How can I find the value of the derivative at a certain value given the graph? 
I need to find the values of $$\lim_{h\rightarrow 0^+} \frac{f(2+h)-f(2)}{h}$$
I know that the value of $f(2)=-1$
I think I need to come up with an equation for the graph, which I can't come up with. I thought it was $|x-2|-1$ but that isn't it.
How can I solve this?
 A: You were on the right track when you tried to develop an expression for $f(x)$.  But you really need not do that.
The limit, $\lim_{h\to 0^+}\frac{f(x+h)-f(x)}{h}$, is not equal to the derivative of $f$ at $x$.  It is called, however, the right-sided derivative.
Here, we need the right-sided derivative at $x=2$, which is given by $\lim_{h\to 0^+}\frac{f(2+h)-f(2)}{h}$.  Note that $h>0$ in this limit.
So to evaluate this right-sided derivative, simply develop the equation for $f(x)$ for $x\ge 2$, which is the equation of the straight line that passes through the points $(2,-1)$ and $(3,1)$.
Can you finish now?
A: You are only concerned with the value of the function when $x\ge 2$. And it's easy to come up with a simple equation that agrees with $f(x)$ when $x\ge 2$.
A: First consider the simpler function $f(x)=|x|.$ It is not differentiable at $0$ since you have $$\lim_{h\rightarrow 0^+}\frac{f(h)-f(0)}{h}=\lim_{h\rightarrow 0^+}\frac{|h|}{h}=1$$
but $$\lim_{h\rightarrow 0^-}\frac{f(h)-f(0)}{h}=\lim_{h\rightarrow 0^-}\frac{|h|}{h}=-1.$$
So the left and right derivatives don't agree.
You can see the gradients of the two slopes are different from a sketch. So we have $$f'(x)=\begin{cases} 1 \text{$\space$ if $x>0$}\\ -1 \text{ $\space$ if $x<0$}\end{cases}.$$
Now to considering your function, you don't need to compute the equation of the graph to find the right hand derivative. It is just the slope of the line for $x>2. $ (You can draw a triangle and find $\frac{\Delta y}{\Delta x}$ or pick two points).
You will find that $$\lim_{h\rightarrow 0^+} \frac{f(2+h)-f(2)}{h}=2$$
so we say that the right hand derivative for $h\rightarrow 0^+$ is 2 and similarly you can show that  $\lim_{h\rightarrow 0^-} \frac{f(2+h)-f(2)}{h}=-2.$ Again since the limits don't agree, the function is not differentiable at $2$.
Alternatively you can find the equation of the graph. First consider the line through points $(2,-1)$ and $(3,1)$, which is $y=2x-5.$ Now $f(x)=|2x-5|-1$ doesn't give exactly what we want since we have $f(2.5)=-1$ but we want $f(2)=-1$, so we shift $x$ to the left by $0.5$ (see this). This gives the required equation of the graph to be $$f(x)=|2(x+0.5)-5|-1=|2x-4|-1.$$
Now you can find $\lim_{h\rightarrow 0^+} \frac{f(2+h)-f(2)}{h}.$
