Verified answer

f'(x) = 2x e^(-x) - (x^2) e^(-x)

= (x e^(-x))(2 - x)

= 0 when x = 0 or 2.

f"(x) = 2 e^(-x) - 2x e^(-x) - 2x e^(-x) + (x^2) e^(-x)

= (e^(-x))(2 - 4x + x^2)

When x = 0, f" = 2 > 0

a) Therefore f(0) = 0 is the minimum value.

When x = 2, f" = -2 e^(-2) < 0

b) Therefore f(2) = 4 e^(-2) is the maximum value.

c) f increases most rapidly where f' has its maximum value,

This is where f" = 0

therefore 2 - 4x + x^2 = 0

x = 2 - sqrt(2) [another solution is 2+sqrt(2) but then f' < 0 so f is decreasing, not increasing]