-1
$\begingroup$

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.

Thanks!

$\endgroup$

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

This question appears to be off-topic. The users who voted to close gave this specific reason:

  • "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – anomaly, 3SAT, Community, John B
If this question can be reworded to fit the rules in the help center, please edit the question.

2
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ 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

Not the answer you're looking for? Browse other questions tagged or ask your own question.