A function that is differentiable at exactly two points












0












$begingroup$


As homework we were given 2 questions:




  1. Find a function that is continuous at exactly one point and not differentiable there.


  2. Find a function that is differentiable at exactly two points.



The answer to the first one was somewhat simple - $ f(x) = x*D(x) $, wherein $ D(x) $ stands for Dirichlet function.



I've been thinking about the second question for a while, and someone suggested me the following:
$ g(x) = x^2*D(x) $



$ f(x) = (x-1)^2 * g(x) $



Claiming that $f(x)$ is only differentiable in $ x = 0, x = 1 $.
Why is that?






calculus derivatives







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  • 1




    $begingroup$
    It follows the exactly same proof to that $x^2D(x)$ is only differentiable in $x=0$. Note that $g(x-1) = (x-1)^2 D(x-1)$ and $D(x-1)=D(x)$.
    $endgroup$
    – Yanko
    Jan 12 at 17:39
















0












$begingroup$


As homework we were given 2 questions:




  1. Find a function that is continuous at exactly one point and not differentiable there.


  2. Find a function that is differentiable at exactly two points.



The answer to the first one was somewhat simple - $ f(x) = x*D(x) $, wherein $ D(x) $ stands for Dirichlet function.



I've been thinking about the second question for a while, and someone suggested me the following:
$ g(x) = x^2*D(x) $



$ f(x) = (x-1)^2 * g(x) $



Claiming that $f(x)$ is only differentiable in $ x = 0, x = 1 $.
Why is that?






calculus derivatives







share|cite|improve this question









$endgroup$








  • 1




    $begingroup$
    It follows the exactly same proof to that $x^2D(x)$ is only differentiable in $x=0$. Note that $g(x-1) = (x-1)^2 D(x-1)$ and $D(x-1)=D(x)$.
    $endgroup$
    – Yanko
    Jan 12 at 17:39














0












0








0





$begingroup$


As homework we were given 2 questions:




  1. Find a function that is continuous at exactly one point and not differentiable there.


  2. Find a function that is differentiable at exactly two points.



The answer to the first one was somewhat simple - $ f(x) = x*D(x) $, wherein $ D(x) $ stands for Dirichlet function.



I've been thinking about the second question for a while, and someone suggested me the following:
$ g(x) = x^2*D(x) $



$ f(x) = (x-1)^2 * g(x) $



Claiming that $f(x)$ is only differentiable in $ x = 0, x = 1 $.
Why is that?






calculus derivatives







share|cite|improve this question









$endgroup$




As homework we were given 2 questions:




  1. Find a function that is continuous at exactly one point and not differentiable there.


  2. Find a function that is differentiable at exactly two points.



The answer to the first one was somewhat simple - $ f(x) = x*D(x) $, wherein $ D(x) $ stands for Dirichlet function.



I've been thinking about the second question for a while, and someone suggested me the following:
$ g(x) = x^2*D(x) $



$ f(x) = (x-1)^2 * g(x) $



Claiming that $f(x)$ is only differentiable in $ x = 0, x = 1 $.
Why is that?






calculus derivatives




calculus derivatives






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Jan 12 at 17:37









TegernakoTegernako

888




888








  • 1




    $begingroup$
    It follows the exactly same proof to that $x^2D(x)$ is only differentiable in $x=0$. Note that $g(x-1) = (x-1)^2 D(x-1)$ and $D(x-1)=D(x)$.
    $endgroup$
    – Yanko
    Jan 12 at 17:39














  • 1




    $begingroup$
    It follows the exactly same proof to that $x^2D(x)$ is only differentiable in $x=0$. Note that $g(x-1) = (x-1)^2 D(x-1)$ and $D(x-1)=D(x)$.
    $endgroup$
    – Yanko
    Jan 12 at 17:39








1




1




$begingroup$
It follows the exactly same proof to that $x^2D(x)$ is only differentiable in $x=0$. Note that $g(x-1) = (x-1)^2 D(x-1)$ and $D(x-1)=D(x)$.
$endgroup$
– Yanko
Jan 12 at 17:39




$begingroup$
It follows the exactly same proof to that $x^2D(x)$ is only differentiable in $x=0$. Note that $g(x-1) = (x-1)^2 D(x-1)$ and $D(x-1)=D(x)$.
$endgroup$
– Yanko
Jan 12 at 17:39










2 Answers
2






active

oldest

votes


















3












$begingroup$

It is simplier to say that$$g(x)=begin{cases}x^2(x-1)^2&text{ if }xinmathbb Q\0&text{ otherwise.}end{cases}$$And, yes, this exmple works. Outside ${0,1}$ $g$ is discontinuous, and therefore non-differentiable. And it follows from the definition of $g'$ that $g'(0)=g'(1)=0$.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    Good example...
    $endgroup$
    – Cloud JR
    Jan 12 at 17:45










  • $begingroup$
    An excellent example, thank you!
    $endgroup$
    – Tegernako
    Jan 12 at 18:04






  • 1




    $begingroup$
    It's your example. I just used a different notation. Anyway, I'm glad I could help.
    $endgroup$
    – José Carlos Santos
    Jan 12 at 18:06



















1












$begingroup$

Clearly, $f$ is not differentiable away from $0$ and $1$: the $x^2$ and $(x-1)^2$ factors don't fix the horribleness of the Dirichlet function there: a rational $y$ near some $x$ is sent near to $x^2(x-1)^2$, while an irrational $y$ near $x$ is sent to $0$, and those differ, so $f$ isn't even continuous away from $0$ and $1$.



