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







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












$begingroup$


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 $x to 1$.






share|cite|improve this answer









$endgroup$













    Your Answer





    StackExchange.ifUsing("editor", function () {
    return StackExchange.using("mathjaxEditing", function () {
    StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
    StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
    });
    });
    }, "mathjax-editing");

    StackExchange.ready(function() {
    var channelOptions = {
    tags: "".split(" "),
    id: "69"
    };
    initTagRenderer("".split(" "), "".split(" "), channelOptions);

    StackExchange.using("externalEditor", function() {
    // Have to fire editor after snippets, if snippets enabled
    if (StackExchange.settings.snippets.snippetsEnabled) {
    StackExchange.using("snippets", function() {
    createEditor();
    });
    }
    else {
    createEditor();
    }
    });

    function createEditor() {
    StackExchange.prepareEditor({
    heartbeatType: 'answer',
    autoActivateHeartbeat: false,
    convertImagesToLinks: true,
    noModals: true,
    showLowRepImageUploadWarning: true,
    reputationToPostImages: 10,
    bindNavPrevention: true,
    postfix: "",
    imageUploader: {
    brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
    contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
    allowUrls: true
    },
    noCode: true, onDemand: true,
    discardSelector: ".discard-answer"
    ,immediatelyShowMarkdownHelp:true
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3071164%2fa-function-that-is-differentiable-at-exactly-two-points%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    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




        4,197919






























            draft saved

            draft discarded




















































            Thanks for contributing an answer to Mathematics Stack Exchange!


            • Please be sure to answer the question. Provide details and share your research!

            But avoid …



            • Asking for help, clarification, or responding to other answers.

            • Making statements based on opinion; back them up with references or personal experience.


            Use MathJax to format equations. MathJax reference.


            To learn more, see our tips on writing great answers.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3071164%2fa-function-that-is-differentiable-at-exactly-two-points%23new-answer', 'question_page');
            }
            );

            Post as a guest















            Required, but never shown





















































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown

































            Required, but never shown














            Required, but never shown












            Required, but never shown







            Required, but never shown







            -

            Popular posts from this blog

            Mario Kart Wii

            Image
            Please consider supporting The Cutting Room Floor on Patreon. Thanks for all your support! Mario Kart Wii From The Cutting Room Floor Jump to: navigation, search Mario Kart Wii Developer : Nintendo Publisher : Nintendo Platform : Wii Released in JP : April 10, 2008 Released in US : April 27, 2008 Released in EU : April 11, 2008 Released in AU : April 24, 2008 Released in KR : April 30, 2009 This game has unused animations. This game has unused areas. This game has unused code. This game has unused objects. This game has unused modes / minigames. This game has unused graphics. This game has unused models. This game has unused items. This game has unused music. This game has regional differences. This game has a notes page This game has a prerelease article Mario Kart Wii is the sixth game in the series (not counting Arcade editions), which intro
            Read more

            What does “Dominus providebit” mean?

            Image
            5 Does "Dominus providebit" mean "The Lord will provide" or "The Lord provides"? I once had a bit of an argument with a guy who studied Latin over that. It's an inscription that appears on the rim of some old Swiss coins that I have. meaning latin-to-english-translation grammar-identification share | improve this question edited Jan 18 at 15:27 V2Blast 105 4 asked Jan 17 at 22:49 Donald K. Donald K. 26 1
            Read more

            Antonio Litta Visconti Arese

            Image
            Antonio Litta Visconti Arese I duca Litta, VII marchese di Gambolò In carica 1797 – 1820 Predecessore Pompeo Litta Visconti Arese Successore Alfonso Litta Visconti Arese Trattamento Sua grazia Altri titoli Duca Litta Conte del Regno napoleonico d'Italia Patrizio milanese Don Nascita Milano, 2 maggio 1748 Morte Milano, 14 agosto 1820 Dinastia Litta Visconti Arese Padre Pompeo Litta Visconti Arese Madre Maria Elisabetta Visconti Borromeo Consorte Barbara Barbiano di Belgioioso Religione cattolicesimo Antonio Litta Visconti Arese , I duca Litta, VII marchese di Gambolò (Milano, 2 maggio 1748 – Milano, 14 agosto 1820), è stato un nobile e politico italiano. Composizione "Il pastore svizzero" di Pietro Morlacchi, con dedica Indice 1 Biografia 2 Onorificenze 3 Araldica 4 Albero genealogico 5 Note 6 Altri progetti Biografia | Antonio Litta Visco
            Read more