$f, g$ continuous in $[a, b]$, differentiable in (a, b), then there exists $c in (a, b)$ such that $[f(b) -...












0












$begingroup$


If $f, g$ are continuous real functions in $[a, b]$, differentiable in (a, b), then there exists $c in (a, b)$ such that $[f(b) - f(a)]g'(c) = [g(b) - g(a)]f'(c)$



I tried to define $h(x) = f(x) - g(x)$ and apply the mean theorem, but it didn't work... I also tried to define $h(x) = f(x)g(x)$ and apply the mean theorem and it didn't work either. I guess I do not have to use the mean theorem... Actually, the mean theorem is a consequence of this result, right?



Does someone know how to prove this exercise? It would be really helpful!



Thanks!






calculus real-analysis analysis







share|cite|improve this question









$endgroup$












  • $begingroup$
    This is Cauchy's mean value theorem. It's given as an exercise?
    $endgroup$
    – zhw.
    Nov 29 '15 at 22:46










  • $begingroup$
    And yes, the Mean Value Theorem which can commonly be found in Stewart's Calculus textbooks is a consequence of this statement (if you take $g:[a,b]rightarrow [a,b]$ given by $g(x)=x$).
    $endgroup$
    – Sinister Cutlass
    Nov 29 '15 at 22:48












  • $begingroup$
    here is a nice simple video explaining it: larsoncalculus.com/calc10/content/proof-videos/chapter-8/…
    $endgroup$
    – Fede Poncio
    Nov 29 '15 at 22:55










  • $begingroup$
    Use $h(x) = [f(b)-f(a)]g'(x) - [g(b)-g(a)]f'(x)$. (In other words, replace the $c$ with the independent variable to define $h$, so that when you apply the Mean Value Theorem to $h$, the theorem will give you the desired $c$.)
    $endgroup$
    – Toby Bartels
    Jan 14 at 6:12
















0












$begingroup$


If $f, g$ are continuous real functions in $[a, b]$, differentiable in (a, b), then there exists $c in (a, b)$ such that $[f(b) - f(a)]g'(c) = [g(b) - g(a)]f'(c)$



I tried to define $h(x) = f(x) - g(x)$ and apply the mean theorem, but it didn't work... I also tried to define $h(x) = f(x)g(x)$ and apply the mean theorem and it didn't work either. I guess I do not have to use the mean theorem... Actually, the mean theorem is a consequence of this result, right?



Does someone know how to prove this exercise? It would be really helpful!



Thanks!






calculus real-analysis analysis







share|cite|improve this question









$endgroup$












  • $begingroup$
    This is Cauchy's mean value theorem. It's given as an exercise?
    $endgroup$
    – zhw.
    Nov 29 '15 at 22:46










  • $begingroup$
    And yes, the Mean Value Theorem which can commonly be found in Stewart's Calculus textbooks is a consequence of this statement (if you take $g:[a,b]rightarrow [a,b]$ given by $g(x)=x$).
    $endgroup$
    – Sinister Cutlass
    Nov 29 '15 at 22:48












  • $begingroup$
    here is a nice simple video explaining it: larsoncalculus.com/calc10/content/proof-videos/chapter-8/…
    $endgroup$
    – Fede Poncio
    Nov 29 '15 at 22:55










  • $begingroup$
    Use $h(x) = [f(b)-f(a)]g'(x) - [g(b)-g(a)]f'(x)$. (In other words, replace the $c$ with the independent variable to define $h$, so that when you apply the Mean Value Theorem to $h$, the theorem will give you the desired $c$.)
    $endgroup$
    – Toby Bartels
    Jan 14 at 6:12














0












0








0


1



$begingroup$


If $f, g$ are continuous real functions in $[a, b]$, differentiable in (a, b), then there exists $c in (a, b)$ such that $[f(b) - f(a)]g'(c) = [g(b) - g(a)]f'(c)$



I tried to define $h(x) = f(x) - g(x)$ and apply the mean theorem, but it didn't work... I also tried to define $h(x) = f(x)g(x)$ and apply the mean theorem and it didn't work either. I guess I do not have to use the mean theorem... Actually, the mean theorem is a consequence of this result, right?



Does someone know how to prove this exercise? It would be really helpful!



