If $f(x) =\int\limits_0^x(t+1) (e^t –1) (t – 2) (t + 4) dt$ then f(x) would assume the local minima at

x = - 4 x = 0 x = -1 none of these

x = -1

Here, $f'(x)= (x + 1) (e^x – 1) (x – 2) ( x + 4)$ Clearly x = -1 and x = 2 are the points of local minima. Hence (C) is the correct answer.