Question

Find the intervals in which the function f given by f(x) = 2x3 − 3x2 − 36x + 7 is (a) strictly increasing (b) strictly decreasing

Answer 1

The given function is f(x) = 2x3 − 3x2 − 36x + 7

$f'(x)=6x^2-6x-36=6(x^2-x-6)=6(x+2)(x-3)$

$\therefore f'(x)=0$⇒ x=-2, 3

The points x = −2 and x = 3 divide the real line into three disjoint intervals i.e.,
$(-\infin, -2),(-2,3)$ and $(3,\infin)$

In interval $(-∞,-2)$ and $(3,∞)$, $f'(x)$ is positive while in interval (-2,3), $f'(x)$ is negative.

Hence, the given function (f) is strictly increasing in intervals $(-∞,-2)$ and $(3,∞)$, while function (f) is strictly decreasing in interval (−2, 3).