Given:- Function f(x) = 2x³ – 24x + 7

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 + 7

⇒

⇒ f’(x) = 6x² – 24

For f(x) to be increasing, we must have

⇒ f’(x) > 0

⇒ 6x² – 24 > 0

⇒

⇒ x² < 4

⇒ x < –2, +2

⇒ x ∈ (–∞,–2) and x ∈ (2,∞)

Thus f(x) is increasing on interval (–∞, –2) ∪ (2, ∞)

Again, For f(x) to be increasing, we must have

f’(x) < 0

⇒ 6x² – 24< 0

⇒

⇒ x² < 4

⇒ x> –1

⇒ x ∈ (–1,∞)

Thus f(x) is decreasing on interval x ∈ (–1, ∞)