Prove or disprove: If $f$ is increasing and differentiable on $(a,b)$ then $f'(x)\ge 0$ on $(a,b)$

Question

Here's the question again:

Prove or disprove: If $f$ is increasing and differentiable on $(a,b)$ then $f'(x)\ge 0$ on $(a,b)$.

I could not find any counterexamples to it so here's my attempt at a proof:

Suppose there is a point $c\in (a,b)$ such that $f'(c)<0$. Let $\varepsilon = - f'(c)/2$. Then there exists a $\delta > 0$ such that $\frac{f(x)-f(c)}{x-c} < f'(c)/2 < 0$ for all $0<|x-c|< \delta $.

Let $x_1 \in (c-\delta , c )$ then $x_1 -c<0$ thus we have that $\frac{f(x_1)-f(c)}{x_1-c} <0$ implies $f(x_1) > f(c)$. Now, let $x_2 \in (c, c+\delta )$ then in a similar way, $f(x_2) < f(c)$. We note that $x_1 < x_2$ but we obtain $f(x_2)< f(c) < f(x_1)$, a contradiction since $f$ is increasing on $(a,b)$.

Is this attempt correct? Could there be any direct proofs?

Answer 1

Your proof is correct and almost a direct proof, let me rephrase it: Let $x<y$. Then $ f(y)-f(x) > 0$. Therefore $$ \frac{f(y) - f(x)}{y-x} > 0.$$ Sending $y\to x$ yields $$ f'(x) \ge 0.$$

Answer 2

Your proof is correct. A direct proof goes like this:

For any $c\in (a,b)$, since $f$ is differentiable at $c$, we have $$\lim_{x\to c}\frac{f(x)-f(c)}{x-c}=f'(c).$$ The expression inside the limit is always nonnegative since $f$ is increasing. Thus, the limit has to be nonnegative.