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?

Firstly, let $f(x) = x^{2} + \cos (x) - 10 x$, where $f(x_{0}) = 0$. Then
\begin{align*}
0 & = x_{0}^{2} + \cos (x_{0}) - 10 x_{0} \\
\Rightarrow 10 x_{0} & = x_{0}^{2} + \cos (x_{0})\\
\Rightarrow x_{0} & = \frac{x_{0}^{2} + \cos (x_{0})}{10} \\
 & = g(x_{0}).
\end{align*}
