Rolle's theorem problem Let $f$ be continuous on an interval $[a, b]$ and differentiable on $(a, b)$ with a derivative that never is zero. Show that $f$ maps $[a, b]$ one-to-one onto some other interval.
I can prove that $f$ must map to other intervals, because if for some $x_1,x_2$ in $[a,b]$ such that $f(x_1)=f(x_2)$, then there will be $c$ in $[a,b]$ such that $f'(c)=0$, which contradicts with the condition. But I don't know how to proceed from there.
 A: Any continuous function maps a closed interval to a closed interval. This is because any continuous function has a maximum and a minimum, and reaches all values between the max and min.
You already proved that $f$ is one to one, since you proved that if $f(x_1)=f(x_2)$. then $x_1=x_2$.
A: The image of $[a,b]$ by $f$ is a closed interval (because compactness and connectedness are preserved by continuous funcitons).
To see one-to-one, suppose $f(c)=f(d)$ for some $c,d\in [a,b]$ with $c \neq d$, say $c<d$. Then, by Rolle's theorem, there exists $\alpha \in (c,d)\subset [a,b]$ such that $f'(\alpha)=0$.
That proves by contradiction that $f(c)=f(d)\Rightarrow c=d$, or equivalently, $c\neq d \Rightarrow f(c)\neq f(d)$.
That is, $f$ is one-to-one.
A: $f$ is continuous, so it maps intervals to intervals by the intermediate value theorem. So once you have two distinct images, the image of an interval is a non-singleton interval.
You seem to know how to show the function is one-to-one. So you are done aren't you?
By the Weierstrass theorem you know that a continuous function on a closed and bounded interval has a maximum and minimum, so you can conclude the interval $f$ maps your closed $[a,b]$ one-to-one to is closed and bounded, provided $[a,b]$ is bounded. If it is not bounded, you cannot conclude anything on the boundedness, though you can still conclude it will not be open otherwise its preimage $[a,b]$ would also have to be open. So if the image is unbounded, then it is closed. If it is bounded, it is at least semiclosed.
