How can I show $max(f,g)$ is differentiable at $c$?












1












$begingroup$


Let $f,g : (a;b) to mathbb{R}$ differentiable, and $max(f,g) : (a;b) to mathbb{R} $ defined by $$max(f,g)(x) := max(f(x),g(x)).$$



Show that $max(f,g)$ is differentiable at $ c in (a;b)$, for all $c in (a;b)$ such that $f(c) neq g(c)$.






real-analysis derivatives







share|cite|improve this question











$endgroup$








  • 5




    $begingroup$
    In a neighbourhood of such $c$, $max(f,g)$ is either equal to $f$ or equal to $g$. In either case, it's differentiable.
    $endgroup$
    – user3482749
    Nov 29 '18 at 15:38






  • 1




    $begingroup$
    What have you tried yourself so far, and what jumps out as the roadblock in the proof to you? Also, how does your course define “differentiable”? (There are a few equivalent definitions.)
    $endgroup$
    – Lynn
    Nov 29 '18 at 15:40
















1












$begingroup$


Let $f,g : (a;b) to mathbb{R}$ differentiable, and $max(f,g) : (a;b) to mathbb{R} $ defined by $$max(f,g)(x) := max(f(x),g(x)).$$



Show that $max(f,g)$ is differentiable at $ c in (a;b)$, for all $c in (a;b)$ such that $f(c) neq g(c)$.






real-analysis derivatives







share|cite|improve this question











$endgroup$








  • 5




    $begingroup$
    In a neighbourhood of such $c$, $max(f,g)$ is either equal to $f$ or equal to $g$. In either case, it's differentiable.
    $endgroup$
    – user3482749
    Nov 29 '18 at 15:38






  • 1




    $begingroup$
    What have you tried yourself so far, and what jumps out as the roadblock in the proof to you? Also, how does your course define “differentiable”? (There are a few equivalent definitions.)
    $endgroup$
    – Lynn
    Nov 29 '18 at 15:40














1












1








1


0



$begingroup$


Let $f,g : (a;b) to mathbb{R}$ differentiable, and $max(f,g) : (a;b) to mathbb{R} $ defined by $$max(f,g)(x) := max(f(x),g(x)).$$



Show that $max(f,g)$ is differentiable at $ c in (a;b)$, for all $c in (a;b)$ such that $f(c) neq g(c)$.






real-analysis derivatives







share|cite|improve this question











$endgroup$




Let $f,g : (a;b) to mathbb{R}$ differentiable, and $max(f,g) : (a;b) to mathbb{R} $ defined by $$max(f,g)(x) := max(f(x),g(x)).$$



Show that $max(f,g)$ is differentiable at $ c in (a;b)$, for all $c in (a;b)$ such that $f(c) neq g(c)$.






real-analysis derivatives




real-analysis derivatives






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 29 '18 at 15:42









Lynn

2,5851526




2,5851526










asked Nov 29 '18 at 15:36









DadaDada

7510




7510








  • 5




    $begingroup$
    In a neighbourhood of such $c$, $max(f,g)$ is either equal to $f$ or equal to $g$. In either case, it's differentiable.
    $endgroup$
    – user3482749
    Nov 29 '18 at 15:38






  • 1




    $begingroup$
    What have you tried yourself so far, and what jumps out as the roadblock in the proof to you? Also, how does your course define “differentiable”? (There are a few equivalent definitions.)
    $endgroup$
    – Lynn
    Nov 29 '18 at 15:40














  • 5




    $begingroup$
    In a neighbourhood of such $c$, $max(f,g)$ is either equal to $f$ or equal to $g$. In either case, it's differentiable.
    $endgroup$
    – user3482749
    Nov 29 '18 at 15:38






  • 1




    $begingroup$
    What have you tried yourself so far, and what jumps out as the roadblock in the proof to you? Also, how does your course define “differentiable”? (There are a few equivalent definitions.)
    $endgroup$
    – Lynn
    Nov 29 '18 at 15:40








5




5




$begingroup$
In a neighbourhood of such $c$, $max(f,g)$ is either equal to $f$ or equal to $g$. In either case, it's differentiable.
$endgroup$
– user3482749
Nov 29 '18 at 15:38




$begingroup$
In a neighbourhood of such $c$, $max(f,g)$ is either equal to $f$ or equal to $g$. In either case, it's differentiable.
$endgroup$
– user3482749
Nov 29 '18 at 15:38




