Showing a function does not have a local max or min












3












$begingroup$


The function is



$f(x) = begin{cases} x^2sin(1/x) & text{if $x neq 0$} \
0 & text{if $x = 0$} end{cases} $



I have proved that $f$ is differentiable at $0$, and $f^{'}(0) = 0$.



Now I have to show that $f$ has neither a local max nor local min at $0$.
I know that by showing that if there is no sign change around $0$, then $f$ has no local max or min.



However, I are there other ways to prove this? I was thinking about how if proving if $f$ is either increasing, decreasing, or constant around $0$, then $f$ does not have any local max or min at $0$ .






real-analysis







share|cite|improve this question









$endgroup$












  • $begingroup$
    If there is no sign change in $f(x)$ than you either have a maximum or a minimum in your case
    $endgroup$
    – Andrei
    Dec 3 '18 at 22:44
















3












$begingroup$


The function is



$f(x) = begin{cases} x^2sin(1/x) & text{if $x neq 0$} \
0 & text{if $x = 0$} end{cases} $



I have proved that $f$ is differentiable at $0$, and $f^{'}(0) = 0$.



Now I have to show that $f$ has neither a local max nor local min at $0$.
I know that by showing that if there is no sign change around $0$, then $f$ has no local max or min.



However, I are there other ways to prove this? I was thinking about how if proving if $f$ is either increasing, decreasing, or constant around $0$, then $f$ does not have any local max or min at $0$ .






real-analysis







share|cite|improve this question









$endgroup$












  • $begingroup$
    If there is no sign change in $f(x)$ than you either have a maximum or a minimum in your case
    $endgroup$
    – Andrei
    Dec 3 '18 at 22:44














3












3








3


1



$begingroup$


The function is



$f(x) = begin{cases} x^2sin(1/x) & text{if $x neq 0$} \
0 & text{if $x = 0$} end{cases} $



I have proved that $f$ is differentiable at $0$, and $f^{'}(0) = 0$.



Now I have to show that $f$ has neither a local max nor local min at $0$.
I know that by showing that if there is no sign change around $0$, then $f$ has no local max or min.



However, I are there other ways to prove this? I was thinking about how if proving if $f$ is either increasing, decreasing, or constant around $0$, then $f$ does not have any local max or min at $0$ .






real-analysis







share|cite|improve this question









$endgroup$




The function is



$f(x) = begin{cases} x^2sin(1/x) & text{if $x neq 0$} \
0 & text{if $x = 0$} end{cases} $



I have proved that $f$ is differentiable at $0$, and $f^{'}(0) = 0$.



Now I have to show that $f$ has neither a local max nor local min at $0$.
I know that by showing that if there is no sign change around $0$, then $f$ has no local max or min.



However, I are there other ways to prove this? I was thinking about how if proving if $f$ is either increasing, decreasing, or constant around $0$, then $f$ does not have any local max or min at $0$ .






real-analysis




real-analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Dec 3 '18 at 22:26









Jeffry SantosaJeffry Santosa

384




384












  • $begingroup$
    If there is no sign change in $f(x)$ than you either have a maximum or a minimum in your case
    $endgroup$
    – Andrei
    Dec 3 '18 at 22:44


















  • $begingroup$
    If there is no sign change in $f(x)$ than you either have a maximum or a minimum in your case
    $endgroup$
    – Andrei
    Dec 3 '18 at 22:44
















$begingroup$
If there is no sign change in $f(x)$ than you either have a maximum or a minimum in your case
$endgroup$
– Andrei
Dec 3 '18 at 22:44




$begingroup$
If there is no sign change in $f(x)$ than you either have a maximum or a minimum in your case
$endgroup$
– Andrei
Dec 3 '18 at 22:44










2 Answers
2






active

oldest

votes


















1












$begingroup$

You can prove this simple by the definition of the local extrema. Function does have a local extrema at the point $x in X$ if there is a neighbourhood $V$, such that $f(x)$ is either the minimum of $f(V)$ or the maximum of $f(V)$.



Let $S_delta(x) = { y mid |y_i - x_i| leq delta }$.



