Application of Derivatives — Class 12 Maths Solution

We have, $f(x) = 2{x^3} - 3{x^2} - 36x + 7$ …(i)
$f(x)$ is a polynomial function.

Hence, $f\left( x \right)$ is continuous and derivable on $R$.
Differentiating (i) w.r.t. $x$, we get
$f'(x) = 6{x^2} - 6x - 36 = 6({x^2} - x - 6) = 6(x - 3)(x + 2)$

(a) For function to be increasing, $f'(x\rangle > 0$
$\Rightarrow$ 6 $(x - 3)(x + 2) > 0$

$\Rightarrow (x - 3)(x - ( - 2)) > 0 \Rightarrow x < - 2$ or $x > 3$.
$\Rightarrow x \in ( - \infty ,\; - 2) \cup (3,\;\infty )$

Therefore, $f$ is strictly increasing on $( - \infty ,\; - 2) \cup (3,\infty )$

(b) For function to be decreasing, $f'(x) < 0$
$\Rightarrow$ 6 ($x - 3\rangle (x + 2) < 0$

$\Rightarrow (x - 3)(x - ( - 2)) < 0 \Rightarrow x < 3,x > - 2$

$\Rightarrow - 2 < x < 3 \Rightarrow x \in ( - 2,3)$

Therefore, $f$ is strictly decreasing on $( - 2,3)$