Equation of secant line in mean value theorem proof












3












$begingroup$


I'm going through a proof for the mean value theorem.



We have a function $f(x)$ continuous on $[a, b]$ and differentiable on $(a, b)$.



Then we define a function $g(x)$ to be the secant line passing through $(a, f(a))$ and $(b, f(b))$.



The slope of said secant is:



$$m=frac{f(b)-f(a)}{b-a}$$



That is clear. Now the proof I'm following defines $g(x)$ like so:



$$g(x) = left[ frac{f(b)-f(a)}{b-a} right](x-a)+f(a)$$



What confuses me: why is the coefficient defined to be $(x-a)$ and not simply $(x)$.






calculus derivatives proof-explanation







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$endgroup$

















    3












    $begingroup$


    I'm going through a proof for the mean value theorem.



    We have a function $f(x)$ continuous on $[a, b]$ and differentiable on $(a, b)$.



    Then we define a function $g(x)$ to be the secant line passing through $(a, f(a))$ and $(b, f(b))$.



    The slope of said secant is:



    $$m=frac{f(b)-f(a)}{b-a}$$



    That is clear. Now the proof I'm following defines $g(x)$ like so:



    $$g(x) = left[ frac{f(b)-f(a)}{b-a} right](x-a)+f(a)$$



    What confuses me: why is the coefficient defined to be $(x-a)$ and not simply $(x)$.






    calculus derivatives proof-explanation







    share|cite|improve this question









    $endgroup$















      3












      3








      3





      $begingroup$


      I'm going through a proof for the mean value theorem.



      We have a function $f(x)$ continuous on $[a, b]$ and differentiable on $(a, b)$.



      Then we define a function $g(x)$ to be the secant line passing through $(a, f(a))$ and $(b, f(b))$.



      The slope of said secant is:



      $$m=frac{f(b)-f(a)}{b-a}$$



      That is clear. Now the proof I'm following defines $g(x)$ like so:



      $$g(x) = left[ frac{f(b)-f(a)}{b-a} right](x-a)+f(a)$$



      What confuses me: why is the coefficient defined to be $(x-a)$ and not simply $(x)$.






      calculus derivatives proof-explanation







      share|cite|improve this question









      $endgroup$




      I'm going through a proof for the mean value theorem.



      We have a function $f(x)$ continuous on $[a, b]$ and differentiable on $(a, b)$.



      Then we define a function $g(x)$ to be the secant line passing through $(a, f(a))$ and $(b, f(b))$.



      The slope of said secant is:



      $$m=frac{f(b)-f(a)}{b-a}$$



      That is clear. Now the proof I'm following defines $g(x)$ like so:



      $$g(x) = left[ frac{f(b)-f(a)}{b-a} right](x-a)+f(a)$$



      What confuses me: why is the coefficient defined to be $(x-a)$ and not simply $(x)$.






      calculus derivatives proof-explanation




      calculus derivatives proof-explanation






      share|cite|improve this question













      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked Dec 16 '18 at 9:13









      MaxMax

      638519




      638519






















          2 Answers
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          active

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          4












          $begingroup$

          The equation of the straight line passing through $(a,f(a)),(b,f(b))$ is given as:



          $displaystylefrac{y-f(a)}{x-a}=frac{f(b)-f(a)}{b-a}$



          $displaystyleimplies g(x)=y=Big[frac{f(b)-f(a)}{b-a}Big](x-a)+f(a)$






          share|cite|improve this answer









          $endgroup$





















            4












            $begingroup$

            First of all, it is not correct to assert that $g(x)$ is the secant line passing through $bigl(a,f(a)bigr)$ and $bigl(b,f(b)bigr)$. It is the function whose graph is the line segment uniting those two points.



            Now, concerning your question: it is so that $g(a)=f(a)$ and $g(b)=f(b)$.






            share|cite|improve this answer











            $endgroup$













            • $begingroup$
              You probably meant $g(a)=f(a)$ and $g(b)=f(b)$.
              $endgroup$
              – Taladris
              Dec 16 '18 at 14:21










            • $begingroup$
              I've edited my answer. Thank you.
              $endgroup$
              – José Carlos Santos
              Dec 16 '18 at 14:23











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            2 Answers
            2






            active

            oldest

            votes








            2 Answers
            2






            active

            oldest

            votes









            active

            oldest

            votes






            active

            oldest

            votes









            4












            $begingroup$

            The equation of the straight line passing through $(a,f(a)),(b,f(b))$ is given as:



            $displaystylefrac{y-f(a)}{x-a}=frac{f(b)-f(a)}{b-a}$



            $displaystyleimplies g(x)=y=Big[frac{f(b)-f(a)}{b-a}Big](x-a)+f(a)$






            share|cite|improve this answer









            $endgroup$


















              4












              $begingroup$

              The equation of the straight line passing through $(a,f(a)),(b,f(b))$ is given as:



