#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.