Proving that a positive derivative means the function is smaller "to the left" and larger "to the right" for certain values I was trying to prove that if $g$ is differentiable on an open interval $I$ with $a\in I$ and $g'(a)>0$ then we can find $x<a$  for which $g(x)<g(a)$ and $y>a$ for which $g(y)>g(a)$, I think I understand limits correctly but it confuses me when it comes to derivative so I just wanted to make sure I understand it correctly.
My proof:
Let $g'(a)=K>0$, that means that $$\lim_{x\to a}\frac{f(x)-f(a)}{x-a}=K$$
Therefore given $\epsilon=\frac{k}{2}$ we can find $\delta >0$ such that for $x\in I$ and $x\in (a-\delta , a+\delta )\setminus \{a\}$
$$\left|\frac{f(x)-f(a)}{x-a}-K\right|<\frac{K}{2}$$
(This is the part I'm not sure about)..
hence
$$-\frac{K}{2}<\frac{f(x)-f(a)}{x-a}-K<\frac{K}{2}$$
hence,
$$\frac{K}{2}(x-a)<f(x)-f(a)<\frac{3K}{2}(x-a)$$
hence,
$$\frac{K}{2}(x-a)+f(a)<f(x)<\frac{3K}{2}(x-a)+f(a)$$
Therefore if we choose $x\in (a-\delta ,a)$ and $y\in (a,a+\delta)$ (which we can find since $I$ is open) we get that
$$f(x)<\frac{3K}{2}(x-a)+f(a)<f(a)\qquad \text{As }x-a<0$$
$$f(a)<\frac{K}{2}(y-a)+f(a)<f(y)\qquad \text{As }y-a>0$$
Is this proof correct?
Also, I really got lost in trying to write the title :P

You have $$-\frac{K}{2}<\frac{f(x)-f(a)}{x-a}-K<\frac{K}{2}.$$ Add $K$ to everything to see the difference quotient is positive for this range of $x.$ How could that quotient be positive to the right of $a$ unless the numerator is positive? Same to the left of $a.$
