This is an exercise from Abbott's second edition of Understanding Analysis.
Let $f$ be differentiable on an interval $A$. Show that if $f'(x) \neq 0$ on $A$, show that $f$ is one-to-one on $A$. Provide an example to show that the converse statement need not be true.
Is my solution (below) correct?
Let $x_{1},x_{2} \in A $ such that $x_1 \neq x_2$. Since the function satisfies all the conditions of the mean value theorem, there exists $c \in (x_1,x_2)$ [without loss of generality we consider $x_1 < x_2$] such that $f(x_2) – f(x_1) = (x_2 -x_1) f'(c) \neq 0$ as it is given that $f'(x)$ is nonzero on $A$ and $x_1 \neq x_2$ is our assumption. Therefore $x_1 \neq x_2$ implies $f(x_1) \neq f(x_2)$ for every $x_1,x_2 \in A$. Thus $f$ is one-to one on $A$.
The converse may not be true. Consider $A=[-1,1)$ and $f(x)= x^3$. Therefore $f$ is injective on $A$. Again $f'(x)= 2x^2$. Therefore $f'(0)=0, 0 \in A$. So the converse need not be true.
Best Answer
If the function is differentiable and $f'(x)\neq0$, by the Darboux Theorem it has the intermediate value property.
Suppose by contradiction that there are two real numbers $x_1,x_2 \in A$ such that $f'(x_1)<0$ and $f'(x_2)>0$. This implies the existence of $x_3\in A$ s.t. $f'(x_3)=0$, contradicting the hypothesis $f'(x)\neq 0$.
Because the derivative $f'$ is either strictly positive or strictly negative on $A$, $f$ is monotonic and hence injective.