Equation of secant line in mean value theorem proof











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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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    up vote
    1
    down vote

    favorite












    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
























      up vote
      1
      down vote

      favorite









      up vote
      1
      down vote

      favorite











      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













      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






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      share|cite|improve this question











      share|cite|improve this question




      share|cite|improve this question










      asked 3 hours ago









      Max

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      585519






















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          down vote



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          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




























            up vote
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            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 $g(a)=a$ and $g(b)=b$.






            share|cite|improve this answer





















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              2 Answers
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              active

              oldest

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              2 Answers
              2






              active

              oldest

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              active

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              active

              oldest

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              up vote
              2
              down vote



              accepted










              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

























                up vote
                2
                down vote



                accepted










                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























                  up vote
                  2
                  down vote



                  accepted







                  up vote
                  2
                  down vote



                  accepted






                  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












                  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 3 hours ago









                  Shubham Johri

                  2,547413




                  2,547413






















                      up vote
                      2
                      down vote













                      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 $g(a)=a$ and $g(b)=b$.






                      share|cite|improve this answer

























                        up vote
                        2
                        down vote













                        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 $g(a)=a$ and $g(b)=b$.






                        share|cite|improve this answer























                          up vote
                          2
                          down vote










                          up vote
                          2
                          down vote









                          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 $g(a)=a$ and $g(b)=b$.






                          share|cite|improve this answer












                          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 $g(a)=a$ and $g(b)=b$.







                          share|cite|improve this answer












                          share|cite|improve this answer



                          share|cite|improve this answer










                          answered 3 hours ago









                          José Carlos Santos

                          146k22117217




                          146k22117217






























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