Question

Find the intervals in which f(x)=2x³-3x²-36x+25 is (1) strictly increasing (2) strictly decreasing.

Answer 1

In the attachment I have answered this problem.     Concept:    1. If Derivative of f(x) is positive for all values of x in interval I , then f(x) is strictly increasing in I.     2. If Derivative of f(x) is negative for all values of x in interval I , then f(x) is strictly decreasing in I.          See the attachment for detailed solution.

Answer 2

It is given that function f(x) = 2x³ – 3x² – 36x + 25

⇒ f’(x) = 6x² – 6x + 36

⇒ f’(x) = 6(x² – x + 6)

⇒ f’(x) = 6(x + 2)(x – 3)

If f’(x) = 0, then we get,

⇒ x = -2, 3

So, the points x = -2 and x = 3 divides the real line into two disjoint intervals, (-∞,2), (-2,3) and (3,∞)

So, in interval (-∞ , -2) , (3, ∞)
f’(x) = 6(x + 2)(x – 3) > 0.
Therefore, the given function (f) is strictly increasing in interval.

So, in interval (-2, 3)
f’(x) = 6(x + 2)(x – 3) < 0
Therefore, the given function (f) is strictly decreasing in interval.