# Absolute Value functions => Piecewise Function

Gold Member
Given the function:

$$f(x) = \left| {x + 1} \right| + \left| {x + 2} \right|$$

How can i write that as a piecewise function? If i was given something in the form of $$f(x) = \left| {g(x)} \right|$$, i know to write it as:

$$f(x) = \left\{ {\begin{array}{*{20}c} {g(x),} & {g(x) \ge 0} \\ { - g(x),} & {g(x) < 0} \\ \end{array}} \right.$$

but im a little stuck with the other one.

Dan.

consider the cases, x+1<0, x+2<0 and other combinations separately:

x+1<0, x+2<0
x+1<0, x+2>0
x+1>0, x+2<0
x+1>0, x+2>0

notice that some of these conditions are impossible.

Gold Member
Hmm ok. So when both are positive $$x + 1,x + 2 \ge 0$$, the piece is $$(x + 1) + (x + 2) = 2x + 3$$. They are both positive only when $$x \ge - 1$$, so my piece ($$2x + 3$$) would only be existant over the domain $$x \ge - 1$$? Is that the way i should go about it?

yes, that should be correct.

Gold Member
Ok thanks for that :)

Gold Member
Just one more thing. Using $$y = \left| x \right|$$ for example. At x=0, it could go either way. y could equal x or -x, so either of the following could be true:

$$\begin{array}{l} f(x) = \left\{ {\begin{array}{*{20}c} {x,} & {x \ge 0} \\ { - x,} & {x < 0} \\ \end{array}} \right. \\ f(x) = \left\{ {\begin{array}{*{20}c} {x,} & {x > 0} \\ { - x,} & {x \le 0} \\ \end{array}} \right. \\ \end{array}$$

Does it matter which way i do it? Or is there some convention i should follow?

HallsofIvy
The convention is that |x|= x for $x\ge 0$ but, in fact, either that or |x|= -x for $x\le 0$ give the same function.