Given:- Function f(x) = 2x³ – 24x + 107

Theorem:- Let f be a differentiable real function defined on an open interval (a,b).

(i) If f’(x) > 0 for all , then f(x) is increasing on (a, b)

(ii) If f’(x) < 0 for all , then f(x) is decreasing on (a, b)

Algorithm:-

(i) Obtain the function and put it equal to f(x)

(ii) Find f’(x)

(iii) Put f’(x) > 0 and solve this inequation.

For the value of x obtained in (ii) f(x) is increasing and for remaining points in its domain it is decreasing.

Here we have,

f(x) = 2x³ – 24x + 107

⇒

⇒ f’(x) = 6x² – 24

For f(x) lets find critical point, we must have

⇒ f’(x) = 0

⇒ 6x² – 24 = 0

⇒ 6(x² – 4) = 0

⇒ (x – 2)(x + 2) = 0

⇒ x = –2, 2

clearly, f’(x) > 0 if x < –2 and x > 2

and f’(x) < 0 if –2 < x < 2

Thus, f(x) increases on (–∞, –2) ∪ (2, ∞)

and f(x) is decreasing on interval x ∈ (–2,2)