Answer
A. The domain is $(-\infty, \infty)$
B. The y-intercept is $0$
The x-intercepts are $\pi~n,$ where $n$ is an integer
C. The function is an odd function.
The function is periodic with a period of $2\pi$
D. There are no asymptotes.
E. The function is decreasing on the intervals $(\frac{\pi}{2}, \pi)\cup (\pi, \frac{3\pi}{2})$
The function is increasing on the intervals $(0, \frac{\pi}{2})\cup (\frac{3\pi}{2}, 2\pi)$
F. The local maxima are $(\frac{\pi}{2} +2\pi~n, 1)$
The local minima are $(\frac{3\pi}{2} +2\pi~n, -1)$
G. The graph is concave down on the intervals $(0.96, 2.19)\cup (\pi, 4.10)\cup (5.33, 2\pi)$
The graph is concave up on the intervals $(0, 0.96)\cup (2.19, \pi)\cup (4.10, 5.33)$
The points of inflection are $(0, 0), (0.96, 0.55), (2.19, 0.55),(\pi,0), (4.10, -0.55),$ and $(5.35,-0.55)$
H. We can see a sketch of the curve below.
Work Step by Step
$y = sin^3~x$
A. The function is defined for all real numbers.
The domain is $(-\infty, \infty)$
B. When $x=0$, then $y = sin^3~0 = 0$
The y-intercept is $0$
When $y = 0$:
$sin^3~x = 0$
$sin~x = 0$
$x = \pi~n,$ where $n$ is an integer
The x-intercepts are $\pi~n,$ where $n$ is an integer
C. $sin^3~(-x) = -sin^3~x$
The function is an odd function.
Since $~~sin~x~~$ is periodic with a period of $2\pi$, then $~~y = sin^3~x~~$ is periodic with a period of $2\pi$
D. We can consider the limits $\lim\limits_{x \to -\infty} sin^3~x$ and $\lim\limits_{x \to \infty} sin^3~x$
The function does not converge to one value as $~~x\to -\infty~~$ and $~~x \to \infty,~~$ so these limits do not exist.
There are no asymptotes.
E. We can find values of $x$ such that $y' = 0$:
$y' =3~sin^2~x~cos~x = 0$
$sin~x = 0$ or $cos~x = 0$
$x = \pi~n,$ where $n$ is an integer
$x = \frac{\pi}{2}+\pi~n,$ where $n$ is an integer
Therefore:
$x = \frac{\pi}{2}~n,$ where $n$ is an integer
When $\frac{\pi}{2} \lt x \lt \pi$ or $\pi \lt x \lt \frac{3\pi}{2}$, then $y' \lt 0$
The function is decreasing on the intervals $(\frac{\pi}{2}, \pi)\cup (\pi, \frac{3\pi}{2})$
When $0 \lt x \lt \frac{\pi}{2}$ or $\frac{3\pi}{2} \lt x \lt 2\pi$, then $y' \gt 0$
The function is increasing on the intervals $(0, \frac{\pi}{2})\cup (\frac{3\pi}{2}, 2\pi)$
F. When $x = \frac{\pi}{2} +2\pi~n$ then $y = sin^3~(\frac{\pi}{2}+2\pi~n) = 1$
The local maxima are $(\frac{\pi}{2} +2\pi~n, 1)$
When $x = \frac{3\pi}{2} +2\pi~n$ then $y = sin^3~(\frac{3\pi}{2}+2\pi~n) = -1$
The local minima are $(\frac{3\pi}{2} +2\pi~n, -1)$
G. We can find the values of $x$ such that $y'' = 0$:
$y'' =6~sin~x~cos^2~x - 3~sin^3~x = 0$
$6~sin~x~cos^2~x = 3~sin^3~x$
$sin~x = 0$ or $2~cos^2~x = sin^2~x$
$sin~x = 0$ or $tan^2~x = 2$
$sin~x = 0$ or $tan~x = \pm~\sqrt{2}$
$x = \pi n, 0.96+2\pi~n, 2.19+2\pi~n, 4.10+2\pi~n, 5.33+2\pi~n$ where $n$ is an integer
The graph is concave down on the intervals $(0.96, 2.19)\cup (\pi, 4.10)\cup (5.33, 2\pi)$
The graph is concave up on the intervals $(0, 0.96)\cup (2.19, \pi)\cup (4.10, 5.33)$
When $x= 0$, then $y = sin^3~0 = 0$
When $x= 0.96$, then $y = sin^3~0.96 = 0.55$
When $x= 2.19$, then $y = sin^3~2.19 = 0.55$
When $x= \pi$, then $y = sin^3~\pi = 0$
When $x= 4.10$, then $y = sin^3~4.10 = -0.55$
When $x= 5.33$, then $y = sin^3~5.33 = -0.55$
The points of inflection are $(0, 0), (0.96, 0.55), (2.19, 0.55),(\pi,0), (4.10, -0.55),$ and $(5.35,-0.55)$
Note that the points of inflection repeat periodically with a period of $2\pi$
H. We can see a sketch of the curve below.
