Answer
A. The domain is $(-\infty, \infty)$
B. The y-intercept is $0$
The x-intercept is $0$
C. The function is an odd function.
D. $\lim\limits_{x \to -\infty} (\frac{1}{5}x^5-\frac{8}{3}x^3+16x) = -\infty$
$\lim\limits_{x \to \infty} (\frac{1}{5}x^5-\frac{8}{3}x^3+16x) = \infty$
There are no asymptotes.
E. The function is increasing on the intervals $(-\infty, -2)\cup (-2, 2)\cup(2,\infty)$
F. There is no local maximum or local minimum.
G. The graph is concave down on the intervals $(-\infty, -2)\cup (0,2)$
The graph is concave up on the intervals $(-2, 0) \cup (2, \infty)$
The points of inflection are $(-2, -17.1),(0,0),$ and $(2,17.1)$
H. We can see a sketch of the curve below.
Work Step by Step
$y = \frac{1}{5}x^5-\frac{8}{3}x^3+16x$
A. The function is defined for all real numbers. The domain is $(-\infty, \infty)$
B. When $x = 0,$ then $y = \frac{1}{5}(0)^5-\frac{8}{3}(0)^3+16(0)= 0$
The y-intercept is $0$
When $y = 0$:
$\frac{1}{5}x^5-\frac{8}{3}x^3+16x = 0$
$x(\frac{1}{5}x^4-\frac{8}{3}x^2+16)=0$
$x = 0$
The x-intercept is $0$
C. $\frac{1}{5}(-x)^5-\frac{8}{3}(-x)^3+16(-x) = -(\frac{1}{5}x^5-\frac{8}{3}x^3+16x)$
The function is an odd function.
D. $\lim\limits_{x \to -\infty} (\frac{1}{5}x^5-\frac{8}{3}x^3+16x) = -\infty$
$\lim\limits_{x \to \infty} (\frac{1}{5}x^5-\frac{8}{3}x^3+16x) = \infty$
There are no asymptotes.
E. We can find the values of $x$ such that $y' = 0$:
$y' = x^4-8x^2+16= (x^2-4)^2 = 0$
$x=-2,2$
When $x \lt -2$ or $-2 \lt x \lt 2$ or $x \gt 2$, then $y' \gt 0$
The function is increasing on the intervals $(-\infty, -2)\cup (-2, 2)\cup(2,\infty)$
F. There is no local maximum or local minimum because the curve is increasing on all intervals.
G. We can find the values of $x$ such that $y'' = 0$:
$y'' = 2(x^2-4)(2x) = (4x)(x^2-4) = 0$
$x = -2, 0, 2$
When $x \lt -2~~$ or $~~0 \lt x \lt 2~~$, then $y'' \lt 0$
The graph is concave down on the intervals $(-\infty, -2)\cup (0,2)$
When $-2 \lt x \lt 0$ or $x \gt 2$, then $y'' \gt 0$
The graph is concave up on the intervals $(-2, 0) \cup (2, \infty)$
When $x = -2$, then $y = \frac{1}{5}(-2)^5-\frac{8}{3}(-2)^3+16(-2) = -17.1$
When $x = 0$, then $y = \frac{1}{5}(0)^5-\frac{8}{3}(0)^3+16(0) = 0$
When $x = 2$, then $y = \frac{1}{5}(2)^5-\frac{8}{3}(2)^3+16(2) = 17.1$
The points of inflection are $(-2, -17.1), (0,0),$ and $(2,17.1)$
H. We can see a sketch of the curve below.
