# 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 $$[0,2]$$...

• $$2x^2=x\iff x(2x-1)=0\implies\ldots$$ Commented Apr 23, 2021 at 20:12
• A fixed point is where $f(x)=x$. Commented Apr 23, 2021 at 20:12
• Whoever downvoted this, we have a new user who provided a question with several thoughts/attempts at the problem. This is exactly what we want to be encouraging, not discouraging!
– Alan
Commented Apr 23, 2021 at 20:38
• @copper.hat. Yes that is true. But there must also be a fixed point for a given function such as this in its domain of definition. This if from my numerical analysis worksheet. Commented Apr 23, 2021 at 20:55
• The title has a typo, the image should be [0,2] Commented Apr 24, 2021 at 2:04

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.

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...

• If by $\phi[-1,1] \to [0,1]"$ you mean to say that $[-1,1]$ is the domain and $[0,1]$ is the range of $\phi$, then we have a problem. since $\phi(1) = 2$. Commented May 7, 2021 at 13:48
• @stevengregory No problem at all for the very proof. I only copied the question's title, and correcting that the proof continues the same. Let us let the OP to do the appropiate corrections. Commented May 8, 2021 at 17:10

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, $$\phi(1) = 2$$.