Exercise: Let $f(x) = x^2$. Show that $1$ is a repelling fixed point of $f(x)$.
What I've tried: I think I need to show that $\left|x^2 - 1\right| >\left|x - 1\right|$ whenever $\left|x - 1\right| <\delta$ for some $\delta>0$, because this would mean that $f^n(x)\not\to 1$ for $x \neq 1$. We have that $\left|x^2 - 1\right| =\left|(x-1)(x+1)\right| = \left|x-1\right|\left|x+1\right| > \left|x-1\right|$ whenever $\left|x+1\right|> 1$ and whenever $\left|x - 1\right| >\dfrac{\left|x-1\right|}{\left|x+1\right|}\Leftrightarrow \left|x-1\right| <\left|x+1\right|$. I don't really what to do from here. I know that $\left|x+1\right| > 1$ holds if $x\geq 0$ or when $x\leq -2$ which would mean that $\left|x-1\right|<1$ or $\left|x - 1\right|\geq 3$, if I'm not mistaken.
Question: How should I approach (in general) inequalities with absolute value signs? How should I solve this exercise? I'm often unable to show where inequalities with absolute values hold, even though they should be rather easy...
Thanks!
There is no general way to approach inequalities with absolute values. In this particular case, you did all the necessary computations: $$ |f(x)-1|>|x-1| $$ whenever $|x+1|>1$ and $x\neq1$. This means, in particular, that whenever $x>0$, you know that iterating $f$ takes you further away from $1$. If you start with any value $x>0$ which is not exactly $1$, the iterates will take you further away from $1$ and convergence to $1$ is impossible.