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Calculus/absolute value limits

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Expert: - 9/22/2010


Question
f(x) = { |x^3 - 4x| x<1
        x^2 - 2x - 2 x>=1

Answer
To consier x<1, the function has the value |x³ - 4x|.
This means take any x value and put into the equation |(x²-4)x|.
If the answer is negative, take the positive.

For example, if x = 1/2, the result is |(1/2² - 4)/2| =
|(1/4 - 16/4)/2| = |(-15/4)/2| = |-15/8| = 15/8 = 1 7/8.

If x <= 1, use f(x) = x(x-2).

Note that at 1, the functions left side would approach |(x²-4)x| for
x = 1, so this would approach |(-3)1| = |-3| = 3.  The true value at 1, however, is given by x² - 2x where x = 1, and that is 1 - 2 = -1.

From this, it can be seen that the function is discontinuous at x=1.
I'm not sure if you're worried about this, however.

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