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Math Help - continuous function

  1. #1
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    continuous function

    suppose that f:[0,1]-->[0,1] is continuous. By considering the function g(x)=(f(x))^2-x, or otherwise, prove that there must be a point c\in[0,1] with f(c)=\sqrt{c}.
    I have no idea so can you help please?
    Thanks
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  2. #2
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by charikaar View Post
    suppose that f:[0,1]-->[0,1] is continuous. By considering the function g(x)=(f(x))^2-x, or otherwise, prove that there must be a point c\in[0,1] with f(c)=\sqrt{c}.
    I have no idea so can you help please?
    Thanks
    This reverts to a problem I have seen you do earlier! You have proven before that in the above that f(c)=c for some c\in[0,1] but f^2(x) is just another continuous function that maps [0,1]\to[0,1]! so there exists some c\in[0,1] such that f^2(c)=c\implies f(c)=\sqrt{c} taking the positive root since |f(c)|=f(c) since f(c)\geqslant 0
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  3. #3
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    Thanks for that. How do we know f(c)=c. Do you have a link to the thread where this was solved previously?
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  4. #4
    MHF Contributor Drexel28's Avatar
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    Quote Originally Posted by charikaar View Post
    Thanks for that. How do we know f(c)=c. Do you have a link to the thread where this was solved previously?
    Maybe it wasn't you... there is another poster with a similar name.

    Note that if g(x)=f(x)-x then g(1)=f(1)-1\leqslant 1-1=0 and g(0)=f(0)-0\geqslant 0-0=0. So, if either g(1)=0 or g(0)=0 we're done, otherwise the above inequalities are strict and we can apply the IVT.
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