Question

The graph of a differentiable function f on the closed interval [-4,4] is shown at the right. the graph of f has horizontal tangents at x = -3, -1, and 2.

Let G(x) = (integral from x to -4)f(t)dt for -4<=x<=4

G(-4) = 0
G(-1) = 2
G is concave down (4,-3) and (-1,2) because G prime is negative in these intervals and f is decreasing.

Question:

Find the value of x at which G has its maximum on the closed interval [-4,4]. Justify your answer please.