A point where the derivative is $infty$












1












$begingroup$



Let $f:I to mathbb{R}$, where $I$ is an interval and $a in I$ such that $f'(a)= infty$. Prove that there is a neighborhood $(a-epsilon, a+epsilon)$ such that $f$ is convex on $(a-epsilon, a)$ and $f$ is concave on $(a, a+epsilon)$.




I tried to use the definition of the derivative, but I only got that there is a left-neighborhood of $a$ where $f(x)< f(a)$ and a right-neighborhood of $a$ where $f(x)>f(a)$.



I suppose that the statement is false, but I couldn't construct an elementary counter example. Is there any?






real-analysis calculus limits derivatives







share|cite|improve this question









$endgroup$












  • $begingroup$
    Did you try the second derivative?
    $endgroup$
    – J. W. Tanner
    Jan 14 at 13:56










  • $begingroup$
    Depending on what $f(a)=infty$ means, I think the statement might be true. Informally: If the function assumes an infinite value at some point, it must be increasing to the left and decreasing to the right of the point, that comes quite close to the definition of being convex/concave at those intervals.
    $endgroup$
    – Henrik
    Jan 14 at 13:56










  • $begingroup$
    @Henrik, it is $f'(a)=infty$ not $f(a)=infty.$ I.e. $f'(a) = lim_{x to a} frac{f(x)-f(a)}{x-a} = infty$.
    $endgroup$
    – Shroud
    Jan 14 at 13:59












  • $begingroup$
    @J.W.Tanner, yes, but apart from the definition, I don't know how to approach it.
    $endgroup$
    – Shroud
    Jan 14 at 14:01










  • $begingroup$
    Can you show f’ is increasing (so f’’ is positive) on $(a-epsilon, a)?$
    $endgroup$
    – J. W. Tanner
    Jan 14 at 14:08
















1












$begingroup$



Let $f:I to mathbb{R}$, where $I$ is an interval and $a in I$ such that $f'(a)= infty$. Prove that there is a neighborhood $(a-epsilon, a+epsilon)$ such that $f$ is convex on $(a-epsilon, a)$ and $f$ is concave on $(a, a+epsilon)$.




I tried to use the definition of the derivative, but I only got that there is a left-neighborhood of $a$ where $f(x)< f(a)$ and a right-neighborhood of $a$ where $f(x)>f(a)$.



I suppose that the statement is false, but I couldn't construct an elementary counter example. Is there any?






real-analysis calculus limits derivatives







share|cite|improve this question









$endgroup$












  • $begingroup$
    Did you try the second derivative?
    $endgroup$
    – J. W. Tanner
    Jan 14 at 13:56










  • $begingroup$
    Depending on what $f(a)=infty$ means, I think the statement might be true. Informally: If the function assumes an infinite value at some point, it must be increasing to the left and decreasing to the right of the point, that comes quite close to the definition of being convex/concave at those intervals.
    $endgroup$
    – Henrik
    Jan 14 at 13:56










  • $begingroup$
    @Henrik, it is $f'(a)=infty$ not $f(a)=infty.$ I.e. $f'(a) = lim_{x to a} frac{f(x)-f(a)}{x-a} = infty$.
    $endgroup$
    – Shroud
    Jan 14 at 13:59












  • $begingroup$
    @J.W.Tanner, yes, but apart from the definition, I don't know how to approach it.
    $endgroup$
    – Shroud
    Jan 14 at 14:01










  • $begingroup$
    Can you show f’ is increasing (so f’’ is positive) on $(a-epsilon, a)?$
    $endgroup$
    – J. W. Tanner
    Jan 14 at 14:08














1












1








1





$begingroup$



Let $f:I to mathbb{R}$, where $I$ is an interval and $a in I$ such that $f'(a)= infty$. Prove that there is a neighborhood $(a-epsilon, a+epsilon)$ such that $f$ is convex on $(a-epsilon, a)$ and $f$ is concave on $(a, a+epsilon)$.




