How to find local maximum and minimum of function $f_{n} = x^{n} \sin x$ at $x=0$

How to find local maximum and minimum of function $f_{n} = x^{n} \sin x$ at $x=0$? Where $n≥2$.

I tried to find local Maxima or minima by finding the critical points, but I'm getting no critical points since $f'_{n}$ is $0$ at $x=0$.

Moreover, the local Maxima and local minima should also depend on nature of $n$, it it is odd or even.

What is the method to find local extrema of such functions?

2 Answers

If you draw a plot of $x^2\sin x$, you will see it has no minimum or maximum at $x=0$. Neither $x^{2n} \sin x$. However, $x^{2n+1} \sin x$ reaches minimum at $x=0$

Calculate $f''$ and use property that $f''(x)$ is negative at $x=x_0$ if it's maximum at $x_0$, positive in case of minimum and equals zero in case of inflection point. Note: this is not always true, but in your case it's ok*. (see e.g. )

UPD: *in this case it is not. As Silent pointed, $f''(0)=0$, so other methods should be used (e.g. proving that $f_n(x_1)>f_n(x_0)<f_n(x_2), x_1<x_0<x_2)$) - see answer below.

You should check if $f_n$ is odd or even function. As it turns out, $f_n$ even for $n$ odd and vice versa. Also, note that for $0<x<\pi$, $f_n(x)>0$ for any $n$. And $f_n(0)=0$. Combining this with continuity of $f$, we see that $f_n$ has local min at 0 for $n$ odd, and saddle point at zero for $n$ even.

Calculating second derivative will be fatal, since in any case it is zero at point $0$.