Showing that $\frac{f(x)-f(x-h)}{h} \le f'(x) \le \frac{f(x+h)-f(x)}{h}$ given that f' is increasing We are considering that $h > 0$. I just don't really know how to approach this question; it seems that by taking the limit of the inequality as h goes to 0, each part of the inequality is going to be equal to $f'$, but I don't really understand why the fact that $f'$ is increasing is relevant to finding a solution to this question (and how to answer the question based on that constraint).

The assumption is the $f'$ is an increasing function on $[x-h,x+h]$.  
Now by the mean value theorem, there exist numbers $\xi_1\in (x-h,x)$ and $\xi_2 \in (x,x+h)$ such that 
$$f'(\xi_1)=\frac{f(x)-f(x-h)}{h}$$
$$f'(\xi_2)=\frac{f(x+h)-f(x)}{h}$$ 
Inasmuch as $f'$ is increasing and $\xi_2>x>\xi_1$, we have
$$\frac{f(x)-f(x-h)}{h}=f'(\xi_1)\le f'(x)\le f'(\xi_2)= \frac{f(x+h)-f(x)}{h} \tag 1$$
as was to be shown!
