Local maxima/minima theorem This nice theorem can be easily proved using Taylor's formula.
It gives us sufficient conditions for local extrema.
Th: Let $f(x)$ be defined in some interval $I \subset \mathbb{R}$ and let us assume it has up to
$n$-th derivative in some neighborhood $K = (a-\delta, a+\delta) \subset I$ of the point $a$.
Let also $f^{(n)}(x)$ be continuous at $a$. Let also:
$$f^{(k)}(a) = 0$$ for $k = 1,2,...,(n-1)$
and
$$f^{(n)}(a) \ne 0$$
Then:
$(1)$ if $n$ is even, then $f(x)$ has a local extremum at the point $a$
which is minimum if $f^{(n)}(a) \gt 0$ and is maximum if $f^{(n)}(a) \lt 0$
$(2)$ if $n$ is odd, then $f(x)$ has no local extremum at the point $a$
My question is the following... If $f$ satisfies condition $(2)$ of this theorem (and the other conditions), can we claim that
(A) $f$ is even monotonous in some neighborhood $(a-\theta,a+\theta)$ of the point $a$?
Or maybe... is it true that
(B) we can construct a function $f$ (which satisfies the above mentioned conditions including (2)) such that for any neighborhood $(a-\theta,a+\theta)$  of the point $a$, $f$ has an extremum in that neighbourhood?
In fact when I start thinking deeply about it, I am not even sure if (B) is the opposite of (A).
Or... are there other possible cases here too?
I mean (in the context of all said above), if $f$ has no local extrema in $(a-\theta,a+\theta)$, does it mean that it's monotonous in it? And vice versa?
 A: $A$ is correct under the assumption that $f^{(n)} $ is continuous at $a$ By continuity of nth derivative at $a$, there exists a neighborhood of $a$ such that the nth derivative at $\alpha $ and at $a$ have the same sign. Let $f^{(n)} (a) \gt 0$, where $n=2k+1$ is odd number. 
By Taylor's theorem, there exists $\alpha \in (a, x) $ such that :
$f(x) = f(a) +\sum_{i=1}^{i=n-1}\frac{f^{(i)}(a)} {i!}(x-a) ^i+ \frac{f^{(n)} (\alpha)}{(n)!} (\alpha) (x-a) ^{n} =f(a) +\frac{f^{(2k+1)} (\alpha)}{(2k+1)!}  (x-a) ^{2k+1}\implies f'(x) =\frac{f^{(2k+1)}(\alpha) }{(2k+1)(2k)!}  (2k+1) (x-a) ^{2k}=\frac{f^{(2k+1)}(\alpha) }{(2k)!}  (x-a) ^{2k}\gt 0 \tag{1} $ 
Edit:
Mr. Peter Petrov has pointed out in comments that the problem with $(1)$ is that $\alpha$ depends upon $x$ so differentiating the Taylor series above is going to be complicated. Therefore, Taylor series for $f'(x) $ can be written as above whence it will be clear that $f'(x) $ is positive in some neighborhood of $a$, which proves the monotonicity of $f$. 
The same process as above can be employed for the case when $f^{(n)} (a)  \lt 0$.
A: Is this proof of (A) correct?
Let $g(x) = f'(x)$
Then we use Taylor's formula for $g$ i.e. for $f'$ in the given neighborhood $K = (a-\delta, a+\delta)$
$$g(x) = (\sum_{k=0}^{n-2} \frac{g^{(k)}(a)}{k!} (x-a)^{k}) + \frac{g^{(n-1)}(\mu)}{(n-1)!} (x-a)^{n-1}$$
where $\mu$ lies between $x$ and $a$.
And then we use the facts that
i) all terms in the $\Sigma$ sum are zeros and
ii) the function $g^{(n-1)} = f^{(n)}$ is continuous at $a$
The latter means there is a neighborhood $T$ of $a$ in which $g^{(n-1)} = f^{(n)}$ is either positive or negative (I mean in the whole neighborhood $T$).
And then from
$$g(x) =  \frac{g^{(n-1)}(\mu)}{(n-1)!} (x-a)^{n-1}$$
$($which is valid for all $x \in V = K \cap T$$)$
and taking into account that $(n-1)$ is even, we conclude that the sign of $$g(x) = f'(x)$$ doesn't change in that neighborhood $V$.
And that means $f$ is either increasing or decreasing in the neighborhood $V$ of $a$.
