I need some help with this question:

Let $f:[0,1] \rightarrow [0,1]$ be a continuous function. Show that there exists a $c \in [0,1]$ such that $f(c)^2 = c$.

Any help will be really appreciated.



closed as off-topic by anomaly, 3SAT, user228113, user147263, John B Feb 20 '16 at 0:02

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Since $f(x)$ is continuous in $[0,1]$ also $f(x)^2$ is continuous on the same domain since the square function is continuous, so the existence of $c \in [0,1]$ such that $f(c)^2=c$ is a consequence of the intermediate value theorem for the function $g(x)=f(x)^2-x$. See: Show that a continuous function has a fixed point.

  • $\begingroup$ Where I am wrong? $\endgroup$ – Emilio Novati Feb 19 '16 at 20:38
  • $\begingroup$ I have no idea why you were down voted. But don't use $g(x) = f(x)^2 -x$ use $g(x) = f(x)^2$. And, I suppose, you must show that the range of f([0,1]) = g([0,1]) = [0,1]. But otherwise you answer is exactly correct. $\endgroup$ – fleablood Feb 19 '16 at 20:46
  • $\begingroup$ Oh, ... you are using g(x) = f(x)^2 -x to show that g(c) = 0 for some c. That's good. But the intermediate value theorem can be assumed to be proven so that isn't necessary. $\endgroup$ – fleablood Feb 19 '16 at 20:48
  • $\begingroup$ Maybe it's not necessary, but it is not wrong. Sometimes the downvotes are not inetellegible ! $\endgroup$ – Emilio Novati Feb 19 '16 at 20:52
  • $\begingroup$ Definitely NOT wrong. I'll upvote to counter it because ... it doesn't make sense that it was downvoted. $\endgroup$ – fleablood Feb 19 '16 at 20:54

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