Prove or disprove: If $f$ is increasing and differentiable on $(a,b)$ then $f'(x)ge 0$ on $(a,b)$












0












$begingroup$


Here's the question again:




Prove or disprove: If $f$ is increasing and differentiable on $(a,b)$
then $f'(x)ge 0$ on $(a,b)$.




I could not find any counterexamples to it so here's my attempt at a proof:



Suppose there is a point $cin (a,b)$ such that $f'(c)<0$. Let $varepsilon = - f'(c)/2$. Then there exists a $delta > 0$ such that $frac{f(x)-f(c)}{x-c} < f'(c)/2 < 0$ for all $0<|x-c|< delta $.



Let $x_1 in (c-delta , c )$ then $x_1 -c<0$ thus we have that $frac{f(x_1)-f(c)}{x_1-c} <0$ implies $f(x_1) > f(c)$. Now, let $x_2 in (c, c+delta )$ then in a similar way, $f(x_2) < f(c)$. We note that $x_1 < x_2$ but we obtain $f(x_2)< f(c) < f(x_1)$, a contradiction since $f$ is increasing on $(a,b)$.



Is this attempt correct? Could there be any direct proofs?






real-analysis proof-verification alternative-proof monotone-functions







share|cite|improve this question









$endgroup$

















    0












    $begingroup$


    Here's the question again:




    Prove or disprove: If $f$ is increasing and differentiable on $(a,b)$
    then $f'(x)ge 0$ on $(a,b)$.




    I could not find any counterexamples to it so here's my attempt at a proof:



    Suppose there is a point $cin (a,b)$ such that $f'(c)<0$. Let $varepsilon = - f'(c)/2$. Then there exists a $delta > 0$ such that $frac{f(x)-f(c)}{x-c} < f'(c)/2 < 0$ for all $0<|x-c|< delta $.



    Let $x_1 in (c-delta , c )$ then $x_1 -c<0$ thus we have that $frac{f(x_1)-f(c)}{x_1-c} <0$ implies $f(x_1) > f(c)$. Now, let $x_2 in (c, c+delta )$ then in a similar way, $f(x_2) < f(c)$. We note that $x_1 < x_2$ but we obtain $f(x_2)< f(c) < f(x_1)$, a contradiction since $f$ is increasing on $(a,b)$.



    Is this attempt correct? Could there be any direct proofs?






    real-analysis proof-verification alternative-proof monotone-functions







    share|cite|improve this question









    $endgroup$















      0












      0








      0





      $begingroup$


      Here's the question again:




      Prove or disprove: If $f$ is increasing and differentiable on $(a,b)$
      then $f'(x)ge 0$ on $(a,b)$.




      I could not find any counterexamples to it so here's my attempt at a proof:



      Suppose there is a point $cin (a,b)$ such that $f'(c)<0$. Let $varepsilon = - f'(c)/2$. Then there exists a $delta > 0$ such that $frac{f(x)-f(c)}{x-c} < f'(c)/2 < 0$ for all $0<|x-c|< delta $.



      Let $x_1 in (c-delta , c )$ then $x_1 -c<0$ thus we have that $frac{f(x_1)-f(c)}{x_1-c} <0$ implies $f(x_1) > f(c)$. Now, let $x_2 in (c, c+delta )$ then in a similar way, $f(x_2) < f(c)$. We note that $x_1 < x_2$ but we obtain $f(x_2)< f(c) < f(x_1)$, a contradiction since $f$ is increasing on $(a,b)$.



      Is this attempt correct? Could there be any direct proofs?






      real-analysis proof-verification alternative-proof monotone-functions







      share|cite|improve this question









      $endgroup$




      Here's the question again:




      Prove or disprove: If $f$ is increasing and differentiable on $(a,b)$
      then $f'(x)ge 0$ on $(a,b)$.




      I could not find any counterexamples to it so here's my attempt at a proof:



      Suppose there is a point $cin (a,b)$ such that $f'(c)<0$. Let $varepsilon = - f'(c)/2$. Then there exists a $delta > 0$ such that $frac{f(x)-f(c)}{x-c} < f'(c)/2 < 0$ for all $0<|x-c|< delta $.



      Let $x_1 in (c-delta , c )$ then $x_1 -c<0$ thus we have that $frac{f(x_1)-f(c)}{x_1-c} <0$ implies $f(x_1) > f(c)$. Now, let $x_2 in (c, c+delta )$ then in a similar way, $f(x_2) < f(c)$. We note that $x_1 < x_2$ but we obtain $f(x_2)< f(c) < f(x_1)$, a contradiction since $f$ is increasing on $(a,b)$.



