This is the converse of Prove that if function f is monotonic, then it is one-to-one. Let $f \colon (a,b) \to \mathbb{R}$ be a continuous function. Prove that if $f$ is one-to-one in $(a,b)$, then $f$ is monotonic.
What I am thinking is assume by contradiction that if $f$ is not monotonic, there exists $x_1, x_2, x_3, x_4$ s.t. $x_1 \le x_2, f(x_1) \le f(x_2), x_3 \le x_4, f(x_3) \ge f(x_4)$. Try to find two points $y_1,y_2$ where $f(y_1) = f(y_2)$. However I find the there are too many cases need to be considered for example the order of $x_1,x_2,x_3, x_4$, the order of $f(x_1),f(x_2),f(x_3),f(x_4)$ and $=$ or $<$.
Best Answer
Hint
Suppose $f$ is not monotonic, then there exists $a,b,c$ such that $a<b<c$ and $f(a) < f(b) >f(c)$ (of course it could be the other way also i.e. $f(a) > f(b)$ and $f(b) < f(c)$ but the argument will be similar to this one). Note that $=$ is not included otherwise it will contradict the injectivity (one-one).
Now pick a number $k$ such that $f(a) < k < f(b)$ and $f(c) < k < f(b)$ and apply the intermediate value theorem.