1. We are asked to find the solution set of the inequality $$3 - \frac{1}{2x} \leq 1$$. 2. Start by isolating the term with $x$: $$3 - \frac{1}{2x} \leq 1 \implies - \frac{1}{2x} \leq 1 - 3 \implies - \frac{1}{2x} \leq -2$$ 3. Multiply both sides by $-1$ to remove the negative sign, remembering to reverse the inequality sign because we multiply by a negative: $$\frac{1}{2x} \geq 2$$ 4. Multiply both sides by $2x$. Since $x$ can be positive or negative, we consider two cases: - Case 1: $x > 0$, then multiplying by $2x > 0$ keeps inequality direction: $$1 \geq 4x \implies x \leq \frac{1}{4}$$ - Case 2: $x < 0$, then multiplying by $2x < 0$ reverses inequality: $$1 \leq 4x \implies x \geq \frac{1}{4}$$ But $x < 0$ and $x \geq \frac{1}{4}$ cannot both be true, so no solution here. 5. Also, $x \neq 0$ because of the denominator. 6. From Case 1, $x > 0$ and $x \leq \frac{1}{4}$, so $0 < x \leq \frac{1}{4}$. 7. Check the inequality for values in this interval to confirm. 8. The solution set is $$\left(0, \frac{1}{4}\right]$$. 9. Among the given options, the interval $\left[\frac{1}{8}, \frac{1}{4}\right]$ fits inside this solution set. Final answer: $\boxed{\left[\frac{1}{8}, \frac{1}{4}\right]}$