Mean Value Theorem Q

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If $h(x) = (fcirc g)(x), xin [a,b]$, $f(x)$ and $g(x)$ are continuous over $[a,b]$ and differentiable over $(a,b)$. And $g(a) = b, g(b) = a$. We need to show that there exists at least one $c in (a,b)$ such that
$$h(a)-h(b) = f'(c)(b-a)$$



I tried to use the mean value theorem and I got
$$h(b)-h(a) = h'(c)(b-a)$$



Do you have any suggestions please






calculus







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  • Make use of the condition about $g$.
    – xbh
    Sep 2 at 9:45










  • Seems that $g([a,b] ) subseteq [a,b] $ is presumed, otherwise $h$ might not be defined.
    – xbh
    Sep 2 at 10:01














up vote
0
down vote

favorite












If $h(x) = (fcirc g)(x), xin [a,b]$, $f(x)$ and $g(x)$ are continuous over $[a,b]$ and differentiable over $(a,b)$. And $g(a) = b, g(b) = a$. We need to show that there exists at least one $c in (a,b)$ such that
$$h(a)-h(b) = f'(c)(b-a)$$



I tried to use the mean value theorem and I got
$$h(b)-h(a) = h'(c)(b-a)$$



Do you have any suggestions please






calculus







share|cite|improve this question





















  • Make use of the condition about $g$.
    – xbh
    Sep 2 at 9:45










  • Seems that $g([a,b] ) subseteq [a,b] $ is presumed, otherwise $h$ might not be defined.
    – xbh
    Sep 2 at 10:01












up vote
0
down vote

favorite









up vote
0
down vote

favorite











If $h(x) = (fcirc g)(x), xin [a,b]$, $f(x)$ and $g(x)$ are continuous over $[a,b]$ and differentiable over $(a,b)$. And $g(a) = b, g(b) = a$. We need to show that there exists at least one $c in (a,b)$ such that
$$h(a)-h(b) = f'(c)(b-a)$$



I tried to use the mean value theorem and I got
$$h(b)-h(a) = h'(c)(b-a)$$



Do you have any suggestions please






calculus







share|cite|improve this question













If $h(x) = (fcirc g)(x), xin [a,b]$, $f(x)$ and $g(x)$ are continuous over $[a,b]$ and differentiable over $(a,b)$. And $g(a) = b, g(b) = a$. We need to show that there exists at least one $c in (a,b)$ such that
$$h(a)-h(b) = f'(c)(b-a)$$



I tried to use the mean value theorem and I got
$$h(b)-h(a) = h'(c)(b-a)$$



Do you have any suggestions please






calculus




calculus






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share|cite|improve this question











share|cite|improve this question




share|cite|improve this question










asked Sep 2 at 9:36









Ahmed

1,259511




1,259511











  • Make use of the condition about $g$.
    – xbh
    Sep 2 at 9:45










  • Seems that $g([a,b] ) subseteq [a,b] $ is presumed, otherwise $h$ might not be defined.
    – xbh
    Sep 2 at 10:01
















  • Make use of the condition about $g$.
    – xbh
    Sep 2 at 9:45










  • Seems that $g([a,b] ) subseteq [a,b] $ is presumed, otherwise $h$ might not be defined.
    – xbh
    Sep 2 at 10:01















Make use of the condition about $g$.
– xbh
Sep 2 at 9:45




Make use of the condition about $g$.
– xbh
Sep 2 at 9:45












Seems that $g([a,b] ) subseteq [a,b] $ is presumed, otherwise $h$ might not be defined.
– xbh
Sep 2 at 10:01




Seems that $g([a,b] ) subseteq [a,b] $ is presumed, otherwise $h$ might not be defined.
– xbh
Sep 2 at 10:01










2 Answers
2






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0
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As I commented, if we assume $g([a,b]) subseteq [a,b]$, then $h$ would be well-defined. Now use the condition about $g$, we write the equation we want to prove as
$$
frac h(a) - h(b) g(a) - g(b) = f'(c).
$$
Now consider Cauchy Mean Value Theorem. If $g'neq 0$ on $[a,b]$ is also presumed, then the equation is easily deduced. Otherwise this question would be much complicated.






