Q. 5

Find the intervals in which the function f given by f (x) = 2x3 – 3x2 – 36x + 7 is

(a) strictly increasing (b) strictly decreasing

Answer :

(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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