Find where the global max and global min of the function occur (on the specified interval).
f (x) = x2ex on the interval [-5,1]
The derivative is
f'(x) = 2xex + x2ex = xex(2 + x)
so there are critical points at x = -2 and at x = 0.
We evaluate f at the critical points and endpoints (it's ok to use approximate y-values here, because we're only comparing the y-values tofigure out which x-values we want):
f has its global max when x = 0 and its global min when x = 1.