# 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?

• Do you mean near x=0? – Peter Foreman Jan 8 at 17:25
• I don't know what you're trying, but the most basic (and typically first learned) test, namely the first derivative test, leads one to consider the intervals on which $x^{n-1}(n\sin x + x \cos x)$ is positive and the intervals on which $x^{n-1}(n\sin x + x \cos x)$ is negative. – Dave L. Renfro Jan 8 at 17:35
• @Peter Foreman, if it is not possible to find the nature of function at a particular point then we can find its nature in its right and left eighborhood. But I'm unable to do this. – Mathsaddict Jan 8 at 17:46
• @Dave L. Renfro I'm stuck on finding such intervals. – Mathsaddict Jan 8 at 17:47
• You'll have to deal with the transcendental equation $\tan x = -\frac{x}{n}.$ You can determine the approximate location of the roots by examining where the graphs of $y = \tan x$ and $y = -\frac{x}{n}$ intersect. It's probably instructive to first consider the specific special cases $n=1,$ $n=2,$ etc. This discussion of the roots of $\tan x = x$ may be helpful. Note the nonzero roots are transcendental (p. 12 of cited slides) and probably can't be expressed in closed form (p. 13). – Dave L. Renfro Jan 8 at 18:05

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. https://en.wikipedia.org/wiki/Inflection_point )
UPD: *in this case it is not. As Silent pointed, $$f''(0)=0$$, so other methods should be used (e.g. proving that $$f_n(x_1)>f_n(x_0)) - see answer below.
You should check if $$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, $$f_n(x)>0$$ 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 $$0$$.