Real continuous function with attractive fixed point has range subset of domain Let $f:\mathbb{R}\to \mathbb{R}$ be $C^1$ with fixed point $p$, such that $|f'(p)|< 1$. Then there exists an interval $I$ containing $p$ such that $f:I\to I$.
I'm trying to prove this rigorously, but there's a roadblock. My best attempt is as follows:
Let $\varepsilon > 0$, such that $\varepsilon$ is small, and consider $f(p+\varepsilon)=f(p)+f'(p)\varepsilon+\mathcal{O}(\varepsilon^2)$ and $f(p-\varepsilon)=f(p)-f'(p)\varepsilon+\mathcal{O}(\varepsilon^2)$.
Then $$|f(p+\varepsilon)-f(p-\varepsilon)|=|\varepsilon(2f'(p))+O(\varepsilon^2)|$$$$\le 2\varepsilon|f'(p)|+|\mathcal{O}(\varepsilon^2)|<2\varepsilon+|\mathcal{O}(\varepsilon^2)|$$
Everything would be as desired if not for the $|\mathcal{O}(\varepsilon^2)|$ term. How does one get rid of it? I could argue that since $\varepsilon$ is small, we can neglect the term, but that is not very rigorous. Would appreciate your insight.

If $f$ is differentiable and $f'$ is continuous at $p$ then there
is some interval $I$ containing $p$ in the interior such that
$|f'(x)| < 1$ for $x \in I$. Without loss of generality we can
presume that $I$ is symmetric about $p$, and so has the form
$I=(p-\delta, p+\delta)$ for some $\delta >0$.
If $x \in I$ then the mean value theorem gives
$f(x) = f(p) + f'(\xi) (x-p)$ for some $\xi \in I$ and so
$|f(x)-p| = |f(x)-f(p) | = |f'(\xi)| |x-p| < |x-p|< \delta$, and so
$f(x) \in I$.