I tried to use the definition of the derivative, but I only got that there is a left-neighborhood of $a$ where $f(x)< f(a)$ and a right-neighborhood of $a$ where $f(x)>f(a)$.



I suppose that the statement is false, but I couldn't construct an elementary counter example. Is there any?






real-analysis calculus limits derivatives







share|cite|improve this question









$endgroup$





Let $f:I to mathbb{R}$, where $I$ is an interval and $a in I$ such that $f'(a)= infty$. Prove that there is a neighborhood $(a-epsilon, a+epsilon)$ such that $f$ is convex on $(a-epsilon, a)$ and $f$ is concave on $(a, a+epsilon)$.




I tried to use the definition of the derivative, but I only got that there is a left-neighborhood of $a$ where $f(x)< f(a)$ and a right-neighborhood of $a$ where $f(x)>f(a)$.



I suppose that the statement is false, but I couldn't construct an elementary counter example. Is there any?






real-analysis calculus limits derivatives




real-analysis calculus limits derivatives






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 14 at 13:48









ShroudShroud

765149




765149












  • $begingroup$
    Did you try the second derivative?
    $endgroup$
    – J. W. Tanner
    Jan 14 at 13:56










  • $begingroup$
    Depending on what $f(a)=infty$ means, I think the statement might be true. Informally: If the function assumes an infinite value at some point, it must be increasing to the left and decreasing to the right of the point, that comes quite close to the definition of being convex/concave at those intervals.
    $endgroup$
    – Henrik
    Jan 14 at 13:56










  • $begingroup$
    @Henrik, it is $f'(a)=infty$ not $f(a)=infty.$ I.e. $f'(a) = lim_{x to a} frac{f(x)-f(a)}{x-a} = infty$.
    $endgroup$
    – Shroud
    Jan 14 at 13:59












  • $begingroup$
    @J.W.Tanner, yes, but apart from the definition, I don't know how to approach it.
    $endgroup$
    – Shroud
    Jan 14 at 14:01










  • $begingroup$
    Can you show f’ is increasing (so f’’ is positive) on $(a-epsilon, a)?$
    $endgroup$
    – J. W. Tanner
    Jan 14 at 14:08


















  • $begingroup$
    Did you try the second derivative?
    $endgroup$
    – J. W. Tanner
    Jan 14 at 13:56










  • $begingroup$
    Depending on what $f(a)=infty$ means, I think the statement might be true. Informally: If the function assumes an infinite value at some point, it must be increasing to the left and decreasing to the right of the point, that comes quite close to the definition of being convex/concave at those intervals.
    $endgroup$
    – Henrik
    Jan 14 at 13:56










  • $begingroup$
    @Henrik, it is $f'(a)=infty$ not $f(a)=infty.$ I.e. $f'(a) = lim_{x to a} frac{f(x)-f(a)}{x-a} = infty$.
    $endgroup$
    – Shroud
    Jan 14 at 13:59












  • $begingroup$
    @J.W.Tanner, yes, but apart from the definition, I don't know how to approach it.
    $endgroup$
    – Shroud
    Jan 14 at 14:01










  • $begingroup$
    Can you show f’ is increasing (so f’’ is positive) on $(a-epsilon, a)?$
    $endgroup$
    – J. W. Tanner
    Jan 14 at 14:08
















$begingroup$
Did you try the second derivative?
$endgroup$
– J. W. Tanner
Jan 14 at 13:56




$begingroup$
Did you try the second derivative?
$endgroup$
– J. W. Tanner
Jan 14 at 13:56












$begingroup$
Depending on what $f(a)=infty$ means, I think the statement might be true. Informally: If the function assumes an infinite value at some point, it must be increasing to the left and decreasing to the right of the point, that comes quite close to the definition of being convex/concave at those intervals.
$endgroup$
– Henrik
Jan 14 at 13:56




$begingroup$
Depending on what $f(a)=infty$ means, I think the statement might be true. Informally: If the function assumes an infinite value at some point, it must be increasing to the left and decreasing to the right of the point, that comes quite close to the definition of being convex/concave at those intervals.
$endgroup$
– Henrik
Jan 14 at 13:56












