If \(3x - (\frac{1}{4})^{-\frac{1}{2}} > \frac{1}{4} - x \), then the interval of values of x is
\(3x - (\frac{1}{4})^{-\frac{1}{2}} > \frac{1}{4} - x\)
= \(3x - 4^{\frac{1}{2}} > \frac{1}{4} - x\)
= \(3x - 2 > \frac{1}{4} - x\)
= \(3x + x > \frac{1}{4} + 2 \implies 4x > \frac{9}{4}\)
\(x > \frac{9}{16}\)
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