1




1




$begingroup$
What have you tried yourself so far, and what jumps out as the roadblock in the proof to you? Also, how does your course define “differentiable”? (There are a few equivalent definitions.)
$endgroup$
– Lynn
Nov 29 '18 at 15:40




$begingroup$
What have you tried yourself so far, and what jumps out as the roadblock in the proof to you? Also, how does your course define “differentiable”? (There are a few equivalent definitions.)
$endgroup$
– Lynn
Nov 29 '18 at 15:40










2 Answers
2






active

oldest

votes


















2












$begingroup$

Since $f(c)neq g(c)$, so assume that $f(c)>g(c)$.



This implies, $max(f,g)(x) = f(x)$, in a neighbourhood of $c$.



So, $$limlimits_{xrightarrow c^-}frac{max(f,g)(x)-max(f,g)(c)}{x-c}=limlimits_{xrightarrow c^-}frac{f(x)-f(c)}{x-c}$$ and $$limlimits_{xrightarrow c^+}frac{max(f,g)(x)-max(f,g)(c)}{x-c}=limlimits_{xrightarrow c^+}frac{f(x)-f(c)}{x-c}.$$



Since $f$ is differentiable at $c$, $limlimits_{xrightarrow c^-}frac{f(x)-f(c)}{x-c}=limlimits_{xrightarrow c^+}frac{f(x)-f(c)}{x-c}$.



Hence,$$limlimits_{xrightarrow c^-}frac{max(f,g)(x)-max(f,g)(c)}{x-c}=limlimits_{xrightarrow c^+}frac{max(f,g)(x)-max(f,g)(c)}{x-c}={max(f,g)(c)}'.$$






share|cite|improve this answer









$endgroup$





















    3












    $begingroup$

    $max(f,g)=frac{1}{2}(f+g+|f-g|)$. So to prove differentiability of $max(f,g)$ at $x=c$, where $f(c)neq g(c)$, it is enough to prove differentiability of $|f-g|$. Note that the modulus function $|;|:mathbb{R}setminus{0}rightarrow mathbb{R}$ defined by $|;|(x):=|x|$ is differentiable, and so is $f-g$. Now $|f-g|=|;|circ(f-g)$, being a composition of two differentiable functions, is differentiable.






    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%2f3018789%2fhow-can-i-show-maxf-g-is-differentiable-at-c%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









      2












      $begingroup$

      Since $f(c)neq g(c)$, so assume that $f(c)>g(c)$.



      This implies, $max(f,g)(x) = f(x)$, in a neighbourhood of $c$.



      So, $$limlimits_{xrightarrow c^-}frac{max(f,g)(x)-max(f,g)(c)}{x-c}=limlimits_{xrightarrow c^-}frac{f(x)-f(c)}{x-c}$$ and $$limlimits_{xrightarrow c^+}frac{max(f,g)(x)-max(f,g)(c)}{x-c}=limlimits_{xrightarrow c^+}frac{f(x)-f(c)}{x-c}.$$



      Since $f$ is differentiable at $c$, $limlimits_{xrightarrow c^-}frac{f(x)-f(c)}{x-c}=limlimits_{xrightarrow c^+}frac{f(x)-f(c)}{x-c}$.



      Hence,$$limlimits_{xrightarrow c^-}frac{max(f,g)(x)-max(f,g)(c)}{x-c}=limlimits_{xrightarrow c^+}frac{max(f,g)(x)-max(f,g)(c)}{x-c}={max(f,g)(c)}'.$$






      share|cite|improve this answer









      $endgroup$


















        2












        $begingroup$

        Since $f(c)neq g(c)$, so assume that $f(c)>g(c)$.



        This implies, $max(f,g)(x) = f(x)$, in a neighbourhood of $c$.



        So, $$limlimits_{xrightarrow c^-}frac{max(f,g)(x)-max(f,g)(c)}{x-c}=limlimits_{xrightarrow c^-}frac{f(x)-f(c)}{x-c}$$ and $$limlimits_{xrightarrow c^+}frac{max(f,g)(x)-max(f,g)(c)}{x-c}=limlimits_{xrightarrow c^+}frac{f(x)-f(c)}{x-c}.$$



        Since $f$ is differentiable at $c$, $limlimits_{xrightarrow c^-}frac{f(x)-f(c)}{x-c}=limlimits_{xrightarrow c^+}frac{f(x)-f(c)}{x-c}$.



