Absolute Value As Statement about DistanceDate: 10/16/2002 at 13:44:28 From: M Subject: Absolute value Hi, The question is: |x|+|x+2|+|x-4| < 5 Thanks. Date: 10/16/2002 at 14:18:02 From: Doctor Schwa Subject: Re: Absolute value Hello, One way to think about this type of question is to interpret absolute value as a statement about distance. |x| means "the distance from 0 on the number line" and similarly |x-4| means "the distance from 4 on the number line" and |x+2| means "the distance from -2 on the number line" Using that interpretation, you can pretty quickly prove that there are no solutions to your equation. You can also do this type of problem by splitting things into cases: what happens when x > 4? When x is between 0 and 4? When x is between -2 and 0? When x < -2? In each case, you can figure out whether what is inside each absolute value sign is negative or positive, and work from there. For instance, if I had a slightly different problem, |x| + |x-4| < 10, I'd say: if x > 4, then both x and x-4 are positive, so x + (x-4) < 10, x < 7. So the numbers between 4 and 7 make it true. If x is between 0 and 4, then x is positive but x-4 is negative, so you have to take its opposite to make it positive, x + -(x-4) < 10, which is always true, so all the numbers between 0 and 4 work. If x is less than 0, then x and (x-4) are both negative, so -x + -(x-4) < 10, which turns out to mean x > -3, so all the numbers between -3 and 0 work. Putting them all together, it turns out that all the numbers between -3 and 7 will work in the original inequality. On the other hand, by the distance approach: The distance from 0 plus the distance from 4 is less than 10. So, where is that? Well, between 0 and 4 the two distances add up to 4, so that works. Outside of that region, one distance is 4 units more than the other, so they add up to 10 when they are 3 + 7, or 7 + 3 ... and the distances work out that way at the points -3 and 7. So the total distance equals 10 at those to points, and is less than 10 when you're between those points. I hope you see now why I prefer the distance approach! - Doctor Schwa, The Math Forum http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994-2015 The Math Forum
http://mathforum.org/dr.math/