So, why is it differentiable at $0$ and $1$: well, let's just check the definition.



$$lim_{xto 0}frac{f(x)-f(0)}{x} = lim_{xto 0}frac{x^2(x-1)^2D(x)}{x} = lim_{xto 0}x(x-1)^2D(x),$$
and $|x(x-1)^2D(x)| leq |x|(x-1)^2 to 0$ as $x to 0$, so the limit exists, and is zero, at $0$. Similarly,
$$lim_{xto 1}frac{f(x)-f(1)}{x-1} = lim_{xto 1}x^2(x-1)D(x),$$
and $|x^2(x-1)D(x)| leq x^2|x-1| to 0$ as $x to 1$.






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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

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    active

    oldest

    votes









    3












    $begingroup$

    It is simplier to say that$$g(x)=begin{cases}x^2(x-1)^2&text{ if }xinmathbb Q\0&text{ otherwise.}end{cases}$$And, yes, this exmple works. Outside ${0,1}$ $g$ is discontinuous, and therefore non-differentiable. And it follows from the definition of $g'$ that $g'(0)=g'(1)=0$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Good example...
      $endgroup$
      – Cloud JR
      Jan 12 at 17:45










    • $begingroup$
      An excellent example, thank you!
      $endgroup$
      – Tegernako
      Jan 12 at 18:04






    • 1




      $begingroup$
      It's your example. I just used a different notation. Anyway, I'm glad I could help.
      $endgroup$
      – José Carlos Santos
      Jan 12 at 18:06
















    3












    $begingroup$

    It is simplier to say that$$g(x)=begin{cases}x^2(x-1)^2&text{ if }xinmathbb Q\0&text{ otherwise.}end{cases}$$And, yes, this exmple works. Outside ${0,1}$ $g$ is discontinuous, and therefore non-differentiable. And it follows from the definition of $g'$ that $g'(0)=g'(1)=0$.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      Good example...
      $endgroup$
      – Cloud JR
      Jan 12 at 17:45










    • $begingroup$
      An excellent example, thank you!
      $endgroup$
      – Tegernako
      Jan 12 at 18:04






    • 1




      $begingroup$
      It's your example. I just used a different notation. Anyway, I'm glad I could help.
      $endgroup$
      – José Carlos Santos
      Jan 12 at 18:06














    3












    3








    3





    $begingroup$

    It is simplier to say that$$g(x)=begin{cases}x^2(x-1)^2&text{ if }xinmathbb Q\0&text{ otherwise.}end{cases}$$And, yes, this exmple works. Outside ${0,1}$ $g$ is discontinuous, and therefore non-differentiable. And it follows from the definition of $g'$ that $g'(0)=g'(1)=0$.






    share|cite|improve this answer









    $endgroup$



    It is simplier to say that$$g(x)=begin{cases}x^2(x-1)^2&text{ if }xinmathbb Q\0&text{ otherwise.}end{cases}$$And, yes, this exmple works. Outside ${0,1}$ $g$ is discontinuous, and therefore non-differentiable. And it follows from the definition of $g'$ that $g'(0)=g'(1)=0$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Jan 12 at 17:41









    José Carlos SantosJosé Carlos Santos

    157k22126227




    157k22126227












    • $begingroup$
      Good example...
      $endgroup$
      – Cloud JR
      Jan 12 at 17:45










    • $begingroup$
      An excellent example, thank you!
      $endgroup$
      – Tegernako
      Jan 12 at 18:04






    • 1




      $begingroup$
      It's your example. I just used a different notation. Anyway, I'm glad I could help.
      $endgroup$
      – José Carlos Santos
      Jan 12 at 18:06


















    • $begingroup$
      Good example...
      $endgroup$
      – Cloud JR
      Jan 12 at 17:45










    • $begingroup$
      An excellent example, thank you!
      $endgroup$
      – Tegernako
      Jan 12 at 18:04






    • 1




      $begingroup$
      It's your example. I just used a different notation. Anyway, I'm glad I could help.
      $endgroup$
      – José Carlos Santos
      Jan 12 at 18:06
















    $begingroup$
    Good example...
    $endgroup$
    – Cloud JR
    Jan 12 at 17:45




    $begingroup$
    Good example...
    $endgroup$
    – Cloud JR
    Jan 12 at 17:45












    $begingroup$
    An excellent example, thank you!
    $endgroup$
    – Tegernako
    Jan 12 at 18:04




