If x is a positive real number, find the range of values for which \(\frac{1}{3x}\) + \(\frac{1}{2}\) > \(\frac{1}{4x}\)
\(\frac{1}{3x}\) + \(\frac{1}{2}\) > \(\frac{1}{4x}\)
= \(\frac{2 + 3x}{6x}\) > \(\frac{1}{4x}\)
= 4x(2 + 3x) > 6x = 12x\(^2\) - 2x = 0
= 2x(6x - 1) > 0 = x(6x - 1) > 0
Case 1 (-, -) = x < 0, 6x -1 < 0
= x < 0, x < \(\frac{1}{6}\) = x < \(\frac{1}{6}\) (solution)
Case 2 (+, +) = x > 0, 6x -1 > 0 = x > 0, x > \(\frac{1}{6}\)
Combining solutions in cases(1) and (2)
= x > 0, x < \(\frac{1}{6}\) = 0 < x < \(\frac{1}{6}\)
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