## Calculus: Early Transcendentals 8th Edition

(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 $(6, \frac{1296}{e^6})$
(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 $(6, \frac{1296}{e^6})$