Q25 of 137 Page 19

#Mark the correct alternative in each of the following

The function f(x) = 2x3 – 15x2 + 36x + 4 is maximum at x =


f(x) = 2x3 – 15x2 + 36x + 4

Differentiating f(x) with respect to x, we get


f’(x)= 6x2 - 30x + 36=6(x-2)(x-3)


Differentiating f’(x) with respect to x, we get


f’’(x)=12x – 30


for maxima at x=c, f’(c)=0 and f’’(c)<0


f’(x)=0 ⇒ x=2 or x=3


f’’(2)=-6<0 and f’’(3)=6>0


Hence x=2 is a point of maxima.

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