Calculus: Early Transcendentals, Chapter 4, 4.3, Section 4.3, Problem 18

f is defined everywhere and is infinite differentiable.
(a) f'(x) = e^(-x)*(4x^3 - x^4) = x^3*e^(-x)*(4-x).
It is positive for x on (0, 4) and f is increasing there, and negative on (-oo, 0) and on (4, +oo) , f is decreasing there.
(b) therefore f has a minimum at x=0, f(0)=0, and naximum at x=4, f(4)=(4/e)^4 approx4,69.
(c) f''(x) = e^(-x)*(12x^2 - 4x^3 - (4x^3 - x^4)) = e^(-x)*(x^4 - 8x^3 + 12x^2) = x^2*e^(-x)*(x^2 - 8x + 12) = x^2*e^(-x)*(x-2)*(x-6).
It is negative for x on (2, 6) where f is concave downward, and positive for x on (-oo, 2) and (6, +oo), f is concave upward there.
2 and 6 are inflection points, 0 is not (f'' doesn't change sign).