# Function maximum with positive derivative

My text book contains the following task which I'm unsure of:

Be $f: [a, b] \rightarrow \mathbb{R}$ differentiable in $b$ and $f\;'(b)>0$. Prove that $f$ contains an isolated local maximum at $b$ (this means there is a $\delta > 0$ with $f(b) > f(x)$ for all $x \in (b- \delta, b)$).

However to my understanding the derivative in $b$ has to be 0 as it contains a maximum at $b$ and the slope is zero. Can it be that this is a error in the book and $f''(b)<0$ or $f(b)>0$ is meant or am I missing something?

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If thinks overlap you can add \; to put a bit of space in. I did it for your f'. – Ross Millikan Jun 29 '11 at 12:42
Do you remember the exact hypotheses that you need in order to conclude that a local maximum $x_0$ has $f\,'(x_0) = 0$? Also, consider $f(x) = x^2$ on $[a,b] = [0,1]$ as an example supporting the task. – t.b. Jun 29 '11 at 12:44
I am not a native speaker of English, but I think derivative instead of derivation should be in the title. Or am I wrong? – Martin Sleziak Jun 29 '11 at 13:03
@Martin: Only if he is asking not about derivation the proof about derivative. – Ilya Jun 29 '11 at 13:08

Because $b$ is the end of a closed interval, if $f\;'(b) \gt 0, f(b)$ will be a maximum. Think of $f(x)=x$ restricted to $[0,1]$. It has a maximum at $1$.

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More precisely, $f$ is left differentiable at $b$, with left derivative $f'_ - (b)$ given by $$f'_ - (b) = \mathop {\lim }\limits_{x \to b - } \frac{{f(b) - f(x)}}{{b - x}} > 0.$$ Put $l = f'_ - (b)$, and let $\varepsilon > 0$. Then there exists a $\delta > 0$ such that $$\bigg|\frac{{f(b) - f(x)}}{{b - x}} - l \bigg| < \varepsilon$$ for any $x \in (b-\delta,b)$. Letting $\varepsilon = l \,(> 0)$ we have, in particular, $$-l < \frac{{f(b) - f(x)}}{{b - x}} - l,$$ so $$\frac{{f(b) - f(x)}}{{b - x}} > 0.$$ Hence $f(b)-f(x) > 0$ (or $f(b) > f(x)$) for any $x \in (b-\delta,b)$.

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It's surprisingly true. Note that due to the definition of derivative $$\lim\limits_{h\downarrow 0}\frac{f(b)-f(b-h)}{h} = f'(b)>0.$$ Consider $g(h) = \frac{f(b)-f(b-h)}{h}$ for $h\geq 0$. It is continuous at all points $h\in (0,b-a]$ and we can define it in $h=0$ by continuity: $g(0):=f'(b)$.

Now $g$ is continuous on $[0,b-a]$ and $g(0)>0$, then there exists $\delta$ such that $g(h)>0$ for $h\in [0,\delta)$, i.e. $f(b) > f(b-h)+f'(b)h>f(b)$. That's what was asked to be proved.

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Since $f$ is only assumed (left) differentiable at $b$, $g$ is not necessarily continuous for $h > 0$. For example, define $f:[0,1] \to \mathbb{R}$ by $f(x)=x$ if $x$ is rational, and $f(x)=x+(1-x)^2$ if $x$ is irrational. – Shai Covo Jun 29 '11 at 15:56
@Shai: you're right, thank you. – Ilya Jun 29 '11 at 15:57

A drawing could help you realize there's no mistake, because this is very easy to see as intuitively true, but there is no mistake in your book. Draw something and the proof should come to you very easily if you have done a little analysis. =)

Hope that helps,

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