$begingroup$
@Henrik, it is $f'(a)=infty$ not $f(a)=infty.$ I.e. $f'(a) = lim_{x to a} frac{f(x)-f(a)}{x-a} = infty$.
$endgroup$
– Shroud
Jan 14 at 13:59






$begingroup$
@Henrik, it is $f'(a)=infty$ not $f(a)=infty.$ I.e. $f'(a) = lim_{x to a} frac{f(x)-f(a)}{x-a} = infty$.
$endgroup$
– Shroud
Jan 14 at 13:59














$begingroup$
@J.W.Tanner, yes, but apart from the definition, I don't know how to approach it.
$endgroup$
– Shroud
Jan 14 at 14:01




$begingroup$
@J.W.Tanner, yes, but apart from the definition, I don't know how to approach it.
$endgroup$
– Shroud
Jan 14 at 14:01












$begingroup$
Can you show f’ is increasing (so f’’ is positive) on $(a-epsilon, a)?$
$endgroup$
– J. W. Tanner
Jan 14 at 14:08




$begingroup$
Can you show f’ is increasing (so f’’ is positive) on $(a-epsilon, a)?$
$endgroup$
– J. W. Tanner
Jan 14 at 14:08










2 Answers
2






active

oldest

votes


















2












$begingroup$

Hint



The result is wrong. Look at



$$f(x)=begin{cases}
-sqrt{-x} + x sin left(frac{1}{x}right) & x <0\
0 & x=0\
sqrt{x} + x sin left(frac{1}{x}right)& x>0
end{cases}$$



And prove that around zero the derivative is not monotonic on each side of zero.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    But on $(-0.2,0)$, $f$ is convex. I think $sin(frac{1}{x})$ instead of $sin x$ works.
    $endgroup$
    – Shroud
    Jan 14 at 14:51












  • $begingroup$
    Thanks for noticing the typo! That was indeed $sin 1/x$ that I had in mind.
    $endgroup$
    – mathcounterexamples.net
    Jan 14 at 15:58



















2












$begingroup$

This is false. If $f$ is continuously differentiable on $Bbb Rsetminus{a}$ and $lim_{xto a}f'(x)=infty$ then it follows from the Mean Value Theorem that $f'(a)=infty$, but $f'$ certainly need not be monotone on $(a-epsilon,a)$.






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2












    $begingroup$

    Hint



    The result is wrong. Look at



    $$f(x)=begin{cases}
    -sqrt{-x} + x sin left(frac{1}{x}right) & x <0\
    0 & x=0\
    sqrt{x} + x sin left(frac{1}{x}right)& x>0
    end{cases}$$



    And prove that around zero the derivative is not monotonic on each side of zero.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      But on $(-0.2,0)$, $f$ is convex. I think $sin(frac{1}{x})$ instead of $sin x$ works.
      $endgroup$
      – Shroud
      Jan 14 at 14:51












    • $begingroup$
      Thanks for noticing the typo! That was indeed $sin 1/x$ that I had in mind.
      $endgroup$
      – mathcounterexamples.net
      Jan 14 at 15:58
















    2












    $begingroup$

    Hint



    The result is wrong. Look at



    $$f(x)=begin{cases}
    -sqrt{-x} + x sin left(frac{1}{x}right) & x <0\
    0 & x=0\
    sqrt{x} + x sin left(frac{1}{x}right)& x>0
    end{cases}$$



    And prove that around zero the derivative is not monotonic on each side of zero.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      But on $(-0.2,0)$, $f$ is convex. I think $sin(frac{1}{x})$ instead of $sin x$ works.
      $endgroup$
      – Shroud
      Jan 14 at 14:51












    • $begingroup$
      Thanks for noticing the typo! That was indeed $sin 1/x$ that I had in mind.
      $endgroup$
      – mathcounterexamples.net
      Jan 14 at 15:58














