To find the range of values of \(x\) for which \(\frac{1}{2}x + \frac{1}{4} > \frac{1}{3}x + \frac{1}{2}\), we need to solve the inequality step by step:

\(\frac{1}{2}x + \frac{1}{4} > \frac{1}{3}x + \frac{1}{2}\)

Subtract \(\frac{1}{3}x\) from both sides:

\(\frac{1}{2}x - \frac{1}{3}x + \frac{1}{4} > \frac{1}{2}\)

Combine the terms on the left side:

\(\frac{1}{6}x + \frac{1}{4} > \frac{1}{2}\)

Subtract \(\frac{1}{4}\) from both sides:

\(\frac{1}{6}x > \frac{1}{2} - \frac{1}{4}\)

Simplify the right side:

\(\frac{1}{6}x > \frac{1}{4}\)

Now, multiply both sides by 6 (which is positive, so the inequality direction doesn't change):

\(x > \frac{1}{4} \cdot 6\)

\(x > \frac{3}{2}\)