Q:

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 $x \in (b-\delta,b)$.

