Find all relative extrema of the function $f ( x ) = - 4 x ^ { 2 } - 32 x - 62$ . Use the Second Derivative Test where applicable.
A) relative $\max : f ( 0 ) = - 62$; no relative $\min$
B) no relative max; no relative $\min$
C) relative min: $f ( - 4 ) = 2$; relative $\max : f ( 0 ) = - 62$
D) relative min: $f ( - 4 ) = 2$; no relative $\max$.
E) relative min: $f ( - 4 ) = 2$; relative $\max : f ( 0 ) = - 62$.

Question

Find all relative extrema of the function $f ( x ) = - 4 x ^ { 2 } - 32 x - 62$ . Use the Second Derivative Test where applicable. 
A) relative $\max : f ( 0 ) = - 62$; no relative $\min$
B) no relative max; no relative $\min$
C) relative min: $f ( - 4 ) = 2$; relative $\max : f ( 0 ) = - 62$
D) relative min: $f ( - 4 ) = 2$; no relative $\max$.
E) relative min: $f ( - 4 ) = 2$; relative $\max : f ( 0 ) = - 62$.

Quizplus · Accepted Answer

To find the relative extrema, we first find the critical points by setting the derivative equal to zero and solving for $x:$

$$f'(x)= -8x-32=0 \Longrightarrow x=-4$$

Thus, $x=-4$ is the only critical point. We can now use the Second Derivative Test to determine the nature of this critical point.

$$f''(x)=-8<0 \quad 	ext{when } x=-4$$

Since $f''(-4) < 0,$ we can conclude that $f(-4)$ is a relative maximum. Additionally, since there are no other critical points, there can be no other relative extrema. Therefore, the answer is $\boxed{	extbf{(D)}}.$

Find All Relative Extrema of the Function f(x)=−4x2−32x−62f ( x ) = - 4 x ^ { 2 } - 32 x - 62f(x)=−4x2−32x−62

Find All Relative Extrema of the Function $f ( x ) = - 4 x ^ { 2 } - 32 x - 62$