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?
 A: 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 $1$.
A: 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 $x \in (b-\delta,b)$.
A: 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.
A: 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,
