Show that a zero of $f$ is a fixed point of $g$

Question

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?

Answer 1

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 $g$.

Answer 2

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*}