Fixed points of $g$?

Question

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 $f$?

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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 $G$?

Answer 1

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