Q. 5

# Find the interval

(a) It is given that function f(x) = 2x3 – 3x2 – 36x + 7

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 f’(x) = 6(x + 2)(x – 3) >0

Therefore, the given function (f) is strictly increasing in interval .

(b) It is given that function f(x) = 2x3 – 3x2 – 36x + 7

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 f’(x) = 6(x + 2)(x – 3) < 0

Therefore, the given function (f) is strictly decreasing in interval .

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