Use the Second Derivative Test to find the relative extrema, if any, of the function $f ( x ) = 2 x ^ { 3 } - 3 x ^ { 2 } - 36 x - 5$
A) Rel. max. $f ( - 3 ) = 22$,
rel. min. $f ( 2 ) = - 73$
B) Rel. max. $f ( - 2 ) = 39$,
rel. min. $f ( 3 ) = - 86$
C) Rel. max. $f ( 3 ) = 86$,
rel. min. $f ( - 2 ) = - 39$
D) Rel. max. $f ( 2 ) = 73$,
rel. min. $f ( - 3 ) = - 22$

Question

Quizplus · Accepted Answer

The critical points are found by setting the first derivative $f'(x) = 6x^2 - 6x - 36$ to zero, giving $x = -2$ and $x = 3$. The second derivative $f''(x) = 12x - 6$ is used to determine concavity. At $x = -2$, $f''(-2) = -30$ (concave down, relative max), and at $x = 3$, $f''(3) = 30$ (concave up, relative min). Thus, the relative max is at $f(-2) = 39$ and the relative min is at $f(3) = -86$.

Use the Second Derivative Test to Find the Relative Extrema f(x)=2x3−3x2−36x−5f ( x ) = 2 x ^ { 3 } - 3 x ^ { 2 } - 36 x - 5f(x)=2x3−3x2−36x−5

Use the Second Derivative Test to Find the Relative Extrema $f ( x ) = 2 x ^ { 3 } - 3 x ^ { 2 } - 36 x - 5$