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?

  • $\begingroup$ Do you mean near x=0? $\endgroup$ – Peter Foreman Jan 8 at 17:25
  • $\begingroup$ 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. $\endgroup$ – Dave L. Renfro Jan 8 at 17:35
  • $\begingroup$ @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. $\endgroup$ – Mathsaddict Jan 8 at 17:46
  • $\begingroup$ @Dave L. Renfro I'm stuck on finding such intervals. $\endgroup$ – Mathsaddict Jan 8 at 17:47
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    $\begingroup$ 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). $\endgroup$ – 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 $f_n(x_1)>f_n(x_0)<f_n(x_2), x_1<x_0<x_2)$) - see answer below.

  • $\begingroup$ Calculating second derivative does not help here!! $\endgroup$ – Silent Feb 9 at 7:14

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 $0$.


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