Thanks!






calculus real-analysis analysis







share|cite|improve this question









$endgroup$




If $f, g$ are continuous real functions in $[a, b]$, differentiable in (a, b), then there exists $c in (a, b)$ such that $[f(b) - f(a)]g'(c) = [g(b) - g(a)]f'(c)$



I tried to define $h(x) = f(x) - g(x)$ and apply the mean theorem, but it didn't work... I also tried to define $h(x) = f(x)g(x)$ and apply the mean theorem and it didn't work either. I guess I do not have to use the mean theorem... Actually, the mean theorem is a consequence of this result, right?



Does someone know how to prove this exercise? It would be really helpful!



Thanks!






calculus real-analysis analysis




calculus real-analysis analysis






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Nov 29 '15 at 22:32









JoanJoan

332




332












  • $begingroup$
    This is Cauchy's mean value theorem. It's given as an exercise?
    $endgroup$
    – zhw.
    Nov 29 '15 at 22:46










  • $begingroup$
    And yes, the Mean Value Theorem which can commonly be found in Stewart's Calculus textbooks is a consequence of this statement (if you take $g:[a,b]rightarrow [a,b]$ given by $g(x)=x$).
    $endgroup$
    – Sinister Cutlass
    Nov 29 '15 at 22:48












  • $begingroup$
    here is a nice simple video explaining it: larsoncalculus.com/calc10/content/proof-videos/chapter-8/…
    $endgroup$
    – Fede Poncio
    Nov 29 '15 at 22:55










  • $begingroup$
    Use $h(x) = [f(b)-f(a)]g'(x) - [g(b)-g(a)]f'(x)$. (In other words, replace the $c$ with the independent variable to define $h$, so that when you apply the Mean Value Theorem to $h$, the theorem will give you the desired $c$.)
    $endgroup$
    – Toby Bartels
    Jan 14 at 6:12


















  • $begingroup$
    This is Cauchy's mean value theorem. It's given as an exercise?
    $endgroup$
    – zhw.
    Nov 29 '15 at 22:46










  • $begingroup$
    And yes, the Mean Value Theorem which can commonly be found in Stewart's Calculus textbooks is a consequence of this statement (if you take $g:[a,b]rightarrow [a,b]$ given by $g(x)=x$).
    $endgroup$
    – Sinister Cutlass
    Nov 29 '15 at 22:48












  • $begingroup$
    here is a nice simple video explaining it: larsoncalculus.com/calc10/content/proof-videos/chapter-8/…
    $endgroup$
    – Fede Poncio
    Nov 29 '15 at 22:55










  • $begingroup$
    Use $h(x) = [f(b)-f(a)]g'(x) - [g(b)-g(a)]f'(x)$. (In other words, replace the $c$ with the independent variable to define $h$, so that when you apply the Mean Value Theorem to $h$, the theorem will give you the desired $c$.)
    $endgroup$
    – Toby Bartels
    Jan 14 at 6:12
















$begingroup$
This is Cauchy's mean value theorem. It's given as an exercise?
$endgroup$
– zhw.
Nov 29 '15 at 22:46




$begingroup$
This is Cauchy's mean value theorem. It's given as an exercise?
$endgroup$
– zhw.
Nov 29 '15 at 22:46












$begingroup$
And yes, the Mean Value Theorem which can commonly be found in Stewart's Calculus textbooks is a consequence of this statement (if you take $g:[a,b]rightarrow [a,b]$ given by $g(x)=x$).
$endgroup$
– Sinister Cutlass
Nov 29 '15 at 22:48






$begingroup$
And yes, the Mean Value Theorem which can commonly be found in Stewart's Calculus textbooks is a consequence of this statement (if you take $g:[a,b]rightarrow [a,b]$ given by $g(x)=x$).
$endgroup$
– Sinister Cutlass
Nov 29 '15 at 22:48














$begingroup$
here is a nice simple video explaining it: larsoncalculus.com/calc10/content/proof-videos/chapter-8/…
$endgroup$
– Fede Poncio
Nov 29 '15 at 22:55




$begingroup$
here is a nice simple video explaining it: larsoncalculus.com/calc10/content/proof-videos/chapter-8/…
$endgroup$
– Fede Poncio
Nov 29 '15 at 22:55












$begingroup$
Use $h(x) = [f(b)-f(a)]g'(x) - [g(b)-g(a)]f'(x)$. (In other words, replace the $c$ with the independent variable to define $h$, so that when you apply the Mean Value Theorem to $h$, the theorem will give you the desired $c$.)
$endgroup$
– Toby Bartels
Jan 14 at 6:12




