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












1












$begingroup$


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?






calculus







share|cite|improve this question









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  • $begingroup$
    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 $x^{n-1}(nsin x + x cos x)$ is positive and the intervals on which $x^{n-1}(nsin x + x cos x)$ 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






  • 1




    $begingroup$
    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


















1












$begingroup$


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?






calculus







share|cite|improve this question









$endgroup$












  • $begingroup$
    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 $x^{n-1}(nsin x + x cos x)$ is positive and the intervals on which $x^{n-1}(nsin x + x cos x)$ 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






  • 1




    $begingroup$
    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
















1












1








1





$begingroup$


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?






calculus







share|cite|improve this question









$endgroup$




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?






calculus




calculus






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 8 at 17:09









MathsaddictMathsaddict

3459




3459












  • $begingroup$
    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 $x^{n-1}(nsin x + x cos x)$ is positive and the intervals on which $x^{n-1}(nsin x + x cos x)$ 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






  • 1




    $begingroup$
    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




















  • $begingroup$
    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 $x^{n-1}(nsin x + x cos x)$ is positive and the intervals on which $x^{n-1}(nsin x + x cos x)$ 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






  • 1




    $begingroup$
    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


















$begingroup$
Do you mean near x=0?
$endgroup$
– Peter Foreman
Jan 8 at 17:25




$begingroup$
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 $x^{n-1}(nsin x + x cos x)$ is positive and the intervals on which $x^{n-1}(nsin x + x cos x)$ is negative.
$endgroup$
– Dave L. Renfro
Jan 8 at 17:35






$begingroup$
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 $x^{n-1}(nsin x + x cos x)$ is positive and the intervals on which $x^{n-1}(nsin x + x cos x)$ 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




$begingroup$
@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




$begingroup$
@Dave L. Renfro I'm stuck on finding such intervals.
$endgroup$
– Mathsaddict
Jan 8 at 17:47




1




1




$begingroup$
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






$begingroup$
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












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$begingroup$

If you draw a plot of $x^2sin x$, 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 )






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    $begingroup$

    If you draw a plot of $x^2sin x$, 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 )






    share|cite|improve this answer









    $endgroup$


















      3












      $begingroup$

      If you draw a plot of $x^2sin x$, 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 )






      share|cite|improve this answer









      $endgroup$
















        3












        3








        3





        $begingroup$

        If you draw a plot of $x^2sin x$, 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 )






        share|cite|improve this answer









        $endgroup$



        If you draw a plot of $x^2sin x$, 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 )







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 8 at 17:43









        Mike_Mike_

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