## Calculus: Early Transcendentals 8th Edition

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.
$y = sin^3~x$ 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.