root of the function $f(x)=\sqrt{2}-x$ by using fixed point iteration How can I find the approximate value of $\sqrt{2}$ by using the fixed point iteration? I have tried
$x-\sqrt{2}=0$,
$x^2=2$,
$x^2-2+x=x$,
$g(x)=x^2-2+x$,
$g\prime(x)=2x+1$
And i choose $x_0=-\frac{1}{2}$. 
But i cant find the approximate value correctly. Where am i wrong?

You cannot use fixed point iteration to approximate $\sqrt{2}$ using this map (Meaning the map the OP has at the title: $f(x)=\sqrt{2}-x$).
The fixed point of the map is found by setting:
$$f(x)=x\implies x_0=\frac{\sqrt{2}}{2}$$
But this map satisfies:
$$|f'(x)|_{x=x_0}=1$$
consequently the fixed point ($x_0$) is $\mathbf{neutral}$, so in particular i's not an attractor, which means your iteration in an open interval $(x_0-\epsilon,x_0+\epsilon)$ around $x_0$ will fail.
So you need to look at a different map, like those pointed out by Nicolas and Cipra.
