Find the intervals in which the function f given by $f (x) = 2 x^{3} - 3 x^{2} - 36 x + 7$ is

a) strictly increasing

b) strictly decreasing

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Solution

Given, $f (x) = 2 x^{3} - 3 x^{2} - 36 x + 7$
$\Rightarrow$ Differentiating w.r.t. x, $f^{'} (x) = \frac{d}{d x} (2 x^{3} - 3 x^{2} - 36 x + 7)$
$= 2.3 x^{2} - 3.2 x - 36.1 + 0 = 6 x^{2} - 6 x - 36 = 6 (x^{2} - x - 6)$
$\Rightarrow f^{'} (x) = 6 (x - 3) (x + 2)$
On putting f'(x)=0 we get $6 (x - 3) (x + 2) = 0 \Rightarrow x = 3$ and -2
which divides real line into three intervals namely $(- \infty, - 2) (- 2, 3)$ and $(3, \infty)$ .

$\begin{matrix} I n t e r v a l s & S i g n o f f^{'} (x) & N a t u r e o f f (x) (- \infty, - 2) & (-) (-) = + v e & S t r i c t l y i n c r e a s i n g (∵ x < - 2 i . e ., x i s - 3, - 4, - 5 \dots f o r t h e s e v a l u e s, (x - 3) a n d (x - 2) b o t h w i l l n e g a t i v e) (- 2, 3) & (-) (+) = - v e & S t r i c t l y d e c r e a s i n g (∵ 3 > x > - 2 i . e, x i s 1, 2,, 0, - 1 \dots f o r t h e s e v a l u e s (x - 3) w i l l b e n e g a t i v e a n d (x + 2) w i l l b e p o s i t i v e) (3, \infty) & (+) (+) = + v e & S t r i c t l y i n c r e s i n g (∵ x > 3 i . e ., x i s 4, 5, 6, \dots f o r t h e s e) v a l u e s (x - 3) a n d (x + 2) b o t h a r e p o s i t i v e) \end{matrix}$
Hence, the given function f is strictly increasing in intervals $(- \infty, - 2)$ and $(3, \infty)$ , while function f is strictly decreasing in the interval (-2, 3).

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