Some doubts regarding local minimums of twice-differentiable functions (Sorry if this question, along with some other recent questions, is trivial or there's errors in how I worded it. I'm a beginner calculus student)
Let $f:[a,b] \rightarrow \mathbb{R}$ be a continuous function twice-differentiable on the open interval $(a,b)$.
Proposition: if $c \in (a,b)$ is a strict local minimum of  $f$, there exists a $\delta >0$ such that $f'$ is monotonically increasing in $(c, c + \delta)$. 
Question 1: does this proposition hold? 
Question 2: Let us consider a new function $g$ defined similarly to how $f$ was. Let $x_0$ satisfy $g'(x_0)=0$ and $g''(x_0)>0$. Moreover, there exists no $\delta >0$ such that $g'$ is monotonically increasing in $(x_0, x_0 + \delta)$. Then per contraposition on our original proposition, $x_0$ is not a local minimum of $g$. But this contradicts the 2nd derivative test. What's going on? 
Note for $g$ to satisfy what we want, $g'$ could be some pathological differentiable function . Probably something you get after playing around with $\sin(\frac{1}{x})$; $g$ would of course be its primitive, surely not something expressible in elementary terms. 
 A: Your proposition does not hold. As an example, define $f$ by $f(0) = 0$
and
$$
 f(x) = x^4 ( 2 + \sin \frac 1x)
$$
for $x \ne 0$. $f$ is differentiable on $\Bbb R$ and has a minimum at $x = 0$. 
The derivative is given by $f'(0) = 0$ and 
$$
 f'(x) = 4x^3 ( 2 + \sin \frac 1x) - x^2 \cos \frac 1x
 = x^2 \bigl( 4x ( 2 + \sin \frac 1x) - \cos \frac 1x \bigr)
$$
for $x \ne 0$.
$f'$ is also differentiable on $\Bbb R$ so that $f$ is twice
differentiable.
But $f'$ is not monotonically increasing in any interval $(0, \delta)$ because
it takes both positive and negative values arbitrary close to $x=0$.

It does not even hold if $f$ has derivatives of all order. 
As an example, define $f$ by
$f(0) = 0$, 
$$
  f(x) = e^{-\frac 1x} ( 2 + \sin \frac {1}{x^2})
$$
for $x > 0$ and $f(-x) = f(x)$. $f$ has a minimum at $x = 0$
and has derivatives of all order (roughly because $e^{-\frac 1x}$
converges to zero for $x \to 0^{+}$ faster than any power of $x$).
For $x > 0$,
$$
 f'(x) = \frac{1}{x^3 }e^{-\frac 1x} \bigl( x( 2 + \sin \frac {1}{x^2})
 -  2  \cos \frac {1}{x^2}  \bigr)
$$
which again takes both positive and negative values arbitrarily
close to $x = 0$.
