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Thread: continuous function

  1. #1
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    continuous function

    suppose that $\displaystyle f:[0,1]-->[0,1]$ is continuous. By considering the function $\displaystyle g(x)=(f(x))^2-x$, or otherwise, prove that there must be a point $\displaystyle c\in[0,1]$ with $\displaystyle f(c)=\sqrt{c}$.
    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 $\displaystyle f:[0,1]-->[0,1]$ is continuous. By considering the function $\displaystyle g(x)=(f(x))^2-x$, or otherwise, prove that there must be a point $\displaystyle c\in[0,1]$ with $\displaystyle f(c)=\sqrt{c}$.
    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 $\displaystyle f(c)=c$ for some $\displaystyle c\in[0,1]$ but $\displaystyle f^2(x)$ is just another continuous function that maps $\displaystyle [0,1]\to[0,1]$! so there exists some $\displaystyle c\in[0,1]$ such that $\displaystyle f^2(c)=c\implies f(c)=\sqrt{c}$ taking the positive root since $\displaystyle |f(c)|=f(c)$ since $\displaystyle f(c)\geqslant 0$
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    Thanks for that. How do we know $\displaystyle f(c)=c$. 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 $\displaystyle f(c)=c$. 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 $\displaystyle g(x)=f(x)-x$ then $\displaystyle g(1)=f(1)-1\leqslant 1-1=0$ and $\displaystyle g(0)=f(0)-0\geqslant 0-0=0$. So, if either $\displaystyle g(1)=0$ or $\displaystyle g(0)=0$ we're done, otherwise the above inequalities are strict and we can apply the IVT.
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