# Let $f$ be an continuous function and show that it exists. [closed]

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!

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

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• "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
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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.
• 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. – fleablood Feb 19 '16 at 20:46