Prove that $\phi(x) = 2x^2$ : $\phi[-1,1] \to [0,1]$ has two fixed points in its domain of definition

Solution 1:

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.

Solution 2:

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...

Solution 3:

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, $\phi(1) = 2$.