prove function is surjective /analysis proofs!!

Question

Suppose $f:(a,b)\longrightarrow\mathbb R$, differentiable, where $(a,b)\subseteq\mathbb R$ is an open interval. Assume that $f'(x)$ is not $=0$.

1. Show that there is an open interval $(c,d)\subseteq\mathbb R$ such that $f((a,b))=(c,d)$ (hint: $f$ is surjective on $(c,d)$).

2. Since $f$ bijectively maps $(a,b)$ into $(c,d)$ (it is injective and surjective), we may define an inverse function $g:(c, d)\rightarrow(a, b)$ where $g(f(x)) = x$. Prove that $g'(f(x))= 1/f'(x)$ for all $x\in(a,b)$.

Answer 1

This is the simplest case of the Inverse function theorem.

Answer 2

1. Let $c=\inf \{\,f(x)\|a<x<b\,\}$, $d=\sup \{\,f(x)\|a<x<b\,\}$,. The assumption that $f(x)=c$ for some $x\in(a,b)$ leads to $f'(x)=0$, similarly for $d$. The rest is IVT.

2. Apply $\frac d{dx}$ to both sides of $g(f(x))=x$.