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Find the Derivative at Each Critical Point and Determine the Local

Question 47

Multiple Choice

Find the derivative at each critical point and determine the local extreme values.
- y=x(16x2) y = x \left( 16 - x ^ { 2 } \right)


A)
 Critical Pt.  derivative  Extremum  Value x=2.310 local max 24.63x=2.310 local min 49.27\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)
 Critical Pt.  derivative  Extremum  Value x=2.310 local max 49.27x=2.310 local min 24.63\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)
 Critical Pt.  derivative  Extremum  Value x=2.310 local max 49.27x=2.310 local min 24.63\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)
 Critical Pt.  derivative  Extremum  Value x=2.310 local max 24.63x=2.310 local min 24.63\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}

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