Find the derivative at each critical point and determine the local extreme values.
-$$y = \left\{ \begin{array} { l l }
- \frac { 1 } { 4 } x ^ { 2 } - \frac { 1 } { 2 } x + \frac { 15 } { 4 } , & x \leq 1 \
x ^ { 3 } - 6 x ^ { 2 } + 8 x , & x 1
\end{array} \right.$$
A)
$\begin{array}{l|l|l|l}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline x=-1 & 0 & \text { local max } & 4 \
x=3.15 & 0 & \text { local min } & -3.08
\end{array}$

B)
$\begin{array}{l|l|l|l}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline x=-1 & \text { undefined } & \text { local min } & 4 \
x=3.15 & 0 & \text { local max } & -3.08
\end{array}$

C)
$\begin{array}{l|l|l|l}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline x=-1 & 0 & \text { local min } & 4 \
x=3.15 & 0 & \text { local max } & -3.08
\end{array}$

D)
$\begin{array}{l|l|l|l}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline x=1 & 0 & \text { local max } & 4 \
x=3.15 & \text { undefined } & \text { local max } & -3.08
\end{array}$

Question

Find the derivative at each critical point and determine the local extreme values.
-$$y = \left\{ \begin{array} { l l } 
- \frac { 1 } { 4 } x ^ { 2 } - \frac { 1 } { 2 } x + \frac { 15 } { 4 } , & x \leq 1 \
x ^ { 3 } - 6 x ^ { 2 } + 8 x , & x > 1
\end{array} \right.$$
A)
$\begin{array}{l|l|l|l}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline x=-1 & 0 & \text { local max } & 4 \
x=3.15 & 0 & \text { local min } & -3.08
\end{array}$

B)
$\begin{array}{l|l|l|l}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline x=-1 & \text { undefined } & \text { local min } & 4 \
x=3.15 & 0 & \text { local max } & -3.08
\end{array}$

C)
$\begin{array}{l|l|l|l}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline x=-1 & 0 & \text { local min } & 4 \
x=3.15 & 0 & \text { local max } & -3.08
\end{array}$

D)
$\begin{array}{l|l|l|l}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline x=1 & 0 & \text { local max } & 4 \
x=3.15 & \text { undefined } & \text { local max } & -3.08
\end{array}$

Quizplus · Accepted Answer

For the first piece, $y = -\frac{1}{4}x^2 - \frac{1}{2}x + \frac{15}{4}$, the derivative is $y' = -\frac{1}{2}x - \frac{1}{2}$. Setting $y' = 0$ gives $x = -1$, which is a critical point within the domain of this piece. The second derivative test or analyzing the sign of the derivative around $x = -1$ would show it's a local maximum. The value at $x = -1$ is $4$.For the second piece, $y = x^3 - 6x^2 + 8x$, the derivative is $y' = 3x^2 - 12x + 8$. Setting $y' = 0$ and solving gives critical points, one of which is around $x = 3.15$. The second derivative test or analyzing the sign of the derivative around this point would show it's a local minimum. The value at $x = 3.15$ is approximately $-3.08$.

Find the Derivative at Each Critical Point and Determine the Local