You can show, that for any $delta > 0, f(S_delta(x))$ will contain both strictly positive and strictly negative values?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So, by proving that since, for any $delta gt 0$, $f(S_delta(x))$ contains both strictly positive and negative values, it is possible to say that $f$ does not have either local maxima or minima?
    $endgroup$
    – Jeffry Santosa
    Dec 4 '18 at 0:23












  • $begingroup$
    No, but $f$ is differentiable at 0 and its derivative at 0 is 0. If the function $f$ had an extremum at 0, its value would be equal to 0. Thus, proving the function has both strictly positive and strictly negative values for each neighbourhood of 0 gives you, that 0 is not a local extrema.
    $endgroup$
    – Jonáš Kulhánek
    Dec 4 '18 at 7:51



















0












$begingroup$

Consider $f(frac 2 {(2n+1) pi})=(frac 2 {(2n+1) pi} )^{2} (-1)^{n}$. You can see that every interval around $0$ has points where the value of $f$ is greater than $f(0)$ and points where the value of $f$ is less than $f(0)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I apologize first, but I don't quite understand the answer. So does this imply that if there are points around $0$ where the value of $f$ is both greater and lower than $f(0)$, then $f$ has neither local maxima nor minima?
    $endgroup$
    – Jeffry Santosa
    Dec 4 '18 at 0:19








  • 1




    $begingroup$
    Isn't this clear from definition of local maximum and local minimum? If $f$ has local maximum at $0$ there there exists $r>0$ such that $f(x) leq f(0)$ for all $x in (-r,r)$. I have shown that there is always some $x$ in this interval such that $f(x) >f(0)$ so we have a contradiction.
    $endgroup$
    – Kavi Rama Murthy
    Dec 4 '18 at 0:22











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2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1












$begingroup$

You can prove this simple by the definition of the local extrema. Function does have a local extrema at the point $x in X$ if there is a neighbourhood $V$, such that $f(x)$ is either the minimum of $f(V)$ or the maximum of $f(V)$.



Let $S_delta(x) = { y mid |y_i - x_i| leq delta }$.



You can show, that for any $delta > 0, f(S_delta(x))$ will contain both strictly positive and strictly negative values?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So, by proving that since, for any $delta gt 0$, $f(S_delta(x))$ contains both strictly positive and negative values, it is possible to say that $f$ does not have either local maxima or minima?
    $endgroup$
    – Jeffry Santosa
    Dec 4 '18 at 0:23












  • $begingroup$
    No, but $f$ is differentiable at 0 and its derivative at 0 is 0. If the function $f$ had an extremum at 0, its value would be equal to 0. Thus, proving the function has both strictly positive and strictly negative values for each neighbourhood of 0 gives you, that 0 is not a local extrema.
    $endgroup$
    – Jonáš Kulhánek
    Dec 4 '18 at 7:51
















1












$begingroup$

You can prove this simple by the definition of the local extrema. Function does have a local extrema at the point $x in X$ if there is a neighbourhood $V$, such that $f(x)$ is either the minimum of $f(V)$ or the maximum of $f(V)$.



Let $S_delta(x) = { y mid |y_i - x_i| leq delta }$.



You can show, that for any $delta > 0, f(S_delta(x))$ will contain both strictly positive and strictly negative values?






share|cite|improve this answer











$endgroup$













  • $begingroup$
    So, by proving that since, for any $delta gt 0$, $f(S_delta(x))$ contains both strictly positive and negative values, it is possible to say that $f$ does not have either local maxima or minima?
    $endgroup$
    – Jeffry Santosa
    Dec 4 '18 at 0:23












  • $begingroup$
    No, but $f$ is differentiable at 0 and its derivative at 0 is 0. If the function $f$ had an extremum at 0, its value would be equal to 0. Thus, proving the function has both strictly positive and strictly negative values for each neighbourhood of 0 gives you, that 0 is not a local extrema.
    $endgroup$
    – Jonáš Kulhánek
    Dec 4 '18 at 7:51














1












1








1





$begingroup$

You can prove this simple by the definition of the local extrema. Function does have a local extrema at the point $x in X$ if there is a neighbourhood $V$, such that $f(x)$ is either the minimum of $f(V)$ or the maximum of $f(V)$.



Let $S_delta(x) = { y mid |y_i - x_i| leq delta }$.



You can show, that for any $delta > 0, f(S_delta(x))$ will contain both strictly positive and strictly negative values?






share|cite|improve this answer











$endgroup$



You can prove this simple by the definition of the local extrema. Function does have a local extrema at the point $x in X$ if there is a neighbourhood $V$, such that $f(x)$ is either the minimum of $f(V)$ or the maximum of $f(V)$.



