The interval in which the function f(x)=2 x^3-3 x^2-36 x+7 is strictl…

$f(x) =2 x^3-3 x^2-36 x+7$

differentiating f(x) wrt x

so  $f'(x) =6 x^2-6 x-36$

so $f'(x) =6(x^2-x-6)$

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

$\Rightarrow f'(x)=6(x(x-3)+2(x-3))$

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

finding critical points

$\Rightarrow f'(x)=0 \Rightarrow 6(x+2)(x-3)=0$

$\Rightarrow x = -2, 3$

using wavy curve method

x > 3      (f'(x) > 0)

f(x) (increasing)

-2 < x < 3    (f'(x) < 0)

f(x) (decreasing)

x < -2    (f'(x)) > 0

f(x) (increasing)

So (-2, 3)