#### Answer

(a) $f$ is decreasing on the intervals $(-\infty, 0)\cup (4,\infty)$
$f$ is increasing on the interval $(0,4)$
(b) The local maximum is $f(4) = \frac{256}{e^4}$
The local minimum is $f(0) = 0$
(c) The graph is concave up on the intervals $(-\infty, 2)\cup (6,\infty)$
The graph is concave down on the interval $(2,6)$
The points of inflection are $(2,\frac{16}{e^2})$ and $(6, \frac{1296}{e^6})$

#### Work Step by Step

(a) $f(x) = x^4~e^{-x}$
We can find the points where $f'(x) = 0$:
$f'(x) = 4x^3~e^{-x}-x^4~e^{-x} = 0$
$x^3~e^{-x}~(4-x) = 0$
$x=0~~~$ or $~~~x = 4$
When $x \lt 0~~$ or $~~x \gt 4$ then $f'(x) \lt 0$
$f$ is decreasing on the intervals $(-\infty, 0)\cup (4,\infty)$
When $0 \lt x \lt 4$ then $f'(x) \gt 0$
$f$ is increasing on the interval $(0,4)$
(b) $f(0) = (0)^4~e^{-0} = 0$
$f(4) = (4)^4~e^{-4} = \frac{256}{e^4}$
The local maximum is $f(4) = \frac{256}{e^4}$
The local minimum is $f(0) = 0$
(c) We can find the points where $f''(x) = 0$:
$f''(x) = 12x^2~e^{-x}-4x^3~e^{-x}-4x^3~e^{-x}+x^4~e^{-x} = 0$
$12x^2~e^{-x}-8x^3~e^{-x}+x^4~e^{-x} = 0$
$x^2~e^{-x}~(12-8x+x^2) = 0$
$x^2~e^{-x}~(x-2)(x-6) = 0$
$x = 0, 2, 6$
Note that $x = 0$ is not a point of inflection because the concavity does not change at $x=0$
The graph is concave up when $f''(x) \gt 0$
The graph is concave up on the intervals $(-\infty, 2)\cup (6,\infty)$
The graph is concave down when $f''(x) \lt 0$
The graph is concave down on the interval $(2,6)$
$f(2) = (2)^4~e^{-2} = \frac{16}{e^2}$
$f(6) = (6)^4~e^{-6} = \frac{1296}{e^6}$
The points of inflection are $(2,\frac{16}{e^2})$ and $(6, \frac{1296}{e^6})$