Find the derivative at each critical point and determine the local extreme values.
-$$y = x ^ { 2 } \sqrt { 1 - x }$$
A)
$\begin{array}{l|l|l|l}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline x=\frac{4}{5} & 0 & \text { local max } & \frac{16}{125} \sqrt{5}
\end{array}$

B)
$\begin{array}{l|l|l|l}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline x=0 & 0 & \text { min } & 0 \
x=1 & 0 & \text { min } & 0 \
x=\frac{4}{5} & 0 & \text { local } \max & \frac{16}{125} \sqrt{5}
\end{array}$

C)
$\begin{array}{l|l|l|l}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline x=0 & 0 & \min & 0 \
x=1 & \text { undefined } & \text { min } & 0 \
x=\frac{4}{5} & 0 & \text { local max } & \frac{16}{125} \sqrt{5}
\end{array}$

D)
$\begin{array}{l|l|l|l}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline x=1 & \text { undefined } & \text { min } & 0 \
x=\frac{4}{5} & 0 & \text { local } \max & \frac{16}{125} \sqrt{5}
\end{array}$

Question

Find the derivative at each critical point and determine the local extreme values.
-$$y = x ^ { 2 } \sqrt { 1 - x }$$
A)
$\begin{array}{l|l|l|l}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline x=\frac{4}{5} & 0 & \text { local max } & \frac{16}{125} \sqrt{5}
\end{array}$

B)
$\begin{array}{l|l|l|l}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline x=0 & 0 & \text { min } & 0 \
x=1 & 0 & \text { min } & 0 \
x=\frac{4}{5} & 0 & \text { local } \max & \frac{16}{125} \sqrt{5}
\end{array}$

C)
$\begin{array}{l|l|l|l}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline x=0 & 0 & \min & 0 \
x=1 & \text { undefined } & \text { min } & 0 \
x=\frac{4}{5} & 0 & \text { local max } & \frac{16}{125} \sqrt{5}
\end{array}$

D)
$\begin{array}{l|l|l|l}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline x=1 & \text { undefined } & \text { min } & 0 \
x=\frac{4}{5} & 0 & \text { local } \max & \frac{16}{125} \sqrt{5}
\end{array}$

Quizplus · Accepted Answer

The derivative of $y = x^2\sqrt{1-x}$ will be zero at $x=\frac{4}{5}$, indicating a local maximum at this point with the value $\frac{16}{125}\sqrt{5}$. At $x=0$, the function has a value of 0, which is a minimum. The derivative is undefined at $x=1$, and the function also has a value of 0 here, indicating another point of interest but not a critical point in the traditional sense since the derivative does not exist.

Find the Derivative at Each Critical Point and Determine the Local