Prove if $f$ is increasing then $f'(x) \ge 0$ Let $f:[a, b] \rightarrow \mathbb{R}$ where $a, b \in \mathbb{R}$ with $a <b$. Suppose $f$ is continuous on $[a,b]$ and differentiable on $(a,b)$. Suppose $f$ is monotonically increasing on $[a,b]$. Show that $f'(x) \ge 0$ for all $x \in (a,b)$.
My attempt: I tried using the Mean Value Theorem, but it doesn't quite seem to work. For example, by the MVT we can conclude that there exists a $c \in (a,b)$ such that $f(b) - f(a) = f'(c) (b-a)$. Which implies that $f'(c) = \frac{f(b) - f(a)}{b-a}$. Now since $f$ is monotonically increasing, $f(b) - f(a) \ge 0$ whenever $b>a$, so $f'(c) \ge 0$. But this only shows for one particular $c \in (a,b)$, and the question asks to show this is true for ALL $x \in (a,b)$. What can I do to complete the proof?
 A: For $x \in (a,b)$
$$f'(x)=\lim_{h \to 0}\frac{f(x+h)-f(x)}{h}\geq0$$
A: Your approach using MVT will never guarantee the location of $c$ or may not cover all $c\in(a, b)$. Consider for example $f(x) = x^{3}$ in $[-1,1]$. Since $f$ is strictly increasing it follows that the ratio $(f(b) - f(a)) /(b-a) >0$ for any two distinct points $a, b\in[-1,1]$ and thus the corresponding $c$ guaranteed by MVT will always have $f'(c) >0$. This misses the point $c=0$ where derivative vanishes. Thus your proof can not be salvaged. Even isolating such troublesome points like the one given in above example and then showing that derivative vanishes there is a bit difficult. 
The proper and much simpler approach is to use the definition of derivative as given in Sahiba Arora's answer.
A: Hint: For a given $x$, there exists $a \le c \le x \le d \le b$ such that $$\frac{f(d)-f(c)}{d-c}=f'(x).$$
A: Your proof can be made to work if $f'$ is continuous. Let $x∈ (a,b)$. Then
for each $n\gg 1$, apply MVT as you have on the interval $(x-\frac1n,x+\frac1n)$  to discover an $x_n$ with $|x_n - x|<1/n$ and $f'(x_n)\geq 0$. Just take $n→∞$ to conclude, using continuity of $f'$.
