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Consider the functions $f(x) = 1 - \frac{1}{2x}$

and $g(x) = 2x(1-x)$

How many roots does $f$ have? Are the roots of $f$ fixed-points of $G$ are there more fixed points of $g$ than roots of $f$?

Confused as to how to answer this question

The roots of $f$ is $1 = \frac{1}{2x} \implies x = \frac{1}{2}$

Now how do I find the fixed points of $G$?

ss sss
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  • You're looking for $x_0$ such that $g(x_0) = x_0$. Write that condition down and do some substitutions. – nomen Nov 08 '19 at 00:27
  • I'm looking at my notes about fixed point I have this https://gyazo.com/37985606463d7c6173a92dd7cc968f0a but I'm unsure how to apply any of that. – ss sss Nov 08 '19 at 00:29
  • I told you how. Write down $g(x_0) = x_0$ and solve for $x_0$. – nomen Nov 08 '19 at 00:32
  • $g(1/2) = 2 \cdot (1 - \frac{1}{2}) = \frac{1}{2}$ therefore the root of $f(x)$ is a fix point of $g$? Third part I still don't understand. – ss sss Nov 08 '19 at 00:37
  • Edit: last comment was a mistake. But all you have to do is count the roots of $f$ and the fixed points of $g$. If you really want to, try to figure out the relationship between $f$ and $g$ (I assumed there was none). – nomen Nov 08 '19 at 00:40
  • $2x(1-x)=x \Leftrightarrow x (2(1-x) -1) = 0 \Leftrightarrow x (1-2x) = 0 \cdots$ – PierreCarre Nov 08 '19 at 16:20

1 Answers1

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$2x(1-x)=x$ implies $x=0$ or $2(1-x)=1$ so $x=0$ or $x =\frac 1 2$.

Kavi Rama Murthy
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