given (f(x)=2x^{3}-3x^{2}-36x - 6). (a) find the intervals on which (f) is increasing or decreasing. (b)…

Explanation:

Step1: Find the derivative

Given $f(x)=2x^{3}-3x^{2}-36x - 6$, then $f'(x)=6x^{2}-6x - 36=6(x^{2}-x - 6)=6(x - 3)(x+2)$

Step2: Find critical points

Set $f'(x)=0$, so $6(x - 3)(x + 2)=0$. The critical points are $x=-2$ and $x = 3$.

Step3: Determine intervals of increase and decrease

Test intervals:

• For $x<-2$, let $x=-3$, then $f'(-3)=6(-3 - 3)(-3+2)=36>0$, so $f(x)$ is increasing on $(-\infty,-2)$.
• For $-2<x<3$, let $x = 0$, then $f'(0)=6(0 - 3)(0 + 2)=-36<0$, so $f(x)$ is decreasing on $(-2,3)$.
• For $x>3$, let $x = 4$, then $f'(4)=6(4 - 3)(4+2)=36>0$, so $f(x)$ is increasing on $(3,\infty)$.

Step4: Find relative extrema

• Evaluate $f(x)$ at critical points. $f(-2)=2(-2)^{3}-3(-2)^{2}-36(-2)-6=2(-8)-3(4)+72 - 6=-16-12 + 72-6=38$ $f(3)=2(3)^{3}-3(3)^{2}-36(3)-6=2(27)-3(9)-108-6=54 - 27-108-6=-87$ Since $f(x)$ changes from increasing to decreasing at $x=-2$, $f(-2)=38$ is a relative maximum. Since $f(x)$ changes from decreasing to increasing at $x = 3$, $f(3)=-87$ is a relative minimum.

Answer:

(a) Increasing on $(-\infty,-2)$ and $(3,\infty)$, decreasing on $(-2,3)$ (b) Rel. max. $f(-2)=38$, rel. min. $f(3)=-87$