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Homework Help: Absolute values of variables

  1. Oct 4, 2005 #1
    so if |x|=(x,if x>=0, and -x, if x<0)
    then what would be like |x-7| be equal too and how do you do this i do not understand why |x|equals (x,if x>=0, and -x, if x<0) could you explain it to me?
     
  2. jcsd
  3. Oct 4, 2005 #2

    Hurkyl

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    That's the definition of |x|. Try plugging in a few actual numbers to get more comfortable with it.

    If f(x) = (x-3)²+4, do you know how to get f(x-7)? Why can't you do the same thing with |x|?
     
  4. Oct 4, 2005 #3
    Here are a couple examples.
    |8+x| equals {8+x if 8+x>=0, -(8+x) if 8+x<0}
    |x²-5| equals {x²-5 if x²-5>=0, -(x²-5) if x²-5<0}

    Like Hurkyl said above, plugging in numbers will help your understanding.
     
  5. Oct 4, 2005 #4

    robphy

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    Picture a number line.
    |x-8| is the distance from "8", which you can write in terms of x, as done above.

    Try it for |x-0| first.
     
  6. Oct 4, 2005 #5
    so basically|x| means the answer is like always positive so therefor there will be a possibility of where the x is negative or -x and positvie just x right? so |x| is like F(x) rite? and if you sub x-7 for f(x) you would get like in order for x to be the positive value it would have to be greater than 7 there for x>=7 and -x would be -x<7?

    did i get his right?
     
  7. Oct 5, 2005 #6

    honestrosewater

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    Right, if you include 0 as a positive number. To avoid confusion about 0, you could also say that |x| is non-negative, which means |x| > 0.
    I don't really understand that. Perhaps x appearing both in the definition and your example is confusing - the two x's aren't being used in the same way. So let's just change (x - 7) to (y - 7). If you write |x| as f(x), |y - 7| would be f(y - 7); You're setting x = y - 7. Plug (y - 7) into your definition in place of x. It now says

    |y - 7| = (y - 7) if (y - 7) > 0; -(y - 7) if (y - 7) < 0.

    You already know that |y - 7| will be non-negative, remember. The definition tells you how to turn (y - 7) into a non-negative number.
    Let y = 0. (0 - 7) = -7. Plug this into your definition:

    |-7| = -7 if -7 > 0; -(-7) if -7 < 0.

    -7 < 0. So what does the definition tell you? |-7| = -(-7) = 7.
    Let y = 8. (8 - 7) = 1. Do the same thing.

    |1| = 1 if 1 > 0; -(1) if 1 < 0.

    1 > 0. So what does the definition tell you in this case? |1| = 1.
    The definition tells you more than that. But do you understand this part?
    The rule is pretty simple - you could say informally that if x is already non-negative, don't do anything to it; If x is negative, do what to it in order to make it non-negative?
     
    Last edited: Oct 5, 2005
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