# Absolute Value functions => Piecewise Function

• danago
In summary, Dan is trying to figure out how to write a function that takes in a value between -1 and 1 and outputs a value between 0 and 3. He is using the example of f(x) = \left| {x + 1} \right| + \left| {x + 2} \right| to help him. The first case, x+1<0, x+2<0, is impossible because x+1>0 and x+2<0. The second case, x+1>0, x+2>0, is possible because both x+1 and x+2 are positive. The third case is when both x+1 and x+2 are positive and x+

#### danago

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 I am 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.

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 existent over the domain $$x \ge - 1$$? Is that the way i should go about it?

yes, that should be correct.

Ok thanks for that :)

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?

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.

Last edited by a moderator:
Ok thanks.

Another way to do it is to find your critical points, (x=-1 and x=-2 in this case) and create an x, f(x) table to test points outside of and between those points within the function. Then you can take the points in the three regions and find the equation of the line between them which is the equation that satisfies the x value of the region it was determined from. A calculator that can come up with a table of values for you really helps with this.