Let $S_delta(x) = { y mid |y_i - x_i| leq delta }$.



You can show, that for any $delta > 0, f(S_delta(x))$ will contain both strictly positive and strictly negative values?







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Dec 4 '18 at 0:16









amWhy

1




1










answered Dec 3 '18 at 22:45









Jonáš KulhánekJonáš Kulhánek

729




729












  • $begingroup$
    So, by proving that since, for any $delta gt 0$, $f(S_delta(x))$ contains both strictly positive and negative values, it is possible to say that $f$ does not have either local maxima or minima?
    $endgroup$
    – Jeffry Santosa
    Dec 4 '18 at 0:23












  • $begingroup$
    No, but $f$ is differentiable at 0 and its derivative at 0 is 0. If the function $f$ had an extremum at 0, its value would be equal to 0. Thus, proving the function has both strictly positive and strictly negative values for each neighbourhood of 0 gives you, that 0 is not a local extrema.
    $endgroup$
    – Jonáš Kulhánek
    Dec 4 '18 at 7:51


















  • $begingroup$
    So, by proving that since, for any $delta gt 0$, $f(S_delta(x))$ contains both strictly positive and negative values, it is possible to say that $f$ does not have either local maxima or minima?
    $endgroup$
    – Jeffry Santosa
    Dec 4 '18 at 0:23












  • $begingroup$
    No, but $f$ is differentiable at 0 and its derivative at 0 is 0. If the function $f$ had an extremum at 0, its value would be equal to 0. Thus, proving the function has both strictly positive and strictly negative values for each neighbourhood of 0 gives you, that 0 is not a local extrema.
    $endgroup$
    – Jonáš Kulhánek
    Dec 4 '18 at 7:51
















$begingroup$
So, by proving that since, for any $delta gt 0$, $f(S_delta(x))$ contains both strictly positive and negative values, it is possible to say that $f$ does not have either local maxima or minima?
$endgroup$
– Jeffry Santosa
Dec 4 '18 at 0:23






$begingroup$
So, by proving that since, for any $delta gt 0$, $f(S_delta(x))$ contains both strictly positive and negative values, it is possible to say that $f$ does not have either local maxima or minima?
$endgroup$
– Jeffry Santosa
Dec 4 '18 at 0:23














$begingroup$
No, but $f$ is differentiable at 0 and its derivative at 0 is 0. If the function $f$ had an extremum at 0, its value would be equal to 0. Thus, proving the function has both strictly positive and strictly negative values for each neighbourhood of 0 gives you, that 0 is not a local extrema.
$endgroup$
– Jonáš Kulhánek
Dec 4 '18 at 7:51




$begingroup$
No, but $f$ is differentiable at 0 and its derivative at 0 is 0. If the function $f$ had an extremum at 0, its value would be equal to 0. Thus, proving the function has both strictly positive and strictly negative values for each neighbourhood of 0 gives you, that 0 is not a local extrema.
$endgroup$
– Jonáš Kulhánek
Dec 4 '18 at 7:51











0












$begingroup$

Consider $f(frac 2 {(2n+1) pi})=(frac 2 {(2n+1) pi} )^{2} (-1)^{n}$. You can see that every interval around $0$ has points where the value of $f$ is greater than $f(0)$ and points where the value of $f$ is less than $f(0)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I apologize first, but I don't quite understand the answer. So does this imply that if there are points around $0$ where the value of $f$ is both greater and lower than $f(0)$, then $f$ has neither local maxima nor minima?
    $endgroup$
    – Jeffry Santosa
    Dec 4 '18 at 0:19








  • 1




    $begingroup$
    Isn't this clear from definition of local maximum and local minimum? If $f$ has local maximum at $0$ there there exists $r>0$ such that $f(x) leq f(0)$ for all $x in (-r,r)$. I have shown that there is always some $x$ in this interval such that $f(x) >f(0)$ so we have a contradiction.
    $endgroup$
    – Kavi Rama Murthy
    Dec 4 '18 at 0:22
















