Show that a zero of $f$ is a fixed point of $g$ I want to show that a solution of the equation $x^2+cos(x)-10x=0$ is a fixed point of $g(x)=(x^2+cos(x))/10$. I tried using the quadratic equation but my solution doesn't simplify nicely in $g$. I'm not sure how to deal with the $cos(x)$ term- any ideas?

If $x$ solves $x^2+\cos(x)-10x=0$, then $x^2+\cos(x)=10x$. So 
$$
g(x)=(x^2+\cos(x))/10=(10x)/10=x
$$
No need for the quadradic equation. I have absolutely no idea what values of $x$ even look like here. but certainly if you bothered to find one it would be a fixed point of $g$.