      Is this attempt correct? Could there be any direct proofs?






      real-analysis proof-verification alternative-proof monotone-functions




      real-analysis proof-verification alternative-proof monotone-functions






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Nov 24 '18 at 22:12









      Ashish KAshish K

      831613




      831613






















          2 Answers
          2






          active

          oldest

          votes


















          3












          $begingroup$

          Your proof is correct and almost a direct proof, let me rephrase it: Let $x<y$. Then
          $ f(y)-f(x) > 0$. Therefore
          $$ frac{f(y) - f(x)}{y-x} > 0.$$
          Sending $yto x$ yields
          $$ f'(x) ge 0.$$






          share|cite|improve this answer









          $endgroup$













          • $begingroup$
            Since $f$ is increasing on $(a,b)$, should not it be $f(y)-f(x) ge 0$ albeit it makes no difference?
            $endgroup$
            – Ashish K
            Nov 24 '18 at 22:20










          • $begingroup$
            @AshishK It depends on what you define as increasing. I'm always confused, since its "$>$" for increasing sequences...
            $endgroup$
            – Calvin Khor
            Nov 24 '18 at 22:22



















          3












          $begingroup$

          Your proof is correct. A direct proof goes like this:



          For any $cin (a,b)$, since $f$ is differentiable at $c$, we have
          $$lim_{xto c}frac{f(x)-f(c)}{x-c}=f'(c).$$
          The expression inside the limit is always nonnegative since $f$ is increasing. Thus, the limit has to be nonnegative.






          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%2f3012158%2fprove-or-disprove-if-f-is-increasing-and-differentiable-on-a-b-then-fx%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












            $begingroup$

            Your proof is correct and almost a direct proof, let me rephrase it: Let $x<y$. Then
            $ f(y)-f(x) > 0$. Therefore
            $$ frac{f(y) - f(x)}{y-x} > 0.$$
            Sending $yto x$ yields
            $$ f'(x) ge 0.$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Since $f$ is increasing on $(a,b)$, should not it be $f(y)-f(x) ge 0$ albeit it makes no difference?
              $endgroup$
              – Ashish K
              Nov 24 '18 at 22:20










            • $begingroup$
              @AshishK It depends on what you define as increasing. I'm always confused, since its "$>$" for increasing sequences...
              $endgroup$
              – Calvin Khor
              Nov 24 '18 at 22:22
















            3












            $begingroup$

            Your proof is correct and almost a direct proof, let me rephrase it: Let $x<y$. Then
            $ f(y)-f(x) > 0$. Therefore
            $$ frac{f(y) - f(x)}{y-x} > 0.$$
            Sending $yto x$ yields
            $$ f'(x) ge 0.$$






            share|cite|improve this answer









            $endgroup$













            • $begingroup$
              Since $f$ is increasing on $(a,b)$, should not it be $f(y)-f(x) ge 0$ albeit it makes no difference?
              $endgroup$
              – Ashish K
              Nov 24 '18 at 22:20










            • $begingroup$
              @AshishK It depends on what you define as increasing. I'm always confused, since its "$>$" for increasing sequences...
              $endgroup$
              – Calvin Khor
              Nov 24 '18 at 22:22














            3












            3








            3





            $begingroup$

            Your proof is correct and almost a direct proof, let me rephrase it: Let $x<y$. Then
            $ f(y)-f(x) > 0$. Therefore
            $$ frac{f(y) - f(x)}{y-x} > 0.$$
            Sending $yto x$ yields
            $$ f'(x) ge 0.$$






            share|cite|improve this answer









            $endgroup$



            Your proof is correct and almost a direct proof, let me rephrase it: Let $x<y$. Then
            $ f(y)-f(x) > 0$. Therefore
            $$ frac{f(y) - f(x)}{y-x} > 0.$$
            Sending $yto x$ yields
            $$ f'(x) ge 0.$$







            share|cite|improve this answer












            share|cite|improve this answer



            share|cite|improve this answer










            answered Nov 24 '18 at 22:16









            Calvin KhorCalvin Khor

            11.2k21438




            11.2k21438












            • $begingroup$
              Since $f$ is increasing on $(a,b)$, should not it be $f(y)-f(x) ge 0$ albeit it makes no difference?
              $endgroup$
              – Ashish K
              Nov 24 '18 at 22:20










            • $begingroup$
              @AshishK It depends on what you define as increasing. I'm always confused, since its "$>$" for increasing sequences...
              $endgroup$
              – Calvin Khor
              Nov 24 '18 at 22:22


















            • $begingroup$
              Since $f$ is increasing on $(a,b)$, should not it be $f(y)-f(x) ge 0$ albeit it makes no difference?
              $endgroup$
              – Ashish K
              Nov 24 '18 at 22:20










            • $begingroup$
              @AshishK It depends on what you define as increasing. I'm always confused, since its "$>$" for increasing sequences...
              $endgroup$
              – Calvin Khor
              Nov 24 '18 at 22:22
















            $begingroup$
            Since $f$ is increasing on $(a,b)$, should not it be $f(y)-f(x) ge 0$ albeit it makes no difference?
            $endgroup$
            – Ashish K
            Nov 24 '18 at 22:20




