Question: Consider the following example:

$f'(x)= x^{-3/2}$ and $f(4)=2 $ 

$f'(x)= x^{-3/2}\Rightarrow \frac {x^{(-3/2) + 1}} {-1/2} \Rightarrow$ $\frac{-2}{\sqrt x} +C =f(x)$

This is where the problem arises now if I consider $x= 4$ then I get 

$$\frac{-2}{\sqrt 4} +C \Rightarrow \frac{-2}{\pm 2} +C $$

Obviously this leads to two different antiderivatives if we consider two different cases which doesn't seem right. I was told that we consider the positive number in these situations but why? And what does the negative case represent if we were to "humor" it? Your answers will appreciated. Thank you.

Also I apologize if the question title is not descriptive of the case I'm describing. If you can think of a better title go ahead and change it or let me know, just so if anyone else looks for a similar problem have an easy time finding it?

Answer: 
The equation $y^2\,=\,x$ for $y$ has two roots, namely $y\_+\,=\,+\sqrt{x}$ and $y\_-\,=\,-\sqrt{x}$. That is, by convention we take $\sqrt{x}$ to mean the positive root, unless otherwise stated. 

As Andre said, the derivative of $\frac{-2}{-\sqrt{x}}+C$ is **not** $f'(x)$, but $-f'(x)$, as you can easily check. 

When you write $\sqrt{4}=\pm2$ you are arbitrarily introducing a second solution that actually does not fit the equation for $f'(x)$.
The square root function only gives one result: the positive root. $\sqrt{4}=2$, not $-2$. So, the antiderivative only has one value.

The negative case would simply represent the antiderivative of $-f'(x)$. If you are unsure why the negative root is not the answer, try using your formula to find the area between any two points on the original function. Graph the function out, and realize that the positive square root is the only one that gives the correct area!