Q. 1 G5.0( 2 Votes )

# Find the intervals in which the following functions are increasing or decreasing.

f(x) = 5x^{3} – 15x^{2} – 120x + 3

Answer :

Given:– Function f(x) = 5x^{3} – 15x^{2} – 120x + 3

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) = 5x^{3} – 15x^{2} – 120x + 3

⇒

⇒ f’(x) = 15x^{2} – 30x – 120

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

⇒ f’(x) = 0

⇒ 15x^{2} – 30x – 120 = 0

⇒ 15(x^{2} – 2x – 8) = 0

⇒ 15(x^{2} – 4x + 2x – 8) = 0

⇒ x^{2} – 4x + 2x – 8 = 0

⇒ (x – 4)(x + 2) = 0

⇒ x = 4 , – 2

clearly, f’(x) > 0 if x < –2 and x > 4

and f’(x) < 0 if –2< x < 4

Thus, f(x) increases on (–∞,–2) ∪ (4, ∞)

and f(x) is decreasing on interval x ∈ (–2,4)

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