0












$begingroup$

Consider $f(frac 2 {(2n+1) pi})=(frac 2 {(2n+1) pi} )^{2} (-1)^{n}$. You can see that every interval around $0$ has points where the value of $f$ is greater than $f(0)$ and points where the value of $f$ is less than $f(0)$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I apologize first, but I don't quite understand the answer. So does this imply that if there are points around $0$ where the value of $f$ is both greater and lower than $f(0)$, then $f$ has neither local maxima nor minima?
    $endgroup$
    – Jeffry Santosa
    Dec 4 '18 at 0:19








  • 1




    $begingroup$
    Isn't this clear from definition of local maximum and local minimum? If $f$ has local maximum at $0$ there there exists $r>0$ such that $f(x) leq f(0)$ for all $x in (-r,r)$. I have shown that there is always some $x$ in this interval such that $f(x) >f(0)$ so we have a contradiction.
    $endgroup$
    – Kavi Rama Murthy
    Dec 4 '18 at 0:22














0












0








0





$begingroup$

Consider $f(frac 2 {(2n+1) pi})=(frac 2 {(2n+1) pi} )^{2} (-1)^{n}$. You can see that every interval around $0$ has points where the value of $f$ is greater than $f(0)$ and points where the value of $f$ is less than $f(0)$.






share|cite|improve this answer









$endgroup$



Consider $f(frac 2 {(2n+1) pi})=(frac 2 {(2n+1) pi} )^{2} (-1)^{n}$. You can see that every interval around $0$ has points where the value of $f$ is greater than $f(0)$ and points where the value of $f$ is less than $f(0)$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 4 '18 at 0:14









Kavi Rama MurthyKavi Rama Murthy

54.2k32055




54.2k32055












  • $begingroup$
    I apologize first, but I don't quite understand the answer. So does this imply that if there are points around $0$ where the value of $f$ is both greater and lower than $f(0)$, then $f$ has neither local maxima nor minima?
    $endgroup$
    – Jeffry Santosa
    Dec 4 '18 at 0:19








  • 1




    $begingroup$
    Isn't this clear from definition of local maximum and local minimum? If $f$ has local maximum at $0$ there there exists $r>0$ such that $f(x) leq f(0)$ for all $x in (-r,r)$. I have shown that there is always some $x$ in this interval such that $f(x) >f(0)$ so we have a contradiction.
    $endgroup$
    – Kavi Rama Murthy
    Dec 4 '18 at 0:22


















  • $begingroup$
    I apologize first, but I don't quite understand the answer. So does this imply that if there are points around $0$ where the value of $f$ is both greater and lower than $f(0)$, then $f$ has neither local maxima nor minima?
    $endgroup$
    – Jeffry Santosa
    Dec 4 '18 at 0:19








  • 1




    $begingroup$
    Isn't this clear from definition of local maximum and local minimum? If $f$ has local maximum at $0$ there there exists $r>0$ such that $f(x) leq f(0)$ for all $x in (-r,r)$. I have shown that there is always some $x$ in this interval such that $f(x) >f(0)$ so we have a contradiction.
    $endgroup$
    – Kavi Rama Murthy
    Dec 4 '18 at 0:22
















$begingroup$
I apologize first, but I don't quite understand the answer. So does this imply that if there are points around $0$ where the value of $f$ is both greater and lower than $f(0)$, then $f$ has neither local maxima nor minima?
$endgroup$
– Jeffry Santosa
Dec 4 '18 at 0:19






$begingroup$
I apologize first, but I don't quite understand the answer. So does this imply that if there are points around $0$ where the value of $f$ is both greater and lower than $f(0)$, then $f$ has neither local maxima nor minima?
$endgroup$
– Jeffry Santosa
Dec 4 '18 at 0:19






1




1




$begingroup$
Isn't this clear from definition of local maximum and local minimum? If $f$ has local maximum at $0$ there there exists $r>0$ such that $f(x) leq f(0)$ for all $x in (-r,r)$. I have shown that there is always some $x$ in this interval such that $f(x) >f(0)$ so we have a contradiction.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 0:22




$begingroup$
Isn't this clear from definition of local maximum and local minimum? If $f$ has local maximum at $0$ there there exists $r>0$ such that $f(x) leq f(0)$ for all $x in (-r,r)$. I have shown that there is always some $x$ in this interval such that $f(x) >f(0)$ so we have a contradiction.
$endgroup$
– Kavi Rama Murthy
Dec 4 '18 at 0:22


















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