    $begingroup$
    An excellent example, thank you!
    $endgroup$
    – Tegernako
    Jan 12 at 18:04




    1




    1




    $begingroup$
    It's your example. I just used a different notation. Anyway, I'm glad I could help.
    $endgroup$
    – José Carlos Santos
    Jan 12 at 18:06




    $begingroup$
    It's your example. I just used a different notation. Anyway, I'm glad I could help.
    $endgroup$
    – José Carlos Santos
    Jan 12 at 18:06











    1












    $begingroup$

    Clearly, $f$ is not differentiable away from $0$ and $1$: the $x^2$ and $(x-1)^2$ factors don't fix the horribleness of the Dirichlet function there: a rational $y$ near some $x$ is sent near to $x^2(x-1)^2$, while an irrational $y$ near $x$ is sent to $0$, and those differ, so $f$ isn't even continuous away from $0$ and $1$.



    So, why is it differentiable at $0$ and $1$: well, let's just check the definition.



    $$lim_{xto 0}frac{f(x)-f(0)}{x} = lim_{xto 0}frac{x^2(x-1)^2D(x)}{x} = lim_{xto 0}x(x-1)^2D(x),$$
    and $|x(x-1)^2D(x)| leq |x|(x-1)^2 to 0$ as $x to 0$, so the limit exists, and is zero, at $0$. Similarly,
    $$lim_{xto 1}frac{f(x)-f(1)}{x-1} = lim_{xto 1}x^2(x-1)D(x),$$
    and $|x^2(x-1)D(x)| leq x^2|x-1| to 0$ as $x to 1$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Clearly, $f$ is not differentiable away from $0$ and $1$: the $x^2$ and $(x-1)^2$ factors don't fix the horribleness of the Dirichlet function there: a rational $y$ near some $x$ is sent near to $x^2(x-1)^2$, while an irrational $y$ near $x$ is sent to $0$, and those differ, so $f$ isn't even continuous away from $0$ and $1$.



      So, why is it differentiable at $0$ and $1$: well, let's just check the definition.



      $$lim_{xto 0}frac{f(x)-f(0)}{x} = lim_{xto 0}frac{x^2(x-1)^2D(x)}{x} = lim_{xto 0}x(x-1)^2D(x),$$
      and $|x(x-1)^2D(x)| leq |x|(x-1)^2 to 0$ as $x to 0$, so the limit exists, and is zero, at $0$. Similarly,
      $$lim_{xto 1}frac{f(x)-f(1)}{x-1} = lim_{xto 1}x^2(x-1)D(x),$$
      and $|x^2(x-1)D(x)| leq x^2|x-1| to 0$ as $x to 1$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Clearly, $f$ is not differentiable away from $0$ and $1$: the $x^2$ and $(x-1)^2$ factors don't fix the horribleness of the Dirichlet function there: a rational $y$ near some $x$ is sent near to $x^2(x-1)^2$, while an irrational $y$ near $x$ is sent to $0$, and those differ, so $f$ isn't even continuous away from $0$ and $1$.



        So, why is it differentiable at $0$ and $1$: well, let's just check the definition.



        $$lim_{xto 0}frac{f(x)-f(0)}{x} = lim_{xto 0}frac{x^2(x-1)^2D(x)}{x} = lim_{xto 0}x(x-1)^2D(x),$$
        and $|x(x-1)^2D(x)| leq |x|(x-1)^2 to 0$ as $x to 0$, so the limit exists, and is zero, at $0$. Similarly,
        $$lim_{xto 1}frac{f(x)-f(1)}{x-1} = lim_{xto 1}x^2(x-1)D(x),$$
        and $|x^2(x-1)D(x)| leq x^2|x-1| to 0$ as $x to 1$.






        share|cite|improve this answer









        $endgroup$



        Clearly, $f$ is not differentiable away from $0$ and $1$: the $x^2$ and $(x-1)^2$ factors don't fix the horribleness of the Dirichlet function there: a rational $y$ near some $x$ is sent near to $x^2(x-1)^2$, while an irrational $y$ near $x$ is sent to $0$, and those differ, so $f$ isn't even continuous away from $0$ and $1$.



        So, why is it differentiable at $0$ and $1$: well, let's just check the definition.



        $$lim_{xto 0}frac{f(x)-f(0)}{x} = lim_{xto 0}frac{x^2(x-1)^2D(x)}{x} = lim_{xto 0}x(x-1)^2D(x),$$
        and $|x(x-1)^2D(x)| leq |x|(x-1)^2 to 0$ as $x to 0$, so the limit exists, and is zero, at $0$. Similarly,
        $$lim_{xto 1}frac{f(x)-f(1)}{x-1} = lim_{xto 1}x^2(x-1)D(x),$$
        and $|x^2(x-1)D(x)| leq x^2|x-1| to 0$ as $x to 1$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 12 at 17:45









        user3482749user3482749

        4,197919




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