Application of mean value and Rolle's theorems - twice differentiable functions












1












$begingroup$


Let $f$ be a function from $[a,b]$ to $Bbb{R}$ that is twice-differentiable (that is, $f'$ and $f''$ exist), and assume that $f(a) = f(b) = 0$ and $f''(x) leq 0$ for every $xin (a,b)$. Show that $f(x) geq 0$ on $[a,b]$.



I think we must use the mean value/ Rolle's theorems There will be a $c$ in $(a,b)$ such that $f'(c) = frac{f(a) - f(b)}{b-a}= 0$. Where do I go from here?



Any help will be appreciated!






calculus rolles-theorem







share|cite|improve this question











$endgroup$

















    1












    $begingroup$


    Let $f$ be a function from $[a,b]$ to $Bbb{R}$ that is twice-differentiable (that is, $f'$ and $f''$ exist), and assume that $f(a) = f(b) = 0$ and $f''(x) leq 0$ for every $xin (a,b)$. Show that $f(x) geq 0$ on $[a,b]$.



    I think we must use the mean value/ Rolle's theorems There will be a $c$ in $(a,b)$ such that $f'(c) = frac{f(a) - f(b)}{b-a}= 0$. Where do I go from here?



    Any help will be appreciated!






    calculus rolles-theorem







    share|cite|improve this question











    $endgroup$















      1












      1








      1





      $begingroup$


      Let $f$ be a function from $[a,b]$ to $Bbb{R}$ that is twice-differentiable (that is, $f'$ and $f''$ exist), and assume that $f(a) = f(b) = 0$ and $f''(x) leq 0$ for every $xin (a,b)$. Show that $f(x) geq 0$ on $[a,b]$.



      I think we must use the mean value/ Rolle's theorems There will be a $c$ in $(a,b)$ such that $f'(c) = frac{f(a) - f(b)}{b-a}= 0$. Where do I go from here?



      Any help will be appreciated!






      calculus rolles-theorem







      share|cite|improve this question











      $endgroup$




      Let $f$ be a function from $[a,b]$ to $Bbb{R}$ that is twice-differentiable (that is, $f'$ and $f''$ exist), and assume that $f(a) = f(b) = 0$ and $f''(x) leq 0$ for every $xin (a,b)$. Show that $f(x) geq 0$ on $[a,b]$.



      I think we must use the mean value/ Rolle's theorems There will be a $c$ in $(a,b)$ such that $f'(c) = frac{f(a) - f(b)}{b-a}= 0$. Where do I go from here?



      Any help will be appreciated!






      calculus rolles-theorem




      calculus rolles-theorem






      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 18:58









      Bernard

      120k740115




      120k740115










      asked Nov 29 '18 at 18:00









      Aishwarya DeoreAishwarya Deore

      324




      324






















          3 Answers
          3






          active

          oldest

          votes


















          0












          $begingroup$

          $f$ is continuous in $[a,b]$ and differentiable in $(a,b), f(a)=f(b)=0$



          By Rolle's Theorem, $exists cin(a,b)$ such that $f'(c)=frac{f(b)-f(a)}{b-a}=0$



          $f''(x)leq0, forall xin(a,b)$



          $impliesforall xin(a,c), int_{x}^{c}f''(x)dx=f'(c)-f'(x)=-f'(x)leq0$



          $impliesforall xin(a,c), f'(x)geq0$



          $implies int_{a}^{x} f'(x)dx=f(x)-f(a)=f(x)geq0$



          Similarly, $forall xin(c,b), f'(x)leq0$



          $impliesint_{x}^{b} f'(x)dx=f(b)-f(x)=-f(x)leq0$



          $implies f(x)geq0$



          As far as $f(c)$ is concerned, continuity requires $f(c)geq0$






          share|cite|improve this answer









          $endgroup$





















            0












            $begingroup$

            Assume that $f(c) < 0$ for some $c in (a, b)$, and apply Rolle's theorem
            repeatedly:




            • There is an $x_1 in (a, c)$ with $f'(x_1) = frac{f(c)-f(a)}{c-a} < 0$.


            • There is an $x_2 in (c, b)$ with $f'(x_2) = frac{f(b)-f(c)}{b-c} > 0$.