            $begingroup$
            Since $f$ is increasing on $(a,b)$, should not it be $f(y)-f(x) ge 0$ albeit it makes no difference?
            $endgroup$
            – Ashish K
            Nov 24 '18 at 22:20












            $begingroup$
            @AshishK It depends on what you define as increasing. I'm always confused, since its "$>$" for increasing sequences...
            $endgroup$
            – Calvin Khor
            Nov 24 '18 at 22:22




            $begingroup$
            @AshishK It depends on what you define as increasing. I'm always confused, since its "$>$" for increasing sequences...
            $endgroup$
            – Calvin Khor
            Nov 24 '18 at 22:22











            3












            $begingroup$

            Your proof is correct. A direct proof goes like this:



            For any $cin (a,b)$, since $f$ is differentiable at $c$, we have
            $$lim_{xto c}frac{f(x)-f(c)}{x-c}=f'(c).$$
            The expression inside the limit is always nonnegative since $f$ is increasing. Thus, the limit has to be nonnegative.






            share|cite|improve this answer









            $endgroup$


















              3












              $begingroup$

              Your proof is correct. A direct proof goes like this:



              For any $cin (a,b)$, since $f$ is differentiable at $c$, we have
              $$lim_{xto c}frac{f(x)-f(c)}{x-c}=f'(c).$$
              The expression inside the limit is always nonnegative since $f$ is increasing. Thus, the limit has to be nonnegative.






              share|cite|improve this answer









              $endgroup$
















                3












                3








                3





                $begingroup$

                Your proof is correct. A direct proof goes like this:



                For any $cin (a,b)$, since $f$ is differentiable at $c$, we have
                $$lim_{xto c}frac{f(x)-f(c)}{x-c}=f'(c).$$
                The expression inside the limit is always nonnegative since $f$ is increasing. Thus, the limit has to be nonnegative.






                share|cite|improve this answer









                $endgroup$



                Your proof is correct. A direct proof goes like this:



                For any $cin (a,b)$, since $f$ is differentiable at $c$, we have
                $$lim_{xto c}frac{f(x)-f(c)}{x-c}=f'(c).$$
                The expression inside the limit is always nonnegative since $f$ is increasing. Thus, the limit has to be nonnegative.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 24 '18 at 22:16









                Eclipse SunEclipse Sun

                6,9841437




                6,9841437






























                    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%2f3012158%2fprove-or-disprove-if-f-is-increasing-and-differentiable-on-a-b-then-fx%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

                    Biblatex bibliography style without URLs when DOI exists (in Overleaf with Zotero bibliography)

                    Image
                    2 2 I have a Zotero bibliography in Overleaf. I use biblatex . usepackage[backend=biber,citestyle=authoryear]{biblatex} addbibresource{zoteroALH.bib} Bibliography entries look like: @book{de_saint-gervais_uniformization_2016, title = {Uniformization of {Riemann} {Surfaces}}, isbn = {978-3-03719-145-3}, url = {https://www.ems-ph.org/books/book.php?proj_nr=198&srch=series%7Chem}, urldate = {2019-02-03}, author = {de Saint-Gervais, Paul Henri}, year = {2016}, doi = {10.4171/145} } In the bibliography down the paper the reference shows like this: There is the DOI, great, but also the URL which is unnecessary when the DOI is there. How can I remove the URL when the DOI is present? biblatex bibliographies overleaf zote
                    Read more

                    How to change which sound is reproduced for terminal bell?

                    Image
                    3 Recently upgrade to 18.10 from 18.04, and using the terminal is becoming annoying: the new alert sound is a drip ( /usr/share/sounds/gnome/default/alerts/drip.ogg ), and the previous one is still located at ( /usr/share/sounds/ubuntu/stereo/bell.ogg ). My issue is that at the gnome-terminal, when autocompleting or pressing left BOTH sounds are reproduced, at the same time. I want to just have ubuntu/stereo/bell.ogg active, not both . At the moment I have been only able to reduce the volume of the alerts (but still both are played), or turn the alerts down (but I dislike both "solutions"). gnome-terminal 18.10 teminal-bell share | improve this question
                    Read more

                    Can I use Tabulator js library in my java Spring + Thymeleaf project?

                    Image
                    0 I want to add Tabulator js to my java project. I added tabulator.min.js, tabulator.min.css to resources folder. After that added to my html page these dependencies as recommending in readme.md. <link rel="stylesheet" th:href="@{/css/tabulator.min.css}"> <script type="text/javascript" th:src="@{/js/tabulator.min.js}"></script> Now my tables view all elements, that I have in the html page, but not correct. In the documentation of Tabulator is example how to create header and table, var table = new Tabulator("#example-table", { columns:[ {title:"Name", field:"name", sorter:"string", width:200, editor:true}, {title:"Age", field:"age", sorter:"
                    Read more