Show that $f$ is increasing on $(a, b)$ if and only if $f^{\prime}(x) \geq0$ for all $x \in (a, b)$. A function $f : (a, b) \rightarrow \mathbb{R}$ is said to be increasing on $(a, b)$ if $f(x)\leq f(y)$ for all $x < y \in (a, b)$. Suppose that $f$ is differentiable on $(a, b)$. Show that $f$ is increasing on $(a, b)$ if and only if $f^{\prime}(x) \geq0$ for all $x \in (a, b)$.
For the backward direction, I try to use Mean Value Theorem to show but I don't know how to show that $f$ is continuous on $[a,b]$. Can anyone guide me ?
 A: You don't need continuity on $[a,b]$, you only need it on $[x,y]$. And you have that, since $x$ and $y$ are in $(a,b)$, and our function is differentiable and therefore continuous on $(a,b)$. So it is continuous on $[x,y]$. 
Now apply the MVT to conclude that $\frac{f(y)-f(x)}{y-x}=f'(c)$ for some $c$ in the interval $(x,y)$, and use the fact that $f'(c)\ge 0$.
You did not ask about the other direction, so I assume that is under control.
A: *

*Let $f$ be increasing on $(a,b).$ Choose $x\in(a,b).$ Then for all $y\neq x,$ $\frac{f(y)-f(x)}{y-x}>0$ since $f$ is incresing on $(a,b)$. Consequently $f'(x)=\lim_{~y\to x}\frac{f(y)-f(x)}{y-x}\geq0.$ (Note that $x$ is a limit point of $(a,b)$ and $\lim_{~y\to x}\frac{f(y)-f(x)}{y-x}$ is deined)

*Conversely,  let $f'(x)\geq0~\forall~x\in(a,b).$ Choose $x,y\in(a,b)$ such that $a<x<y<b.$ Then by MVT, $\exists~\xi\in(x,y)$ such that $\frac{f(y)-f(x)}{y-x}=f'(\xi)\geq0$ whence $f(y)\geq f(x).$
A: You won't be able to show that differentiability on $(a,b)$ will imply continuity on $[a,b]$, since it is not true. Consider the following function defined on $[0,1]$:
$$
f(x)=\begin{cases}1&\text{if }x=0\\x&\text{if }0<x<1\\0&\text{if }x=1\end{cases}
$$
This function is differentiable in $(0,1)$. In fact its derivative is $f'(x)=1$, but obviously it is not continuous on $[0,1]$.
The backward direction is: If a function $f$ is differentiable in $(a,b)$, and $f'(x)\geq 0$ then it is increasing on $(a,b)$. 
If you accept that every function differentiable at $c$ is continuous at $c$, then by the definition of differentiability on an open interval, you have that the function is continuous at the open interval. Then you can use the Mean Value Theorem to conclude the backward direction.
Now you are left with proving that differentiability at $c$ implies continuity at $c$, but this is simple:
Differentiability at $c$, implies that: $f'(c)=\lim_{h\to 0}\dfrac{f(c+h)-f(c)}{h}$, then for $\epsilon>0$ sufficiently small, we can find $\delta>0$ such that if $|h|<\delta$ then
$$
\left|\dfrac{f(c+h)-f(c)}{h}-f'(c)\right|<\epsilon
$$
That is, if $|h|<\delta$ then $-|h|\epsilon+f'(c)h<f(c+h)-f(c)<|h|\epsilon+f'(c)h$, which proves that:
$$
\lim_{h\to 0}f(c+h)=f(c),
$$
that is $f$ is continuous at $c$.
A: if a function is differentiable on (a, b), then it is continuous on (a,b). so the condition's given, you don't have to prove it. 
and for the converse part, you have to use the definition, not the mean value theorem cause mvt saying about ONE C in between x,y. the converse part you need to prove f'(x) >0 for ALL c in(a,b).