            • There is an $x_3 in (x_1, x_2)$ with $f''(x_3) = frac{f'(x_2)-f'(x_1)}{x_2-x_1} > 0$.



            This is a contradiction to the assumption that $f''(x) le 0$
            for all $x in (a, b)$.




            An alternative approach: Fix any $c in (a, b)$ and define
            $$
            g(x) = (c-a)(c-b)f(x) - (x-a)(x-b)f(c) , .
            $$
            Then $g(a) = g(c) = g(b) = 0$, and repeated application of Rolle's
            theorem shows that $g''(xi) = 0$ for some $xi in (a, b)$.
            Then
            $$
            0 = g''(xi) = (c-a)(c-b)f''(xi) - 2f(c) \
            implies f(c) = frac 12 (c-a)(c-b)f''(xi) ge 0 , .
            $$




            Yet another approach: Assume that $f$ attains its minimum at $c in (a, b)$ and $f(c) < 0$. Then $f'(c) = 0$ and Taylor's formula gives
            a contradiction:
            $$
            f(b) = f(c) + frac 12 (b-c)^2 f''(xi) < 0
            $$
            for some $xi in (c, b)$.



            The real idea here is that $f$ is concave,
            and a continuous concave function on a compact interval attains its
            minimum at one of the boundary points: Therefore
            $$
            f(x) ge min (f(a), f(b)) = 0
            $$
            for $x in (a, b)$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              Thank you! I get the first method
              $endgroup$
              – Aishwarya Deore
              Nov 29 '18 at 18:31



















            0












            $begingroup$

            By Rolle's theorem, there exists $cin(a,b)$ such that $f'(c)=0$.



            Now $f''(x)le 0$ in $(a,b)$ implies $f'(x)$ is non-increasing in $[a,b]$, so




            • if $ale xle c$, then $f'(x)ge f'(c)=0$, in other words $f(x)$ is non-decreasing in $[a,c]$, so that for all $xin[a,c]$, $f(x)ge f(a)=0$;

            • if $cle xle b$, then $f'(x)le f'(c)=0$, in other words $f(x)$ is non-increasing in $[c,b]$, so that for all $xin[c,b]$, $f(x)ge f(b)=0$.


            Thus, in each case, $f(x)ge 0$.






            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%2f3018957%2fapplication-of-mean-value-and-rolles-theorems-twice-differentiable-functions%23new-answer', 'question_page');
              }
              );