$begingroup$
Use $h(x) = [f(b)-f(a)]g'(x) - [g(b)-g(a)]f'(x)$. (In other words, replace the $c$ with the independent variable to define $h$, so that when you apply the Mean Value Theorem to $h$, the theorem will give you the desired $c$.)
$endgroup$
– Toby Bartels
Jan 14 at 6:12










1 Answer
1






active

oldest

votes


















0












$begingroup$

Use $$ h(x) = [f(b) - f(a)] g(x) - [g(b) - g(a)] f(x) .$$ (In other words, replace the $c$ with the independent variable to define $h$, so that when you apply the Mean Value Theorem to $h$, the theorem will give you the desired $c$.) Notice that $h(a) = h(b)$ (both equal $f(a) g(b) - f(b) g(a)$), so the Mean Value Theorem (or Rolle's Theorem) give you $c in (a,b)$ such that $h'(c) = 0$. But $h'(c) = [f(b) - f(a)] g'(c) - [g(b) - g(a)] f'(c)$, so there you go.



Incidentally, people often phrase this theorem (Cauchy's Mean Value Theorem) with the conclusion that $$ frac{f'(c)} {g'(c)} = frac{f(b) - f(a)} {g(b) - g(a)} .$$ For this, you need some fine print to guarantee that this won't involve division by zero (such as that $g' ne 0$ on $(a,b)$), but your version without division is nice because it works regardless of that.






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%2f1552237%2ff-g-continuous-in-a-b-differentiable-in-a-b-then-there-exists-c-i%23new-answer', 'question_page');
    }
    );

    Post as a guest















    Required, but never shown

























    1 Answer
    1






    active

    oldest

    votes








    1 Answer
    1






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    0












    $begingroup$

    Use $$ h(x) = [f(b) - f(a)] g(x) - [g(b) - g(a)] f(x) .$$ (In other words, replace the $c$ with the independent variable to define $h$, so that when you apply the Mean Value Theorem to $h$, the theorem will give you the desired $c$.) Notice that $h(a) = h(b)$ (both equal $f(a) g(b) - f(b) g(a)$), so the Mean Value Theorem (or Rolle's Theorem) give you $c in (a,b)$ such that $h'(c) = 0$. But $h'(c) = [f(b) - f(a)] g'(c) - [g(b) - g(a)] f'(c)$, so there you go.



    Incidentally, people often phrase this theorem (Cauchy's Mean Value Theorem) with the conclusion that $$ frac{f'(c)} {g'(c)} = frac{f(b) - f(a)} {g(b) - g(a)} .$$ For this, you need some fine print to guarantee that this won't involve division by zero (such as that $g' ne 0$ on $(a,b)$), but your version without division is nice because it works regardless of that.






    share|cite|improve this answer









    $endgroup$


















      0












      $begingroup$

      Use $$ h(x) = [f(b) - f(a)] g(x) - [g(b) - g(a)] f(x) .$$ (In other words, replace the $c$ with the independent variable to define $h$, so that when you apply the Mean Value Theorem to $h$, the theorem will give you the desired $c$.) Notice that $h(a) = h(b)$ (both equal $f(a) g(b) - f(b) g(a)$), so the Mean Value Theorem (or Rolle's Theorem) give you $c in (a,b)$ such that $h'(c) = 0$. But $h'(c) = [f(b) - f(a)] g'(c) - [g(b) - g(a)] f'(c)$, so there you go.



      Incidentally, people often phrase this theorem (Cauchy's Mean Value Theorem) with the conclusion that $$ frac{f'(c)} {g'(c)} = frac{f(b) - f(a)} {g(b) - g(a)} .$$ For this, you need some fine print to guarantee that this won't involve division by zero (such as that $g' ne 0$ on $(a,b)$), but your version without division is nice because it works regardless of that.






      share|cite|improve this answer









      $endgroup$
















        0












        0








        0





        $begingroup$

        Use $$ h(x) = [f(b) - f(a)] g(x) - [g(b) - g(a)] f(x) .$$ (In other words, replace the $c$ with the independent variable to define $h$, so that when you apply the Mean Value Theorem to $h$, the theorem will give you the desired $c$.) Notice that $h(a) = h(b)$ (both equal $f(a) g(b) - f(b) g(a)$), so the Mean Value Theorem (or Rolle's Theorem) give you $c in (a,b)$ such that $h'(c) = 0$. But $h'(c) = [f(b) - f(a)] g'(c) - [g(b) - g(a)] f'(c)$, so there you go.



