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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

Daniel
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2 Answers2

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

Cameron Buie
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  • you are absolutely right – Daniel Jul 10 '12 at 23:02
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    @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
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    @AlexBecker You surely meant $3x^2$. – Pedro Jul 11 '12 at 01:29
  • @PeterTamaroff Indeed. Slip of the fingers. – Alex Becker Jul 11 '12 at 02:26
  • Can we say that Even if $f'(a) = 0$, if $f''(a) = 0$ then f(x) won't become not-Strictly-Increasing? – RishiNandha Vanchi May 17 '20 at 04:29
  • @RishiNandha_M I'm not sure I understand what you're asking. It might be better to post a new question, so that you are able to clarify more than a few comments will allow. – Cameron Buie May 17 '20 at 10:31
  • I thought it's not worth a new question. Say $f'(x)>0$ except at $x = a \ ; \ f'(a)=0$. So if $f''(a)=0$ Can we say that $f(x)$ is strictly increasing? – RishiNandha Vanchi May 17 '20 at 11:47
  • @RishiNanda_M Ah! I see. Well, it's certainly possible, as the function $x\mapsto x^3$ shows, but I'm afraid I don't know the answer in general. Hopefully someone on the site will! – Cameron Buie May 17 '20 at 14:49
  • @RishiNandha_M: As it turns out, we can say more than that. Suppose that $f:\Bbb R\to\Bbb R$ is a differentiable function whose derivative is continuous, and that $f'(x)\ge 0$ for all $x.$ If $f$ isn't increasing everywhere, then there exist $a,b\in\Bbb R$ with $a<b$ such that $f'(x)=0$ for all $x\in[a,b].$ – Cameron Buie May 23 '20 at 17:55
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    As an immediate corollary, if $f:\Bbb R\to\Bbb R$ is a differentiable function, $f'$ is continuous and non-negative everywhere, and $f'(x)=0$ for at most countably-many $x$ (or more generally, on a set of measure $0$), then $f$ is increasing. – Cameron Buie May 23 '20 at 17:57
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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 $\iff f'(x) \geq 0$

ray lin
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  • This is incorrect. When finding derivative, you might take terms like $a \cdot h \to 0$. But what really happens here is, its not $=0$ but $\to 0$. Thus $f(a+h) > f(a)$ may be true even if $f'(a)=0$ – RishiNandha Vanchi May 24 '20 at 02:50