share|cite|improve this answer



























    up vote
    0
    down vote













    It was easier than I expected



    I have to apply the Mean Value Theorem on $f(x)$, then



    there exists at least one $c in (a,b)$ such that



    $$f'(c) = fracf(b) - f(a)b-a$$



    Now, since $h(a) = (f circ g) (a) = f(g(a)) = f(b)$



    $h(b) = (f circ g) (b) = f(g(b)) = f(a)$



    then



    $$f'(c) = frach(a) - h(b)b-a$$






    share|cite|improve this answer




















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      2 Answers
      2






      active

      oldest

      votes








      2 Answers
      2






      active

      oldest

      votes









      active

      oldest

      votes






      active

      oldest

      votes








      up vote
      0
      down vote













      As I commented, if we assume $g([a,b]) subseteq [a,b]$, then $h$ would be well-defined. Now use the condition about $g$, we write the equation we want to prove as
      $$
      frac h(a) - h(b) g(a) - g(b) = f'(c).
      $$
      Now consider Cauchy Mean Value Theorem. If $g'neq 0$ on $[a,b]$ is also presumed, then the equation is easily deduced. Otherwise this question would be much complicated.






      share|cite|improve this answer
























        up vote
        0
        down vote













        As I commented, if we assume $g([a,b]) subseteq [a,b]$, then $h$ would be well-defined. Now use the condition about $g$, we write the equation we want to prove as
        $$
        frac h(a) - h(b) g(a) - g(b) = f'(c).
        $$
        Now consider Cauchy Mean Value Theorem. If $g'neq 0$ on $[a,b]$ is also presumed, then the equation is easily deduced. Otherwise this question would be much complicated.






        share|cite|improve this answer






















          up vote
          0
          down vote










          up vote
          0
          down vote









          As I commented, if we assume $g([a,b]) subseteq [a,b]$, then $h$ would be well-defined. Now use the condition about $g$, we write the equation we want to prove as
          $$
          frac h(a) - h(b) g(a) - g(b) = f'(c).
          $$
          Now consider Cauchy Mean Value Theorem. If $g'neq 0$ on $[a,b]$ is also presumed, then the equation is easily deduced. Otherwise this question would be much complicated.






          share|cite|improve this answer












          As I commented, if we assume $g([a,b]) subseteq [a,b]$, then $h$ would be well-defined. Now use the condition about $g$, we write the equation we want to prove as
          $$
          frac h(a) - h(b) g(a) - g(b) = f'(c).
          $$
          Now consider Cauchy Mean Value Theorem. If $g'neq 0$ on $[a,b]$ is also presumed, then the equation is easily deduced. Otherwise this question would be much complicated.







          share|cite|improve this answer












          share|cite|improve this answer



          share|cite|improve this answer










          answered Sep 2 at 10:06









          xbh

          3,515320




          3,515320




















              up vote
              0
              down vote













              It was easier than I expected



              I have to apply the Mean Value Theorem on $f(x)$, then



              there exists at least one $c in (a,b)$ such that



              $$f'(c) = fracf(b) - f(a)b-a$$



              Now, since $h(a) = (f circ g) (a) = f(g(a)) = f(b)$



              $h(b) = (f circ g) (b) = f(g(b)) = f(a)$



              then



              $$f'(c) = frach(a) - h(b)b-a$$






              share|cite|improve this answer
























                up vote
                0
                down vote













                It was easier than I expected



                I have to apply the Mean Value Theorem on $f(x)$, then



                there exists at least one $c in (a,b)$ such that



                $$f'(c) = fracf(b) - f(a)b-a$$



                Now, since $h(a) = (f circ g) (a) = f(g(a)) = f(b)$



                $h(b) = (f circ g) (b) = f(g(b)) = f(a)$



                then



                $$f'(c) = frach(a) - h(b)b-a$$






                share|cite|improve this answer






















                  up vote
                  0
                  down vote










                  up vote
                  0
                  down vote









                  It was easier than I expected



                  I have to apply the Mean Value Theorem on $f(x)$, then



                  there exists at least one $c in (a,b)$ such that



                  $$f'(c) = fracf(b) - f(a)b-a$$



                  Now, since $h(a) = (f circ g) (a) = f(g(a)) = f(b)$



                  $h(b) = (f circ g) (b) = f(g(b)) = f(a)$



                  then



                  $$f'(c) = frach(a) - h(b)b-a$$






                  share|cite|improve this answer












                  It was easier than I expected



                  I have to apply the Mean Value Theorem on $f(x)$, then



                  there exists at least one $c in (a,b)$ such that



                  $$f'(c) = fracf(b) - f(a)b-a$$



                  Now, since $h(a) = (f circ g) (a) = f(g(a)) = f(b)$



                  $h(b) = (f circ g) (b) = f(g(b)) = f(a)$



                  then



                  $$f'(c) = frach(a) - h(b)b-a$$







                  share|cite|improve this answer












                  share|cite|improve this answer



                  share|cite|improve this answer










                  answered Sep 2 at 14:51









                  Ahmed

                  1,259511




                  1,259511



























                       

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