# continuous function

• Apr 12th 2010, 03:23 PM
charikaar
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
• Apr 12th 2010, 03:32 PM
Drexel28
Quote:

Originally Posted by charikaar
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 $f(c)\geqslant 0$
• Apr 12th 2010, 03:41 PM
charikaar
Thanks for that. How do we know $f(c)=c$. Do you have a link to the thread where this was solved previously?
• Apr 12th 2010, 03:44 PM
Drexel28
Quote:

Originally Posted by charikaar
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.