Show $x^2$ in the interval $(0,1/3]$ has no fixed points.
I understand that the range of that domain is always lower than $y=x$, but what is a proper way of showing this? $$\left(0,\frac13\right] \to \left(0,\frac19\right]$$
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1Funny looking intervals! – lhf Apr 12 '12 at 02:24
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@lhf: I now have a pain in the neck, straining to tilt my head to the right... – Aryabhata Apr 12 '12 at 02:25
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Let $f(x)=x^2$. You’re being asked to show that there is no $x\in(0,1/3]$ such that $f(x)=x$. Set up the equation $f(x)=x$, solve it, and discover that its only solutions are outside of $(0,1/3]$.
Brian M. Scott
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The quadratic equation $x^2=x$ can be written as $x^2-x=0$, and then as $x(x-1)=0$. That has two solutions. Neither of them is in the interval $(0,1/3]$.
