#Mark the correct alternative in each of the following
Let f(x) = 2x3 – 3x2 – 12x + 5 on [–2, 4]. The relative maximum occurs at x =
f(x) = 2x3 – 3x2 – 12x + 5, x∈[-2,4]
Differentiating f(x) with respect to x, we get
f’(x)= 6x2 – 6x – 12=6(x+1)(x-2)
Differentiating f’(x) with respect to x, we get
f’’(x)=12x-6
For maxima at x=c, f’(c)=0 and f’’(c)<0
f’(x)=0 ⇒ x=-1 or 2
f’’(-1)=-18<0 and f’’(2)=18>0
Hence, x=-1 is the point of local maxima.