    2












    2








    2





    $begingroup$

    Hint



    The result is wrong. Look at



    $$f(x)=begin{cases}
    -sqrt{-x} + x sin left(frac{1}{x}right) & x <0\
    0 & x=0\
    sqrt{x} + x sin left(frac{1}{x}right)& x>0
    end{cases}$$



    And prove that around zero the derivative is not monotonic on each side of zero.






    share|cite|improve this answer











    $endgroup$



    Hint



    The result is wrong. Look at



    $$f(x)=begin{cases}
    -sqrt{-x} + x sin left(frac{1}{x}right) & x <0\
    0 & x=0\
    sqrt{x} + x sin left(frac{1}{x}right)& x>0
    end{cases}$$



    And prove that around zero the derivative is not monotonic on each side of zero.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Jan 14 at 15:58

























    answered Jan 14 at 14:33









    mathcounterexamples.netmathcounterexamples.net

    26.4k22157




    26.4k22157












    • $begingroup$
      But on $(-0.2,0)$, $f$ is convex. I think $sin(frac{1}{x})$ instead of $sin x$ works.
      $endgroup$
      – Shroud
      Jan 14 at 14:51












    • $begingroup$
      Thanks for noticing the typo! That was indeed $sin 1/x$ that I had in mind.
      $endgroup$
      – mathcounterexamples.net
      Jan 14 at 15:58


















    • $begingroup$
      But on $(-0.2,0)$, $f$ is convex. I think $sin(frac{1}{x})$ instead of $sin x$ works.
      $endgroup$
      – Shroud
      Jan 14 at 14:51












    • $begingroup$
      Thanks for noticing the typo! That was indeed $sin 1/x$ that I had in mind.
      $endgroup$
      – mathcounterexamples.net
      Jan 14 at 15:58
















    $begingroup$
    But on $(-0.2,0)$, $f$ is convex. I think $sin(frac{1}{x})$ instead of $sin x$ works.
    $endgroup$
    – Shroud
    Jan 14 at 14:51






    $begingroup$
    But on $(-0.2,0)$, $f$ is convex. I think $sin(frac{1}{x})$ instead of $sin x$ works.
    $endgroup$
    – Shroud
    Jan 14 at 14:51














    $begingroup$
    Thanks for noticing the typo! That was indeed $sin 1/x$ that I had in mind.
    $endgroup$
    – mathcounterexamples.net
    Jan 14 at 15:58




    $begingroup$
    Thanks for noticing the typo! That was indeed $sin 1/x$ that I had in mind.
    $endgroup$
    – mathcounterexamples.net
    Jan 14 at 15:58











    2












    $begingroup$

    This is false. If $f$ is continuously differentiable on $Bbb Rsetminus{a}$ and $lim_{xto a}f'(x)=infty$ then it follows from the Mean Value Theorem that $f'(a)=infty$, but $f'$ certainly need not be monotone on $(a-epsilon,a)$.






    share|cite|improve this answer









    $endgroup$


















      2












      $begingroup$

      This is false. If $f$ is continuously differentiable on $Bbb Rsetminus{a}$ and $lim_{xto a}f'(x)=infty$ then it follows from the Mean Value Theorem that $f'(a)=infty$, but $f'$ certainly need not be monotone on $(a-epsilon,a)$.






      share|cite|improve this answer









      $endgroup$
















        2












        2








        2





        $begingroup$

        This is false. If $f$ is continuously differentiable on $Bbb Rsetminus{a}$ and $lim_{xto a}f'(x)=infty$ then it follows from the Mean Value Theorem that $f'(a)=infty$, but $f'$ certainly need not be monotone on $(a-epsilon,a)$.






        share|cite|improve this answer









        $endgroup$



        This is false. If $f$ is continuously differentiable on $Bbb Rsetminus{a}$ and $lim_{xto a}f'(x)=infty$ then it follows from the Mean Value Theorem that $f'(a)=infty$, but $f'$ certainly need not be monotone on $(a-epsilon,a)$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 14 at 14:30









        David C. UllrichDavid C. Ullrich

        60.3k43994




        60.3k43994






























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