Bunuel wrote:
Is x negative?
(1) \(x^3(1 - x^2) < 0\)
(2) \(x^2 - 1 < 0\)
My reasoning:
Statement 1: either \(x^3\) is negative or \((1 - x^2)\) is negative. They are opposite signs. However we can have a positive x and a negative x in this case.
Testing values of 2 and \(\frac{-1}{3}\) will prove this.
Statement 2: \(x^2 - 1 < 0\) -> \(x^2 < 1\) so \(x > -1 & x < 1\) Again inconclusive because we can have any positive or negative fraction.
Combining the two statements only "negative fractions" are in our overlapping set. The answer is C. x MUST be less than 0