              Post as a guest















              Required, but never shown

























              3 Answers
              3






              active

              oldest

              votes








              3 Answers
              3






              active

              oldest

              votes









              active

              oldest

              votes






              active

              oldest

              votes









              0












              $begingroup$

              $f$ is continuous in $[a,b]$ and differentiable in $(a,b), f(a)=f(b)=0$



              By Rolle's Theorem, $exists cin(a,b)$ such that $f'(c)=frac{f(b)-f(a)}{b-a}=0$



              $f''(x)leq0, forall xin(a,b)$



              $impliesforall xin(a,c), int_{x}^{c}f''(x)dx=f'(c)-f'(x)=-f'(x)leq0$



              $impliesforall xin(a,c), f'(x)geq0$



              $implies int_{a}^{x} f'(x)dx=f(x)-f(a)=f(x)geq0$



              Similarly, $forall xin(c,b), f'(x)leq0$



              $impliesint_{x}^{b} f'(x)dx=f(b)-f(x)=-f(x)leq0$



              $implies f(x)geq0$



              As far as $f(c)$ is concerned, continuity requires $f(c)geq0$






              share|cite|improve this answer









              $endgroup$


















                0












                $begingroup$

                $f$ is continuous in $[a,b]$ and differentiable in $(a,b), f(a)=f(b)=0$



                By Rolle's Theorem, $exists cin(a,b)$ such that $f'(c)=frac{f(b)-f(a)}{b-a}=0$



                $f''(x)leq0, forall xin(a,b)$



                $impliesforall xin(a,c), int_{x}^{c}f''(x)dx=f'(c)-f'(x)=-f'(x)leq0$



                $impliesforall xin(a,c), f'(x)geq0$



                $implies int_{a}^{x} f'(x)dx=f(x)-f(a)=f(x)geq0$



                Similarly, $forall xin(c,b), f'(x)leq0$



                $impliesint_{x}^{b} f'(x)dx=f(b)-f(x)=-f(x)leq0$



                $implies f(x)geq0$



                As far as $f(c)$ is concerned, continuity requires $f(c)geq0$






                share|cite|improve this answer









                $endgroup$
















                  0












                  0








                  0





                  $begingroup$

                  $f$ is continuous in $[a,b]$ and differentiable in $(a,b), f(a)=f(b)=0$



                  By Rolle's Theorem, $exists cin(a,b)$ such that $f'(c)=frac{f(b)-f(a)}{b-a}=0$



                  $f''(x)leq0, forall xin(a,b)$



                  $impliesforall xin(a,c), int_{x}^{c}f''(x)dx=f'(c)-f'(x)=-f'(x)leq0$



                  $impliesforall xin(a,c), f'(x)geq0$



                  $implies int_{a}^{x} f'(x)dx=f(x)-f(a)=f(x)geq0$



                  Similarly, $forall xin(c,b), f'(x)leq0$



                  $impliesint_{x}^{b} f'(x)dx=f(b)-f(x)=-f(x)leq0$



                  $implies f(x)geq0$



                  As far as $f(c)$ is concerned, continuity requires $f(c)geq0$






                  share|cite|improve this answer









                  $endgroup$



                  $f$ is continuous in $[a,b]$ and differentiable in $(a,b), f(a)=f(b)=0$



                  By Rolle's Theorem, $exists cin(a,b)$ such that $f'(c)=frac{f(b)-f(a)}{b-a}=0$



                  $f''(x)leq0, forall xin(a,b)$



                  $impliesforall xin(a,c), int_{x}^{c}f''(x)dx=f'(c)-f'(x)=-f'(x)leq0$



                  $impliesforall xin(a,c), f'(x)geq0$



                  $implies int_{a}^{x} f'(x)dx=f(x)-f(a)=f(x)geq0$



                  Similarly, $forall xin(c,b), f'(x)leq0$



                  $impliesint_{x}^{b} f'(x)dx=f(b)-f(x)=-f(x)leq0$



                  $implies f(x)geq0$



                  As far as $f(c)$ is concerned, continuity requires $f(c)geq0$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Nov 29 '18 at 18:27









                  Shubham JohriShubham Johri

                  5,137717




                  5,137717























                      0












                      $begingroup$

                      Assume that $f(c) < 0$ for some $c in (a, b)$, and apply Rolle's theorem
                      repeatedly:




                      • There is an $x_1 in (a, c)$ with $f'(x_1) = frac{f(c)-f(a)}{c-a} < 0$.


                      • There is an $x_2 in (c, b)$ with $f'(x_2) = frac{f(b)-f(c)}{b-c} > 0$.


                      • There is an $x_3 in (x_1, x_2)$ with $f''(x_3) = frac{f'(x_2)-f'(x_1)}{x_2-x_1} > 0$.



                      This is a contradiction to the assumption that $f''(x) le 0$
                      for all $x in (a, b)$.




                      An alternative approach: Fix any $c in (a, b)$ and define
                      $$
                      g(x) = (c-a)(c-b)f(x) - (x-a)(x-b)f(c) , .
                      $$
                      Then $g(a) = g(c) = g(b) = 0$, and repeated application of Rolle's
                      theorem shows that $g''(xi) = 0$ for some $xi in (a, b)$.
                      Then
                      $$
                      0 = g''(xi) = (c-a)(c-b)f''(xi) - 2f(c) \
                      implies f(c) = frac 12 (c-a)(c-b)f''(xi) ge 0 , .
                      $$




                      Yet another approach: Assume that $f$ attains its minimum at $c in (a, b)$ and $f(c) < 0$. Then $f'(c) = 0$ and Taylor's formula gives
                      a contradiction:
                      $$
                      f(b) = f(c) + frac 12 (b-c)^2 f''(xi) < 0
                      $$
                      for some $xi in (c, b)$.



