# $f$ strictly increasing does not imply $f'>0$

We know that a function $f: [a,b] \to \mathbb{R}$ continuous on $[a,b]$ and differentiable on $(a,b)$, and if $f'>0 \mbox{ on} (a,b)$ , f is strictly increasing on $[a,b]$. Is there any counterexample that shows the converse fails?

I have been trying to come up with simple examples but they all involve functions that are discontinuous or has derivative $f'=0$ which does not agree with the hypothesis hmmm

• – Cameron Buie Nov 6 '12 at 6:44
• Hint. if all terms of a convergent sequence are positive so the limit is either positive or zero. – user231343 Sep 16 '16 at 8:11
• See related math.stackexchange.com/q/1845927/72031 – Paramanand Singh Jul 18 '18 at 8:35

Consider $f(x)=x^3$ on $[-1,1]$. It is strictly increasing, but has zero derivative at $0$.

• you are absolutely right – Daniel Jul 10 '12 at 23:02
• @jsk Note that you can come up with counterexamples like this by taking a nonnegative continuous function with a $0$ (in this case $x^2/3$) and integrating. – Alex Becker Jul 10 '12 at 23:49
• @AlexBecker You surely meant $3x^2$. – Pedro Tamaroff Jul 11 '12 at 1:29
• @PeterTamaroff Indeed. Slip of the fingers. – Alex Becker Jul 11 '12 at 2:26

Definition of Strictly Increasing: if $f(x)$ is strictly increasing then,

$\forall x\forall y : y>x, f(y)> f(x)$

Definition of Monotonically increasing: if $f(x)$ is monotonically increasing then, $\forall x\forall y : y\geq x, f(y)\geq f(x)$

Definition of Derivative: $f'(x) = \lim_{∆x\to 0} \frac{f(x+∆x)-f(x)}{∆x}$

Postulate: $f(x)$ is a Strictly Increasing function $\iff$ $f'(x) >0$

Proof: WLOG, let $∆x\overset{\underset{\mathrm{def}}{}}{=} y-x$

Thus, $y = x + ∆x$

$f'(x) = \lim_{(y-x) \to 0} \frac{f(y)-f(x)}{y-x}$

$\because y > x, ∆x >0$ and $y-x > 0$

also, $\because f(y) > f(x), f(y) - f(x) > 0$

Thus $\lim_{∆x\to 0} \frac{f(x+∆x)-f(x)}{∆x} > 0$

$\therefore f'(x) > 0$ QED

Using the same logic, $f(x)$ is monotonic $\iff f'(x) \geq 0$