        Hence,$$limlimits_{xrightarrow c^-}frac{max(f,g)(x)-max(f,g)(c)}{x-c}=limlimits_{xrightarrow c^+}frac{max(f,g)(x)-max(f,g)(c)}{x-c}={max(f,g)(c)}'.$$






        share|cite|improve this answer









        $endgroup$
















          2












          2








          2





          $begingroup$

          Since $f(c)neq g(c)$, so assume that $f(c)>g(c)$.



          This implies, $max(f,g)(x) = f(x)$, in a neighbourhood of $c$.



          So, $$limlimits_{xrightarrow c^-}frac{max(f,g)(x)-max(f,g)(c)}{x-c}=limlimits_{xrightarrow c^-}frac{f(x)-f(c)}{x-c}$$ and $$limlimits_{xrightarrow c^+}frac{max(f,g)(x)-max(f,g)(c)}{x-c}=limlimits_{xrightarrow c^+}frac{f(x)-f(c)}{x-c}.$$



          Since $f$ is differentiable at $c$, $limlimits_{xrightarrow c^-}frac{f(x)-f(c)}{x-c}=limlimits_{xrightarrow c^+}frac{f(x)-f(c)}{x-c}$.



          Hence,$$limlimits_{xrightarrow c^-}frac{max(f,g)(x)-max(f,g)(c)}{x-c}=limlimits_{xrightarrow c^+}frac{max(f,g)(x)-max(f,g)(c)}{x-c}={max(f,g)(c)}'.$$






          share|cite|improve this answer









          $endgroup$



          Since $f(c)neq g(c)$, so assume that $f(c)>g(c)$.



          This implies, $max(f,g)(x) = f(x)$, in a neighbourhood of $c$.



          So, $$limlimits_{xrightarrow c^-}frac{max(f,g)(x)-max(f,g)(c)}{x-c}=limlimits_{xrightarrow c^-}frac{f(x)-f(c)}{x-c}$$ and $$limlimits_{xrightarrow c^+}frac{max(f,g)(x)-max(f,g)(c)}{x-c}=limlimits_{xrightarrow c^+}frac{f(x)-f(c)}{x-c}.$$



          Since $f$ is differentiable at $c$, $limlimits_{xrightarrow c^-}frac{f(x)-f(c)}{x-c}=limlimits_{xrightarrow c^+}frac{f(x)-f(c)}{x-c}$.



          Hence,$$limlimits_{xrightarrow c^-}frac{max(f,g)(x)-max(f,g)(c)}{x-c}=limlimits_{xrightarrow c^+}frac{max(f,g)(x)-max(f,g)(c)}{x-c}={max(f,g)(c)}'.$$







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Nov 29 '18 at 15:58









          GoldyGoldy

          42414




          42414























              3












              $begingroup$

              $max(f,g)=frac{1}{2}(f+g+|f-g|)$. So to prove differentiability of $max(f,g)$ at $x=c$, where $f(c)neq g(c)$, it is enough to prove differentiability of $|f-g|$. Note that the modulus function $|;|:mathbb{R}setminus{0}rightarrow mathbb{R}$ defined by $|;|(x):=|x|$ is differentiable, and so is $f-g$. Now $|f-g|=|;|circ(f-g)$, being a composition of two differentiable functions, is differentiable.






              share|cite|improve this answer











              $endgroup$


















                3












                $begingroup$

                $max(f,g)=frac{1}{2}(f+g+|f-g|)$. So to prove differentiability of $max(f,g)$ at $x=c$, where $f(c)neq g(c)$, it is enough to prove differentiability of $|f-g|$. Note that the modulus function $|;|:mathbb{R}setminus{0}rightarrow mathbb{R}$ defined by $|;|(x):=|x|$ is differentiable, and so is $f-g$. Now $|f-g|=|;|circ(f-g)$, being a composition of two differentiable functions, is differentiable.






                share|cite|improve this answer











                $endgroup$
















                  3












                  3








                  3





                  $begingroup$

                  $max(f,g)=frac{1}{2}(f+g+|f-g|)$. So to prove differentiability of $max(f,g)$ at $x=c$, where $f(c)neq g(c)$, it is enough to prove differentiability of $|f-g|$. Note that the modulus function $|;|:mathbb{R}setminus{0}rightarrow mathbb{R}$ defined by $|;|(x):=|x|$ is differentiable, and so is $f-g$. Now $|f-g|=|;|circ(f-g)$, being a composition of two differentiable functions, is differentiable.






