What will be the minimum condition on $f_n(x)=x^n \sin x$

Question

I have this function $f_n(x)=x^n \sin x$. It is required to show that the function will have minimum when $n$ is odd. I took the derivative:

$f'(x)=x^n \cos x+n \sin x \cdot x^{n-1}$ But I am stuck here. Any help would be good. Thanks

Answer 1

if $n$ is odd then $\forall \ x \ f(-x) = f(x)$. Consider the neighbourhood of $(0, 0)$ $f(\frac{\pi}{2}) = f(-\frac{\pi}{2}) = (\frac{\pi}{2})^n$ and since $f(x)$ is continuous and differentiable on $(-\frac{\pi}{2}, \frac{\pi}{2})$ according to Rolle's theorem $\exists \ \xi \in (-\frac{\pi}{2}, \frac{\pi}{2})$ such that $f'(\xi) = 0$ and since $f(\frac{\pi}{2}) > f(0) \rightarrow \xi$ is a local minimum