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 test these points: $f''(-2) = -30$ (concave down, relative max) and $f''(3) = 30$ (concave up, relative min). Thus, the relative maximum is at $f(-2) = 39$ and the relative minimum 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 $f(x)=2 x^{3}-3 x^{2}-36 x-5$