                  share|cite|improve this answer











                  $endgroup$



                  $max(f,g)=frac{1}{2}(f+g+|f-g|)$. So to prove differentiability of $max(f,g)$ at $x=c$, where $f(c)neq g(c)$, it is enough to prove differentiability of $|f-g|$. Note that the modulus function $|;|:mathbb{R}setminus{0}rightarrow mathbb{R}$ defined by $|;|(x):=|x|$ is differentiable, and so is $f-g$. Now $|f-g|=|;|circ(f-g)$, being a composition of two differentiable functions, is differentiable.







                  share|cite|improve this answer














                  share|cite|improve this answer



                  share|cite|improve this answer








                  edited Nov 29 '18 at 16:26

























                  answered Nov 29 '18 at 16:04









                  SurajitSurajit

                  5689




                  5689






























                      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%2f3018789%2fhow-can-i-show-maxf-g-is-differentiable-at-c%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

                      Plaza Victoria

                      Image
                      Plaza Victoria Plaza Victoria en 2017 Otros nombres Plaza de La Victoria Localización El Almendral, Valparaíso, Chile Vías adyacentes Chacabuco, Molina, Plaza Victoria y Edwards Creación 1841 Metro Bellavista Ubicación 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 Coordenadas: 33°02′46″S 71°37′11″O  /  -33.04624167, -71.61980833 [editar datos en Wikidata] La Plaza Victoria , también llamada Plaza de la Victoria , [ 1 ] ​ es una plaza ubicada en el sector El Almendral del plan de la ciudad de Valparaíso, Chile, de importancia patrimonial, que incluye valiosas estatuas y especies vegetales. [ 2 ] ​ La plaza se creó como tal en la primera mitad del siglo XIX, tomando previamente los nombres de Plaza Nueva y Plaza de Orrego . Si bien desde sus inicios tuvo relevancia histórica, no fue sino a partir de fines de los años 1860 cuando comenzó a cobrar mayor importancia y se convirtió en uno de los principales centros sociales de l...
                      Read more

                      Puebla de Zaragoza

                      Image
                      Para otros usos de este término, véase Puebla (desambiguación). Heroica Puebla de Zaragoza Localidad De arriba a abajo de izquierda a derecha: Estrella de Puebla, Volcán Popocatépetl, Catedral de Puebla, Monumento a Ignacio Zaragoza, Estadio Cuauhtémoc, Angelópolis y Vista panorámica de la ciudad. Escudo Otros nombres: El relicario de América/ La ciudad de los Ángeles/ La Angelópolis Lema: Muy noble y muy leal ciudad de Puebla de los Ángeles Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en México Heroica Puebla de Zaragoza Localización de Heroica Puebla de Zaragoza en Puebla Mapa interactivo Coordenadas 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Coordenadas: 19°03′05″N 98°13′04″O  /  19.051388888889, -98.217777777778 Idioma oficial Español Entidad Localidad  • País   México  • Estado Puebla  • Municipio Puebla Presidente municipal Claudia Rivera Vivanc...
                      Read more

                      Musa

                      Image
                      Para el género de plantas, véase Musa (planta). Apolo y la Nueve Musas En la mitología griega, las musas (en griego antiguo μοῦσαι « mousai» ) son, según los escritores más antiguos, las divinidades inspiradoras de las Artes: cada una de ellas está relacionada con ramas artísticas y del conocimiento. Son hijas de Zeus y de Mnemósine, compañeras del séquito de Apolo, dios olímpico de la música y patrón de las bellas artes, quien tuvo romances con cada una de ellas, dejando descendientes. Bajaban a la tierra a susurrar ideas e inspirar a aquellos mortales que las invocaran. En la época más arcaica eran las ninfas inspiradoras de las fuentes, en las cuales eran adoradas. Finalmente, alrededor de los siglos VIII-VII a. C. [ 1 ] ​ prevaleció en todo el territorio de la Hélade la adoración de las nueve Musas, que son Calíope, Clío, Erato, Euterpe, Melpómene, Polimnia, Talía, Terpsícore y Urania. El culto a las musas era originalmente de Tracia y Beocia, y fueron de vital impo...
                      Read more