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Re: DS absolute value [#permalink]
18 Nov 2007, 16:28
If x<0, then sqrt(-x|x|) is
-x -1 1 x sqrt(x)
Quick guess - A -x?
if x < 0, then x is negative, thus -x is positive, and |x| as the absolute value is a positive value - so it's equivalent is positive positive X^2 the square root of which is positive x. If x is negative, then -x is positive.
Guys, does n't this have to be -X instead of X ?
If it were -X and let us say we pick a value of -3 for X (because X < 0), then the result would be -(-3) = 3. This does not agree with what was given in the question.