Positive derivative at $c$, then $f(x)>f(c)$ for close $x>c$ Question: Let $f$ be defined on an open interval $(a, b)$ and assume that for some $c \in (a, b)$ $f^{\prime} (c)> 0$ then show that there is a open ball around $c$, say $B(c) \subseteq (a, b)$ in which $f(x)> f(c)$ if $x>c$ and $f(x)< f(c)$ if $x<c$. Note:  $f^{\prime}$ denotes the derivative of the function $f$.
What I have understood is that I have to show that function $f$ is strictly increasing in the interval $(a, b)$. Given that $f^{\prime} (c)> 0$ which implies that $f$ must be differentiable. Now, I am trying to apply the Lagrange mean value theorem. But I am not sure which interval to take so that I can prove desired inequality. Also, how can I introduce open ball around $c$ so that $B(c) \subseteq (a, b)$ and desired inequality holds true there.
Thank you
 A: No, $f$ need not be increasing in any interval surrounding $c$.  The result here follows directly from the definition of the derivative at $c$.  
$\lim\limits_{x\to c}\dfrac{f(x)-f(c)}{x-c} =f'(c)>0$,
which implies that when $x$ is sufficiently close to $c$, i.e. for $x$ in some interval $(c-\delta,c+\delta)$, $\dfrac{f(x)-f(c)}{x-c}$ is positive.  For $x>c$, this implies $f(x)>f(c)$.  For $x<c$, this implies $f(x)<f(c)$.
To see why you can't show $f$ is increasing in an interval, see Differentiable+Not monotone. Even though the link doesn't work in the answer, hopefully the reference given or the comments or googling will lead you to more info if you're interested.

For just a particular point like you're asking about, $f(x) = x+2x^2\sin(1/x)$ (and $f(0)=0$) provides an example where $f$ is not monotone in any interval containing $0$, although $f'(0)=1$.  When $x\neq 0$ we have $f'(x)=1+4x\sin(1/x)-2\cos(1/x)$, which is negative arbitrarily close to $0$, e.g. when $x=\dfrac{1}{2\pi n}$ for each nonzero integer $n$.  By the same reasoning as in the solution to your problem, this implies that $f$ can't be increasing in any interval containing a $\dfrac{1}{2n\pi}$, while every interval containing $0$ has infinitely many of these.  Without using derivatives, you can verify that when $n$ is sufficiently large, $f\left(\dfrac{1}{2n\pi}\right)-f\left(\dfrac1{2n\pi+\frac\pi2}\right)<0$.
