Piecewise Notation and Absolute ValueDate: 9/9/96 at 13:25:22 From: Anonymous Subject: Piecewise Notation and Absolute Value Using piecewise notation, write x in terms of y for the following equation: "y = 2x + abs(2-x)". I have graphed the equation and I think part of the answer is x = y-2 for y in the range of negative infinity to 4. Is there another way to do this rather than graphing? Please describe the method. Date: 02/04/97 at 22:27:01 From: Doctor Sam Subject: Re: Piecewise Notation and Absolute Value If you already graphed y = 2x + abs(2-x) then you know that it looks like parts of two lines. Those parts are the "pieces" in "piecewise notation." The reason that there are two pieces is that the absolute value function is also in pieces: x when x is positive or zero abs (x) = -x when x is negative To change your absolute value equation into piecewise form ask yourself, "When is 2-x positive or zero? When is it negative?" Since 2-x = 0 when x = 2, this point marks the x-value of the part of the function when the graph changes from one piece to the next. When x < 2 the expression inside abs(...) is a positive number so abs(2 - x) = 2 - x. On the other hand, when x > 2 the expression inside abs( ) is a negative number and so its absolute value is the opposite. That is, abs(2 - x) = - (2 - x) = x - 2 for x-values in this range. To finish the problem, figure out what y is in each part of the graph: / 2x + (2 - x) when x < 2 \ y = | | \ 2x + (x - 2) when x >=2 / Finally, simplify these two expressions to get that y = x + 2 when x < 2 and y = 3x-2 when x >=2. I hope that helps! Write back if you need more help. --Doctors Sam and Sydney, The Math Forum Date: 09/08/2010 at 11:49:04 From: Renee Subject: Piecewise Notation and Absolute Value - explanation on forum Hello, I'm writing to you for clarification on this problem. I found the explanation about absolute values very helpful; yet the one thing I can't figure out is how/when you know to make the plain "greater than" versus the "greater than or equal to" for the final part of the piecewise function: why did the 'equal to' / 2x + (x - 2) when x >= 2 <---- appear in this section ... y = | \ 2x + (2 - x) when x < 2 <---- rather than in this one? I have tried looking up the answer but cannot find an explanation to the specific problem that you have already demonstrated. I was thinking that maybe the 'positive' section would have the equality because you included the question "When is 2 - x positive or zero?" However, in the end, you concluded only the strict inequality x < 2. Thank you so much! Date: 09/08/2010 at 14:02:16 From: Doctor Peterson Subject: Re: Piecewise Notation and Absolute Value - explanation on forum Hi, Renee. Where you put the "or equal" is actually arbitrary in this kind of problem; it makes no difference whether you say / 2x + (2 - x) when x < 2 y = | \ 2x + (x - 2) when x >= 2 or / 2x + (2 - x) when x <= 2 y = | \ 2x + (x - 2) when x > 2 This is because for x = 2, both ... 2x + (2 - x) and 2x + (x - 2) ... have the same value -- namely, 4. In particular, note that |0| = 0, which is equal to both x and -x; the two "pieces" meet, so that one point lies on both "pieces," and can be considered a part of either one. More generally, when a piecewise function is continuous, it doesn't matter which "piece" we put the junctions with. So technically, Doctors Sam and Sydney made a mistake in implicitly answering the question "When is 2 - x positive or zero? When is it negative?" Instead of ... x < 2; x >= 2, ... it should have been as you suggest: x <= 2; x > 2 I suspect they wrote what they did out of habit: when it doesn't matter, we tend to include 0 with the positives rather than the negatives, and they included 2 with the numbers greater than 2. Ideally, they would have been more careful -- or just explained why, in the end, it made no difference. - Doctor Peterson, The Math Forum http://mathforum.org/dr.math/ Date: 09/08/2010 at 16:45:45 From: Renee Subject: Thank you (Piecewise Notation and Absolute Value - explanation on forum) Thank you thank you thank you so much!!! Very very helpful and cleared things up a lot :) you da best! :D Renee |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/