The second derivative will be (3x^2-3*x-6), the function will be convex if f''(x)>0, now this will be possible if either both (x-2) and (x+1) are positive or they are negative, now consider two cases:
Case 1: if both are positive then x>2, and x>-1, the set of points are (2,inf) intersection (-1,inf)=(-1,2).
Case 2: if both are negative then, x<2 and x<-1, the set will be (-inf,2) int (-inf,-1)=(-inf,-1)
Now it is clear that the function is convex on (-1,2) by Case 1, so it cannot be concave in that region so we rule out option b as it says the function is concave on (-1,2)
Now a can be ruled out since in the interval (-inf,2) f''(x) is not always negative, take x=-3,
Also d can be ruled out because in the interval (2,inf) f''(x)>0, so it cannot be negative and hence cannot be concave too.
Option c satisfies all the criterion, in the interval (-1,2) the function will be convex, as given in Case 1, also in the interval (-inf,-1), f''(x)<0.
so answer is option c.
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