Question

Find the intervals in which the function $f$ given by $f\left(x\right)=2x^3-3x^2-36x+7$ is (a) strictly increasing (b) strictly decreasing.

Answer 1

The given function is $f\left(x\right)=2x^3-3x^2-36x+7$

$f^{\prime }\left(x\right)=6x^2-6x-36=6(x^2-x-6)=6(x+2)(x-3)$

$\therefore f^{\prime }\left(x\right)=0\Rightarrow x=-2,3$

The points $x=2$ and $x=3$ divide the real line into three disjoint intervals

i.e., $(-\infty ,-2),(-2,3),$ and $(3,\infty ).$

In intervals $(-\infty ,-2)$ and $(3,\infty ),f^{\prime }\left(x\right)>0$

while in interval $(-2,3),f^{\prime }\left(x\right)<0$

Hence, the given function $\left(f\right)$ is strictly increasing in intervals

$(-\infty ,-2)$ and $(3,\infty )$ while function (f) is strictly decreasing in interval $(-2,3).$