Let $f:[0,1] \rightarrow [0,1]$ be a continuous function. Show that there exists a $c \in [0,1]$ such that $f(c)^2 = c$.
This all the information I have. I am not quite sure of what to do.
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Are you saying $(f(c))^2 = c$ or $f(f(c)) = c$? – A. Thomas Yerger Feb 18 '16 at 22:58
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@Joe, it's a function taking values in the closed unit interval. – A. Thomas Yerger Feb 18 '16 at 22:59
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There I have fixed it. @Joe – apprentice Feb 18 '16 at 22:59
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1Hint: consider $g(x) = f(x)^2 - x$. – BrianO Feb 18 '16 at 23:03
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Possible duplicate of Show that a continuous function has a fixed point – Feb 18 '16 at 23:21
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@Sally I dont think it is. I wouldn't have created an account for just one question – apprentice Feb 18 '16 at 23:30
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Consider the function $g(x)=x-f(x)^2$ on $[0,1]$ and use the intermediate value theorem to argue that there must be a root. Look at $g(0)$ and $g(1)$, we can assume that $f(0)\neq0$ and $f(1)\neq1$ since otherwise we are done.
Sri-Amirthan Theivendran
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Hint: Consider the continuous function $$h:[0,1]\to \Bbb{R}:x\mapsto f(x)^2-x.$$ What can you say about $h(0)$ and $h(1)$? Conclude using the intermediate value theorem.
Nitrogen
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The fact that it is the square of a continuous function ($f(c)^2=c$) doesn't bring any "added value" compared to the raw problem ($f(c)=c$)... – Jean Marie Feb 18 '16 at 23:12
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The endpoints of the interval matter. If the end point was 2 instead of 1, then the image could have points getting mapped outside. – Daniel Akech Thiong Feb 18 '16 at 23:16