I was never really taught how to deal with the absolute value function, but I need it from time to time. So here are a few hints.
Solving Equations ¶
Lets say you want to solve the equation
$$|x - a | = b$$
for $x$. Then you need to realize that this equation is equivalent to two equations:
$$x - a = b \qquad \text{ and } \qquad -(x-a) = b$$
you can solve both of them independantly. You can get 0, 1 or 2 solutions when the absolute function is involved:
$$x = a + b \qquad \text{ and } \qquad x = a - b$$ or shorter $$x = a \pm b$$
Solving Inequalities ¶
Lets say you want to solve the inequality
$$|a - x| \leq b$$
for $x$. Again, this inequality is equivalent to the two inequalities
$$a - x \leq b \qquad \text{ and } \qquad -(a-x) \leq b$$
You can solve both of them independantly for $x$:
$$a - b \leq x \qquad \text{ and } \qquad -x \leq a + b$$
leading to
$$a - b \leq x \leq a + b$$
Note that both inequalities have to be fulfilled at the same time! Just try it for $a = 0$ and $b = -5$!
Derivatives ¶
The function $f(x) = |x|$ is equivalent to $f(x) = \sqrt{x^2}$. Hence you can derive the absolute value by deriving the root of the square function of its argument. And the chain rule, of course:
\begin{align} f'(x) &= (\sqrt{x^2})'\ &= \frac{1}{2 \sqrt{x^2}} \cdot (2 x)\ &= \frac{x}{\sqrt{x^2}}\ &= \frac{x}{|x|}\ &= \text{sign}(x) \end{align}
Note that the derivative is not devined at 0.