        Incidentally, people often phrase this theorem (Cauchy's Mean Value Theorem) with the conclusion that $$ frac{f'(c)} {g'(c)} = frac{f(b) - f(a)} {g(b) - g(a)} .$$ For this, you need some fine print to guarantee that this won't involve division by zero (such as that $g' ne 0$ on $(a,b)$), but your version without division is nice because it works regardless of that.






        share|cite|improve this answer









        $endgroup$



        Use $$ h(x) = [f(b) - f(a)] g(x) - [g(b) - g(a)] f(x) .$$ (In other words, replace the $c$ with the independent variable to define $h$, so that when you apply the Mean Value Theorem to $h$, the theorem will give you the desired $c$.) Notice that $h(a) = h(b)$ (both equal $f(a) g(b) - f(b) g(a)$), so the Mean Value Theorem (or Rolle's Theorem) give you $c in (a,b)$ such that $h'(c) = 0$. But $h'(c) = [f(b) - f(a)] g'(c) - [g(b) - g(a)] f'(c)$, so there you go.



        Incidentally, people often phrase this theorem (Cauchy's Mean Value Theorem) with the conclusion that $$ frac{f'(c)} {g'(c)} = frac{f(b) - f(a)} {g(b) - g(a)} .$$ For this, you need some fine print to guarantee that this won't involve division by zero (such as that $g' ne 0$ on $(a,b)$), but your version without division is nice because it works regardless of that.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Jan 14 at 6:22









        Toby BartelsToby Bartels

        669515




        669515






























            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%2f1552237%2ff-g-continuous-in-a-b-differentiable-in-a-b-then-there-exists-c-i%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

            Questions related to Moebius Transform of Characteristic Function of the Primes

            Image
            2 $begingroup$ Consider the function defined in (1) below related to the fundamental prime counting function $pi(x)$ . Note that A143519(n) is not multiplicative. (1) $quad f(x)=sumlimits_{n=1}^{x}A143519(n)$ https://oeis.org/A143519 The following plot illustrates $f(x)$ defined in formula (1) above. Figure (1) : Illustration of $f(x)$ defined in formula (1) The integer zeros of $f(x)$ for $xle 10,000$ are listed in (2) below. (2) $quad$ {1,6,9,12,19,30,79,80,81,116,193,201,287,288,291,668,673,679,680,685,686,1109} The zero crossings of $f(x)$ for $xle 10,000$ where $f(x)$ doesn't settle at zero are listed in (3) below. (3) $quad$ {14,21,33,114,115,118,195,286,290,295,442,445,665,667,670,671,678,682} Question (1) : Does $f(x)$ have a finite number of integer zeros, and if so ...
            Read more

            List of scandals in India

            Image
            List of scandals in India From Wikipedia, the free encyclopedia Jump to navigation Jump to search This is a dynamic list and may never be able to satisfy particular standards for completeness. You can help by expanding it with reliably sourced entries. This article has multiple issues. Please help improve it or discuss these issues on the talk page . (Learn how and when to remove these template messages) This article has an unclear citation style . The references used may be made clearer with a different or consistent style of citation and footnoting. ( January 2014 ) (Learn how and when to remove this template message) This list is in reverse order. As a consequence it appears to be slanted towards recent events. Please try to keep recent events in historical perspective and add more content related to non-recent events. ( May 2013 ) (Learn how and when to remove this template message) ...
            Read more

            Can not write log (Is /dev/pts mounted?) - openpty in Ubuntu-on-Windows?

            Image
            0 1 What exactly does the error message beginning with E: mean? I assume it has to do with the file structure on Ubuntu-on-Windows, but what exactly? The following package was automatically installed and is no longer required: os-prober Use 'apt-get autoremove' to remove it. The following extra packages will be installed: libxslt1.1 The following NEW packages will be installed: libxslt1.1 xmlstarlet 0 upgraded, 2 newly installed, 0 to remove and 0 not upgraded. Need to get 435 kB of archives. After this operation, 1023 kB of additional disk space will be used. Get:1 http://archive.ubuntu.com/ubuntu/ trusty/main libxslt1.1 amd64 1.1.28-2build1 [145 kB] Get:2 http://archive.ubuntu.com/ubuntu/ trusty/universe xmlstarlet amd64 1.5.0-1 [290 kB] ...
            Read more