              $displaystylefrac{y-f(a)}{x-a}=frac{f(b)-f(a)}{b-a}$



              $displaystyleimplies g(x)=y=Big[frac{f(b)-f(a)}{b-a}Big](x-a)+f(a)$






              share|cite|improve this answer









              $endgroup$
















                4












                4








                4





                $begingroup$

                The equation of the straight line passing through $(a,f(a)),(b,f(b))$ is given as:



                $displaystylefrac{y-f(a)}{x-a}=frac{f(b)-f(a)}{b-a}$



                $displaystyleimplies g(x)=y=Big[frac{f(b)-f(a)}{b-a}Big](x-a)+f(a)$






                share|cite|improve this answer









                $endgroup$



                The equation of the straight line passing through $(a,f(a)),(b,f(b))$ is given as:



                $displaystylefrac{y-f(a)}{x-a}=frac{f(b)-f(a)}{b-a}$



                $displaystyleimplies g(x)=y=Big[frac{f(b)-f(a)}{b-a}Big](x-a)+f(a)$







                share|cite|improve this answer












                share|cite|improve this answer



                share|cite|improve this answer










                answered Dec 16 '18 at 9:17









                Shubham JohriShubham Johri

                4,759717




                4,759717























                    4












                    $begingroup$

                    First of all, it is not correct to assert that $g(x)$ is the secant line passing through $bigl(a,f(a)bigr)$ and $bigl(b,f(b)bigr)$. It is the function whose graph is the line segment uniting those two points.



                    Now, concerning your question: it is so that $g(a)=f(a)$ and $g(b)=f(b)$.






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      You probably meant $g(a)=f(a)$ and $g(b)=f(b)$.
                      $endgroup$
                      – Taladris
                      Dec 16 '18 at 14:21










                    • $begingroup$
                      I've edited my answer. Thank you.
                      $endgroup$
                      – José Carlos Santos
                      Dec 16 '18 at 14:23
















                    4












                    $begingroup$

                    First of all, it is not correct to assert that $g(x)$ is the secant line passing through $bigl(a,f(a)bigr)$ and $bigl(b,f(b)bigr)$. It is the function whose graph is the line segment uniting those two points.



                    Now, concerning your question: it is so that $g(a)=f(a)$ and $g(b)=f(b)$.






                    share|cite|improve this answer











                    $endgroup$













                    • $begingroup$
                      You probably meant $g(a)=f(a)$ and $g(b)=f(b)$.
                      $endgroup$
                      – Taladris
                      Dec 16 '18 at 14:21










                    • $begingroup$
                      I've edited my answer. Thank you.
                      $endgroup$
                      – José Carlos Santos
                      Dec 16 '18 at 14:23














                    4












                    4








                    4





                    $begingroup$

                    First of all, it is not correct to assert that $g(x)$ is the secant line passing through $bigl(a,f(a)bigr)$ and $bigl(b,f(b)bigr)$. It is the function whose graph is the line segment uniting those two points.



                    Now, concerning your question: it is so that $g(a)=f(a)$ and $g(b)=f(b)$.






                    share|cite|improve this answer











                    $endgroup$



                    First of all, it is not correct to assert that $g(x)$ is the secant line passing through $bigl(a,f(a)bigr)$ and $bigl(b,f(b)bigr)$. It is the function whose graph is the line segment uniting those two points.



                    Now, concerning your question: it is so that $g(a)=f(a)$ and $g(b)=f(b)$.







                    share|cite|improve this answer














                    share|cite|improve this answer



                    share|cite|improve this answer








                    edited Dec 16 '18 at 14:23

























                    answered Dec 16 '18 at 9:17









                    José Carlos SantosJosé Carlos Santos

                    154k22124227




                    154k22124227












                    • $begingroup$
                      You probably meant $g(a)=f(a)$ and $g(b)=f(b)$.
                      $endgroup$
                      – Taladris
                      Dec 16 '18 at 14:21










                    • $begingroup$
                      I've edited my answer. Thank you.
                      $endgroup$
                      – José Carlos Santos
                      Dec 16 '18 at 14:23


















                    • $begingroup$
                      You probably meant $g(a)=f(a)$ and $g(b)=f(b)$.
                      $endgroup$
                      – Taladris
                      Dec 16 '18 at 14:21










                    • $begingroup$
                      I've edited my answer. Thank you.
                      $endgroup$
                      – José Carlos Santos
                      Dec 16 '18 at 14:23
















                    $begingroup$
                    You probably meant $g(a)=f(a)$ and $g(b)=f(b)$.
                    $endgroup$
                    – Taladris
                    Dec 16 '18 at 14:21




                    $begingroup$
                    You probably meant $g(a)=f(a)$ and $g(b)=f(b)$.
                    $endgroup$
                    – Taladris
                    Dec 16 '18 at 14:21












                    $begingroup$
                    I've edited my answer. Thank you.
                    $endgroup$
                    – José Carlos Santos
                    Dec 16 '18 at 14:23




                    $begingroup$
                    I've edited my answer. Thank you.
                    $endgroup$
                    – José Carlos Santos
                    Dec 16 '18 at 14:23


















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