Expressing absolute value equations as piecewise functions I'm not sure how to express this function in piecewise form without using absolute values:
$$
f(x) = 3|x-2| - |x+1|$$
I know how to do it when there is just one absolute value, such as: 
$$g(x) = 3+|2x-5|$$
$$
g(x)=
\begin{cases} 2x-2& \text{; }x\ge\frac52\\
    8-2x&\text{; }x<\frac52 \end{cases}
$$
To express $g(x)$ in piecewise form, I made 2 cases. Case 1 positive and Case 2 negative. 
But I can't exactly do that for this problem [$f(x)$]... Could someone tell me how to proceed?
EDIT: Thank you for all the help!
 A: The only places where we need to break the function into "pieces" are those points where the expression inside each absolute value becomes zero. This occurs at $x = -1$ and $x = 2$.
Accordingly, we consider what the function looks like on each of the intervals $(-\infty,-1)$, $(-1,2)$, and $(2,\infty)$. 
For $x < -1$, we have $x + 1 < 0$ and so $|x+1| = -(x+1)$. Similarly, $x - 2 < x + 1 < 0$ and so $|x-2| = -(x-2)$. So on the interval $(-\infty,-1)$ the function can be simplified as $f(x) = -3(x-2) +(x+1) = -2x+7$.
Thus far, we have established
$$f(x) = \begin{cases}
-2x + 7 & \text{if }x < -1 \\
??? & \text{if }-1 \leq x \leq 2 \\
??? & \text{if }x > 2 \\
\end{cases}$$
You can handle the other intervals in exactly the same way.
A: Try expressing each of the parts - $3|x-2|$ and $|x+1|$ in piecewise form first. This tells you where the behaviour of the function is going to change: at $x=-1$ and $x=2$. So you have three intervals to look at: $\{x < -1\}$, $\{-1 \le x \lt 2\}$ and $\{x \ge 2\}$. Then all you have to do is see how the two functions combine in each of those regions.
