I have to determine sup and inf of the following function:
$$\ \frac{(\pi^2-x^2)\sin(x)}{1+\cos(x)} ~~~~~~~~ f:(-\pi, \pi) \rightarrow ~\mathbb{R}$$
1) I thought of derivatiove and comparison to $0$, however this equation seems to be too complicated, there have to be an easier option.
2) I thought of Taylor series in $\ \pi$
3) Result of my thoughts: I'm stuck.
I would appreciate explanation.
Take the derivative. You'll see that you can simplify its numerator with $\sin^2(x)+\cos^2(x)=1$ and then divide the numerator and denominator by $1+\cos(x)$ to arrive at $$f'(x)=\frac{-x^2-2x\sin(x)+\pi^2}{1+\cos(x)} $$ The numerator is $-x^2-2x\sin(x)+\pi^2=\pi^2+\sin^2(x)-(x+\sin(x))^2$. Since $g(x)=x+\sin(x)$ is an increasing function (check by taking the derivative), $-\pi=g(-\pi)<g(x)<g(\pi)=\pi$ for $x\in(-\pi,\pi)$. Thus, $(x+\sin(x))<\pi^2$ which means that the numerator is positive, $f'(x)>0$ for $x\in(-\pi,\pi)$, and $f$ is strictly increasing in the interval. So $$\sup\{f(x):-\pi<x<\pi\}=\lim_{x\to \pi^-}f(x) \\ \inf\{f(x):-\pi<x<\pi\}=\lim_{x\to -\pi^+}f(x) $$ Update: Since $f$ is odd on the interval, the second limit is the negative of the first limit. For the first limit, make the substitution $y=\pi-x\to 0^+$ as $x\to \pi^-$. So $x=\pi-y$, and the limit becomes $$\lim_{x\to \pi^-}f(x)=\lim_{y\to 0^+}\frac{y(2\pi-y)\sin(y)}{1-\cos(y)}=2\pi\lim_{y\to 0^+}\frac{y^2}{1-\cos y}\lim_{y\to 0^+}\frac{\sin y}{y} $$ Can you find the last two limits?