Proof about First order derivative 
Show that if $f'(c)>0$ then there exists $\delta>0$ such that
  $x \in (c,c+\delta) \ \ \implies \ \ f(x)>f(c)$
  $x \in (c-\delta,c) \ \ \implies \ \ f(x)<f(c)$  

My Attempt 
Now for $|x-c|<\delta$ , and $f'(c)>0$;
We know $$f(x)=f(c)+f'(c)(x-c)+R_{2,c}$$
So if;
$\bullet x \in (c,c+\delta) \ \ \implies \ \ x>c$ which means $$f(x)-f(c)=f'(c)(x-c)+R_{2,c}>0$$ so $f(x)>f(c)$
$\bullet x \in (c-\delta,c) \ \ \implies \ \ x<c$ which means $$f(x)-f(c)=f'(c)(x-c)+R_{2,c}<0$$ so $f(x)<f(c)$   
Is this proof correct??

If $f'(c)=:a>0$ then by definition of $f'$, for any $\epsilon>0$, there exists $\delta>0$ such that $|x-c|<\delta$ implies $\left|\frac{f(x)-f(c)}{x-c}-a\right |<\epsilon$. If we let $\epsilon=a$, this implies $\frac{f(x)-f(c)}{x-c}>0$ for alls $x$ with $|x-c|<\delta$. If additionally $x>c$, we conclude $f(x)-f(c)>0$, similarly $f(x)-f(c)<0$ for $x<c$.
