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

Question

Find the derivative at each critical point and determine the local extreme values.
-$$y = \left\{ \begin{array} { l l } 
4 - x , & x < 0 \
4 + 3 x - x ^ { 2 } , & x \geq 0
\end{array} \right.$$

A)
$\begin{array}{l|l|l|l}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline x=4 & \text { undefined } & \text { local min } & 4 \
x=0 & 0 & \text { local max } & \frac{25}{4}
\end{array}$

B)
$\begin{array}{l|l|l|l}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline x=0 & \text { undefined } & \text { local min } & 4 \
x=\frac{5}{2} & 0 & \text { local max } & \frac{41}{4}
\end{array}$

C)
$\begin{array}{l|l|l|l}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline x=0 & \text { undefined } & \text { local min } & -4 \
x=\frac{3}{2} & 0 & \text { local max } & \frac{7}{4}
\end{array}$

D)
$\begin{array}{l|l|l|l}
\text { Critical Pt. } & \text { derivative } & \text { Extremum } & \text { Value } \
\hline x=0 & \text { undefined } & \text { local min } & 4 \
x=\frac{3}{2} & 0 & \text { local max } & \frac{25}{4}
\end{array}$

Quizplus · Accepted Answer

For $x < 0$, the derivative of $y = 4 - x$ is $-1$, which does not have any critical points. For $x \geq 0$, the derivative of $y = 4 + 3x - x^2$ is $3 - 2x$. Setting this equal to zero gives $x = \frac{3}{2}$ as a critical point. At $x = 0$, the function changes its formula, making it a critical point by definition. Evaluating $y$ at $x = \frac{3}{2}$ gives $4 + \frac{9}{2} - \frac{9}{4} = 4 + \frac{18}{4} - \frac{9}{4} = 4 + \frac{9}{4} = \frac{25}{4}$, indicating a local maximum. At $x = 0$, the value of $y$ is $4$, which is a local minimum since the function changes from decreasing to increasing at this point.

Find the Derivative at Each Critical Point and Determine the Local