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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 $[0,2]$...

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    $\begingroup$ $$2x^2=x\iff x(2x-1)=0\implies\ldots$$ $\endgroup$
    – DonAntonio
    Apr 23, 2021 at 20:12
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    $\begingroup$ A fixed point is where $f(x)=x$. $\endgroup$
    – Blitzer
    Apr 23, 2021 at 20:12
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    $\begingroup$ 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! $\endgroup$
    – Alan
    Apr 23, 2021 at 20:38
  • $\begingroup$ @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. $\endgroup$ Apr 23, 2021 at 20:55
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    $\begingroup$ The title has a typo, the image should be [0,2] $\endgroup$ Apr 24, 2021 at 2:04

3 Answers 3

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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.

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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...

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  • $\begingroup$ 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 $\phi(1) = 2$. $\endgroup$ May 7, 2021 at 13:48
  • $\begingroup$ @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. $\endgroup$
    – DonAntonio
    May 8, 2021 at 17:10
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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, $\phi(1) = 2$.

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