TITLE: Prove that $\phi(x) = 2x^2$ : $\phi[-1,1] \to [0,1]$ has two fixed points in its domain of definition
QUESTION [14 upvotes]: Have can one prove that there are two?
I know how to prove when the image set $[a,b]$ is in the domain of definition $[a,b]$, but applying the method to this problem doesn't seem to work.
If we take the derivative of $\phi'(x) = 2x^2$ = $4x$, then the $\phi$ is increasing. Plugging in our values $[a,b] = [-1,1]$ We will get $[4(-1), 4(1)]$ so we end up with $[-4,4]$ which is not in our image set $[0,2]$...

REPLY [2 votes]: Simple, direct meaning of definition: for any function $\;f\;$ , a point $\;a\;$ in its domain of definion is a fixed point if $\;f(a)=a\;$ . For your function, this means that we must have
$$f(x)=x\iff \left(f(x)=\right)2x^2=x\iff x(2x-1)=0\iff x=0,\,x=\frac12$$
and there you have two points in $\;[-1,1]\;$ which are fixed by your function...