Is inverse use of mean value theorem right?












4












$begingroup$


If we have $f$ is differentiable on $(a,b)$, and continuous on $[a,b]$, then



for any $xin (a,b)$, exists $y, z in [a,b]$, such that



$f '(x)=dfrac{f(z)-f(y)}{z-y}$



Is this right?










share|cite|improve this question











$endgroup$

















    4












    $begingroup$


    If we have $f$ is differentiable on $(a,b)$, and continuous on $[a,b]$, then



    for any $xin (a,b)$, exists $y, z in [a,b]$, such that



    $f '(x)=dfrac{f(z)-f(y)}{z-y}$



    Is this right?










    share|cite|improve this question











    $endgroup$















      4












      4








      4


      1



      $begingroup$


      If we have $f$ is differentiable on $(a,b)$, and continuous on $[a,b]$, then



      for any $xin (a,b)$, exists $y, z in [a,b]$, such that



      $f '(x)=dfrac{f(z)-f(y)}{z-y}$



      Is this right?










      share|cite|improve this question











      $endgroup$




      If we have $f$ is differentiable on $(a,b)$, and continuous on $[a,b]$, then



      for any $xin (a,b)$, exists $y, z in [a,b]$, such that



      $f '(x)=dfrac{f(z)-f(y)}{z-y}$



      Is this right?







      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 '12 at 16:39









      amWhy

      192k28225439




      192k28225439










      asked Nov 29 '12 at 16:36









      user46262user46262

      1094




      1094






















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          $begingroup$

          No. Consider $f(x)=x^3$ on the interval $[-1,1]$ with $x=0$.






          share|cite|improve this answer









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            1 Answer
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            1 Answer
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            active

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            active

            oldest

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            12












            $begingroup$

            No. Consider $f(x)=x^3$ on the interval $[-1,1]$ with $x=0$.






            share|cite|improve this answer









            $endgroup$


















              12












              $begingroup$

              No. Consider $f(x)=x^3$ on the interval $[-1,1]$ with $x=0$.






              share|cite|improve this answer









              $endgroup$
















                12












                12








                12





                $begingroup$

                No. Consider $f(x)=x^3$ on the interval $[-1,1]$ with $x=0$.






                share|cite|improve this answer









                $endgroup$



                No. Consider $f(x)=x^3$ on the interval $[-1,1]$ with $x=0$.







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Nov 29 '12 at 16:39









                David MitraDavid Mitra

                63.1k6100164




                63.1k6100164






























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