                      The real idea here is that $f$ is concave,
                      and a continuous concave function on a compact interval attains its
                      minimum at one of the boundary points: Therefore
                      $$
                      f(x) ge min (f(a), f(b)) = 0
                      $$
                      for $x in (a, b)$.






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        Thank you! I get the first method
                        $endgroup$
                        – Aishwarya Deore
                        Nov 29 '18 at 18:31
















                      0












                      $begingroup$

                      Assume that $f(c) < 0$ for some $c in (a, b)$, and apply Rolle's theorem
                      repeatedly:




                      • There is an $x_1 in (a, c)$ with $f'(x_1) = frac{f(c)-f(a)}{c-a} < 0$.


                      • There is an $x_2 in (c, b)$ with $f'(x_2) = frac{f(b)-f(c)}{b-c} > 0$.


                      • There is an $x_3 in (x_1, x_2)$ with $f''(x_3) = frac{f'(x_2)-f'(x_1)}{x_2-x_1} > 0$.



                      This is a contradiction to the assumption that $f''(x) le 0$
                      for all $x in (a, b)$.




                      An alternative approach: Fix any $c in (a, b)$ and define
                      $$
                      g(x) = (c-a)(c-b)f(x) - (x-a)(x-b)f(c) , .
                      $$
                      Then $g(a) = g(c) = g(b) = 0$, and repeated application of Rolle's
                      theorem shows that $g''(xi) = 0$ for some $xi in (a, b)$.
                      Then
                      $$
                      0 = g''(xi) = (c-a)(c-b)f''(xi) - 2f(c) \
                      implies f(c) = frac 12 (c-a)(c-b)f''(xi) ge 0 , .
                      $$




                      Yet another approach: Assume that $f$ attains its minimum at $c in (a, b)$ and $f(c) < 0$. Then $f'(c) = 0$ and Taylor's formula gives
                      a contradiction:
                      $$
                      f(b) = f(c) + frac 12 (b-c)^2 f''(xi) < 0
                      $$
                      for some $xi in (c, b)$.



                      The real idea here is that $f$ is concave,
                      and a continuous concave function on a compact interval attains its
                      minimum at one of the boundary points: Therefore
                      $$
                      f(x) ge min (f(a), f(b)) = 0
                      $$
                      for $x in (a, b)$.






                      share|cite|improve this answer











                      $endgroup$













                      • $begingroup$
                        Thank you! I get the first method
                        $endgroup$
                        – Aishwarya Deore
                        Nov 29 '18 at 18:31














                      0












                      0








                      0





                      $begingroup$

                      Assume that $f(c) < 0$ for some $c in (a, b)$, and apply Rolle's theorem
                      repeatedly:




                      • There is an $x_1 in (a, c)$ with $f'(x_1) = frac{f(c)-f(a)}{c-a} < 0$.


                      • There is an $x_2 in (c, b)$ with $f'(x_2) = frac{f(b)-f(c)}{b-c} > 0$.


                      • There is an $x_3 in (x_1, x_2)$ with $f''(x_3) = frac{f'(x_2)-f'(x_1)}{x_2-x_1} > 0$.



                      This is a contradiction to the assumption that $f''(x) le 0$
                      for all $x in (a, b)$.




                      An alternative approach: Fix any $c in (a, b)$ and define
                      $$
                      g(x) = (c-a)(c-b)f(x) - (x-a)(x-b)f(c) , .
                      $$
                      Then $g(a) = g(c) = g(b) = 0$, and repeated application of Rolle's
                      theorem shows that $g''(xi) = 0$ for some $xi in (a, b)$.
                      Then
                      $$
                      0 = g''(xi) = (c-a)(c-b)f''(xi) - 2f(c) \
                      implies f(c) = frac 12 (c-a)(c-b)f''(xi) ge 0 , .
                      $$




                      Yet another approach: Assume that $f$ attains its minimum at $c in (a, b)$ and $f(c) < 0$. Then $f'(c) = 0$ and Taylor's formula gives
                      a contradiction:
                      $$
                      f(b) = f(c) + frac 12 (b-c)^2 f''(xi) < 0
                      $$
                      for some $xi in (c, b)$.



