Absolute Value functions => Piecewise Function

Click For Summary

Homework Help Overview

The discussion revolves around expressing the function f(x) = |x + 1| + |x + 2| as a piecewise function. Participants explore the implications of absolute value functions and their representation in piecewise form.

Discussion Character

  • Exploratory, Conceptual clarification, Problem interpretation

Approaches and Questions Raised

  • Participants consider different cases based on the conditions of the expressions inside the absolute values, questioning the validity of certain combinations. There is discussion on how to derive the piecewise function based on the positivity or negativity of the components.

Discussion Status

Some participants have provided guidance on how to approach the problem, including identifying critical points and testing intervals. There is acknowledgment of the correctness of certain interpretations, but no explicit consensus on the final piecewise representation has been reached.

Contextual Notes

Participants mention the importance of conventions in defining piecewise functions, particularly regarding the treatment of critical points and the behavior of absolute values at those points.

danago
Gold Member
Messages
1,118
Reaction score
4
Given the function:

<br /> f(x) = \left| {x + 1} \right| + \left| {x + 2} \right|<br />

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:

<br /> f(x) = \left\{ {\begin{array}{*{20}c}<br /> {g(x),} &amp; {g(x) \ge 0} \\<br /> { - g(x),} &amp; {g(x) &lt; 0} \\<br /> \end{array}} \right.<br /> <br />

but I am a little stuck with the other one.

Thanks in advance,
Dan.
 
Physics news on Phys.org
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 <br /> 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:

<br /> \begin{array}{l}<br /> f(x) = \left\{ {\begin{array}{*{20}c}<br /> {x,} &amp; {x \ge 0} \\<br /> { - x,} &amp; {x &lt; 0} \\<br /> \end{array}} \right. \\ <br /> f(x) = \left\{ {\begin{array}{*{20}c}<br /> {x,} &amp; {x &gt; 0} \\<br /> { - x,} &amp; {x \le 0} \\<br /> \end{array}} \right. \\ <br /> \end{array}<br />

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.
 

Similar threads

  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 14 ·
Replies
14
Views
4K
  • · Replies 1 ·
Replies
1
Views
3K
  • · Replies 7 ·
Replies
7
Views
4K
  • · Replies 15 ·
Replies
15
Views
2K
  • · Replies 23 ·
Replies
23
Views
3K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 11 ·
Replies
11
Views
2K
  • · Replies 3 ·
Replies
3
Views
2K
  • · Replies 5 ·
Replies
5
Views
2K