Prove that $\phi(x) = 2x^2$ : $\phi[-1,1] \to [0,1]$ has two fixed points in its domain of definition 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 $[0,2]$...
 A: Let $f$ be a function with domain $D$ and range $R$.  What does it mean for a point $x \in D$ to be a fixed point?  It means that if you plug $x$ into $f$, you get back $x$ again.  In other words, $x$ is a fixed point if $f(x) = x$.
Your function is $\phi(x) = 2x^2$ with domain $[-1,1]$ and range $[0,2]$.  What does it mean for a point $x \in [-1,1]$ to be a fixed point?  It means that when you plug $x$ into $\phi$, you get $x$ back.  In other words, to say that $x$ is a fixed point is to say that
$$2x^2 = x.$$
Thus, a fixed point for $\phi$ is a number between $-1$ and $1$ that solves the quadratic equation $2x^2 - x = 0$.  You can solve this quadratic equation and find the two fixed points.
A: 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...
A: Define $\Phi(x) = \phi(x) - x$. We then want to show that $\Phi$ has two zeros on $[-1,1]$. That $\Phi(-1) > 0$, $\Phi(\frac{1}{3}) < 0$ and $\Phi(1) > 0$ implies that there are at least $2$ (since $\Phi$ is continuous). To see that there are no more, we can take a derivative:
$$\Phi^\prime(x) = 4x - 1.$$
So, $\Phi$ is increasing on $(\frac{1}{4},1]$ and decreasing on $[-1,\frac{1}{4})$. This implies that the only two fixed points of $\phi$ and also narrows down their location.
Alternatively, for a simple function like this, you could just solve
$$0 = 2x^2 - x = x(2x-1) \implies x = 0 \text{ or } x = \frac{1}{2}.$$
Those are both in $[-1,1]$ and so those would be your fixed points.
As a side note, your domain and codomain don't match up. For example, $\phi(1) = 2$.