                      The real idea here is that $f$ is concave,
                      and a continuous concave function on a compact interval attains its
                      minimum at one of the boundary points: Therefore
                      $$
                      f(x) ge min (f(a), f(b)) = 0
                      $$
                      for $x in (a, b)$.






                      share|cite|improve this answer











                      $endgroup$



                      Assume that $f(c) < 0$ for some $c in (a, b)$, and apply Rolle's theorem
                      repeatedly:




                      • There is an $x_1 in (a, c)$ with $f'(x_1) = frac{f(c)-f(a)}{c-a} < 0$.


                      • There is an $x_2 in (c, b)$ with $f'(x_2) = frac{f(b)-f(c)}{b-c} > 0$.


                      • There is an $x_3 in (x_1, x_2)$ with $f''(x_3) = frac{f'(x_2)-f'(x_1)}{x_2-x_1} > 0$.



                      This is a contradiction to the assumption that $f''(x) le 0$
                      for all $x in (a, b)$.




                      An alternative approach: Fix any $c in (a, b)$ and define
                      $$
                      g(x) = (c-a)(c-b)f(x) - (x-a)(x-b)f(c) , .
                      $$
                      Then $g(a) = g(c) = g(b) = 0$, and repeated application of Rolle's
                      theorem shows that $g''(xi) = 0$ for some $xi in (a, b)$.
                      Then
                      $$
                      0 = g''(xi) = (c-a)(c-b)f''(xi) - 2f(c) \
                      implies f(c) = frac 12 (c-a)(c-b)f''(xi) ge 0 , .
                      $$




                      Yet another approach: Assume that $f$ attains its minimum at $c in (a, b)$ and $f(c) < 0$. Then $f'(c) = 0$ and Taylor's formula gives
                      a contradiction:
                      $$
                      f(b) = f(c) + frac 12 (b-c)^2 f''(xi) < 0
                      $$
                      for some $xi in (c, b)$.



                      The real idea here is that $f$ is concave,
                      and a continuous concave function on a compact interval attains its
                      minimum at one of the boundary points: Therefore
                      $$
                      f(x) ge min (f(a), f(b)) = 0
                      $$
                      for $x in (a, b)$.







                      share|cite|improve this answer














                      share|cite|improve this answer



                      share|cite|improve this answer








                      edited Nov 29 '18 at 18:47

























                      answered Nov 29 '18 at 18:17









                      Martin RMartin R

                      28.3k33356




                      28.3k33356












                      • $begingroup$
                        Thank you! I get the first method
                        $endgroup$
                        – Aishwarya Deore
                        Nov 29 '18 at 18:31


















                      • $begingroup$
                        Thank you! I get the first method
                        $endgroup$
                        – Aishwarya Deore
                        Nov 29 '18 at 18:31
















                      $begingroup$
                      Thank you! I get the first method
                      $endgroup$
                      – Aishwarya Deore
                      Nov 29 '18 at 18:31




                      $begingroup$
                      Thank you! I get the first method
                      $endgroup$
                      – Aishwarya Deore
                      Nov 29 '18 at 18:31











                      0












                      $begingroup$

                      By Rolle's theorem, there exists $cin(a,b)$ such that $f'(c)=0$.



                      Now $f''(x)le 0$ in $(a,b)$ implies $f'(x)$ is non-increasing in $[a,b]$, so




                      • if $ale xle c$, then $f'(x)ge f'(c)=0$, in other words $f(x)$ is non-decreasing in $[a,c]$, so that for all $xin[a,c]$, $f(x)ge f(a)=0$;

                      • if $cle xle b$, then $f'(x)le f'(c)=0$, in other words $f(x)$ is non-increasing in $[c,b]$, so that for all $xin[c,b]$, $f(x)ge f(b)=0$.


                      Thus, in each case, $f(x)ge 0$.






                      share|cite|improve this answer









                      $endgroup$


















                        0












                        $begingroup$

                        By Rolle's theorem, there exists $cin(a,b)$ such that $f'(c)=0$.



