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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 $[0,2]$...

Gerry Myerson
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PlatinumMaths
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    $$2x^2=x\iff x(2x-1)=0\implies\ldots$$ – DonAntonio Apr 23 '21 at 20:12
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    A fixed point is where $f(x)=x$. – Blitzer Apr 23 '21 at 20:12
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    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 Apr 23 '21 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. – PlatinumMaths Apr 23 '21 at 20:55
  • @PlatinumMaths Yes, there are exactly two fixed points for $\phi$ both of which lie in the domain of definition. I have no idea why you are looking at $\phi'$. $\phi$ is certainly not increasing. – copper.hat Apr 23 '21 at 21:00
  • @DonAntonio, can you elaborate a little bit? – PlatinumMaths Apr 23 '21 at 21:27
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    The title has a typo, the image should be [0,2] – Calvin Khor Apr 24 '21 at 02:04
  • The up and down votes on this thread have fluctuated like a Brownian motion with upward drift. If you are from the meta post , please vote on the merit of the question and not on your opinion on the meta post, thanks! – Sarvesh Ravichandran Iyer Apr 29 '21 at 07:02

3 Answers3

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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.

Mike Pierce
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D_S
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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...

DonAntonio
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  • 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 $\phi(1) = 2$. – Steven Alexis Gregory May 07 '21 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. – DonAntonio May 08 '21 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, $\phi(1) = 2$.

Gary Moon
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