, and at $x=\frac{36}{5}$, it has a local maximum with the value of $\frac{3888}{125} \sqrt{5}$.

Find the Derivative at Each Critical Point and Determine the Local