                        Now $f''(x)le 0$ in $(a,b)$ implies $f'(x)$ is non-increasing in $[a,b]$, so




                        • if $ale xle c$, then $f'(x)ge f'(c)=0$, in other words $f(x)$ is non-decreasing in $[a,c]$, so that for all $xin[a,c]$, $f(x)ge f(a)=0$;

                        • if $cle xle b$, then $f'(x)le f'(c)=0$, in other words $f(x)$ is non-increasing in $[c,b]$, so that for all $xin[c,b]$, $f(x)ge f(b)=0$.


                        Thus, in each case, $f(x)ge 0$.






                        share|cite|improve this answer









                        $endgroup$
















                          0












                          0








                          0





                          $begingroup$

                          By Rolle's theorem, there exists $cin(a,b)$ such that $f'(c)=0$.



                          Now $f''(x)le 0$ in $(a,b)$ implies $f'(x)$ is non-increasing in $[a,b]$, so




                          • if $ale xle c$, then $f'(x)ge f'(c)=0$, in other words $f(x)$ is non-decreasing in $[a,c]$, so that for all $xin[a,c]$, $f(x)ge f(a)=0$;

                          • if $cle xle b$, then $f'(x)le f'(c)=0$, in other words $f(x)$ is non-increasing in $[c,b]$, so that for all $xin[c,b]$, $f(x)ge f(b)=0$.


                          Thus, in each case, $f(x)ge 0$.






                          share|cite|improve this answer









                          $endgroup$



                          By Rolle's theorem, there exists $cin(a,b)$ such that $f'(c)=0$.



                          Now $f''(x)le 0$ in $(a,b)$ implies $f'(x)$ is non-increasing in $[a,b]$, so




                          • if $ale xle c$, then $f'(x)ge f'(c)=0$, in other words $f(x)$ is non-decreasing in $[a,c]$, so that for all $xin[a,c]$, $f(x)ge f(a)=0$;

                          • if $cle xle b$, then $f'(x)le f'(c)=0$, in other words $f(x)$ is non-increasing in $[c,b]$, so that for all $xin[c,b]$, $f(x)ge f(b)=0$.


                          Thus, in each case, $f(x)ge 0$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered Nov 29 '18 at 19:13









                          BernardBernard

                          120k740115




                          120k740115






























                              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%2f3018957%2fapplication-of-mean-value-and-rolles-theorems-twice-differentiable-functions%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

                              How to send String Array data to Server using php in android

                              Image
                              0 I have a String array which contains value string array = {6,37,38,837,7,...} and I have also used some string values. I want to send that all string values and string array to my database which is located at server side. When I am click on submit button the data will stored in table. On the PHP side I want to read this array as it is,other thing i will do at background means PHP side. See my below code then you will get it what I am says. Here is my code: public class Hotels extends Activity { // Declare Variables JSONArray jsonarray = null; private static final String TAG_ID = "id"; public static final String TAG_NAME = "name"; public static final String TAG_LOCATION = "location"; public static final String TAG_DESC = "description"; String f_date, l_date;...
                              Read more

                              Title Spacing in Bjornstrup Chapter, Removing Chapter Number From Contents

                              Image
                              up vote 5 down vote favorite 2 I would like to make it so that the vertical space below the title of each chapter is the same when using the Bjornstrup package (i.e. so that the descenders do not affect the spacing - as shown by the red arrow in the picture below). I think I'm fairly happy with the spacing around "A chapter", so I'd really like titles without descenders (e.g. "Contents") to have the same vertical spacing as "A chapter" below the title. I would also like to remove the large "0" in the contents title. Any help appreciated, as always. documentclass[11pt,a4paper]{book} % Packages[![enter image description here][1]][1] usepackage[left=3.5cm, right=3.5cm, top=4.3cm, bottom=4.25cm]{geometry} usepackage[utf8]{inputenc} usepackage{libertine} usepackage[libertine]{newtx...
                              Read more

                              Is anime1.com a legal site for watching anime?

                              Image
                              up vote 4 down vote favorite 1 Other anime streaming sites are blocking my IP because I'm Iranian while anime1.com isn't. I wanted to know if it's legal to watch/stream anime from this site. Thanks. anime-production share | improve this question edited Nov 27 at 22:38 W. Are 1,599 2 23 asked Nov 27 at 20:36 mhnajafi 23 3 add a comment ...
                              Read more