If a function has a non-vanishing derivative on an interval, show that it is one-to-one Let $f$ be differentiable on an interval $A$. If $f'(x)\neq 0$ on $A$, show that $f$ is injective (one-to-one) on $A$. 
Is the converse true? Provide an example to show that the converse statement need not be true.
I don't know where to start. Should I use IVT somehow?
 A: Let $x,y\in A$, $x\ne y$. Suppose $x<y$. The function $t\mapsto f(t)$ is differentiable on $(x,y)$ by the assumption. So, there is a $c\in (x,y)$ such that $f(y)-f(x)=f'(c)(y-x)\ne 0$, which implies $f(x)\ne f(y)$. So, $f$ is one-to-one.
A: Proof sketch based on my comment above:
First, note that differentiability of $f$ everywhere in $A$ does not imply that the derivative $f'$ is continuous. However, it does imply that $f'$ has the intermediate value property (also called the Darboux property). What this means is that if $x,y \in A$, with $f'(x) < f'(y)$, then given any $c$ between $f'(x)$ and $f'(y)$, there is some $z$ between $x$ and $y$ such that $f'(z) = c$.
From this fact, we can conclude that either $f'(x) > 0$ everywhere in $A$, or $f'(x) < 0$ everywhere in $A$. If this were not the case, then we would have $f'(x) > 0$ and $f'(y) < 0$ for some $x,y \in A$, hence there would be some $z$ with $f'(z) = 0$.
Without loss of generality, assume that $f'(x) > 0$ everywhere in $A$.
Now let $x,y \in A$ with $x<y$. By the mean value theorem, there is some $z$ such that $x<y<z$ and
$$f'(z) = \frac{f(y)-f(x)}{y-x}$$
From this, conclude that not only is $f$ one-to-one, in fact it is strictly monotone.
