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Absolute value and piecewise functions

  1. Feb 11, 2014 #1

    939

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    1. The problem statement, all variables and given/known data

    Find if continuous and differentiable. I am having problems with the differentiable part.


    2. Relevant equations

    f(x) = x² + 3, |x| ≤ 1
    f(x) = |x| + 3, |x| > 1

    3. The attempt at a solution

    (1)^2 + 3 = 4
    |1| + 3 = 4
    ∴ It is continuous

    Now, rewriting it, you have...
    x² + 3, x ≥ -1 or x ≤ 1
    -x + 3, x < -1 or x > 1
    x + 3, x < -1 or x > 1

    Finding the limits from left and right of 1...
    From the left = 2x, = 2
    From the right: can be both -1 and 1...

    Where did I go wrong?
     
  2. jcsd
  3. Feb 11, 2014 #2

    Dick

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    No, you've got
    x² + 3, x ≥ -1 or x ≤ 1
    -x + 3, x < -1
    x + 3, x > 1

    Worry about the two point x=1 and x=(-1) separately.
     
  4. Feb 11, 2014 #3

    939

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    Thanks a lot!

    Final question can |(2/3)(x)| be turned into (2/3)(-x) and (2/3)(x)?
     
  5. Feb 11, 2014 #4

    Dick

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    Sure. |2x/3| is -2x/3 if x<0 and 2x/3 if x>=0.
     
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