The problem asks us to find the domain of the function $f(x) = \sqrt[5]{2x - 1}$.

1. Problem Description

The problem asks us to find the domain of the function $f(x) = \sqrt[5]{2x - 1}$.

2. Solution Steps

The function $f(x) = \sqrt[5]{2x - 1}$ involves a fifth root. Since the index of the radical is odd (5), the expression inside the radical can be any real number. Thus, there are no restrictions on the value of $2x - 1$. In other words, $2x - 1$ can be any real number. Since $2x - 1$ is a linear expression, it is defined for all real numbers $x$. Therefore, the domain of $f(x)$ is all real numbers. However, the available options are:

(a) $\{x \in \mathbb{R}: x \le \frac{1}{2}\}$

(b) $\{x \in \mathbb{R}: x \ge \frac{1}{2}\}$

(c) $\{x \in \mathbb{R}: x < \frac{1}{2}\}$

(d) $\{x \in \mathbb{R}: x > \frac{1}{2}\}$

It seems there might be a typo in the original function. If the function was $f(x) = \sqrt{2x - 1}$, then we would require $2x - 1 \ge 0$.

$2x - 1 \ge 0$

$2x \ge 1$

$x \ge \frac{1}{2}$

The domain would be $\{x \in \mathbb{R}: x \ge \frac{1}{2}\}$.

Since the original function given is $f(x) = \sqrt[5]{2x - 1}$, and the fifth root is defined for all real numbers, the domain of $f(x)$ is all real numbers. However, since the options provided are inequalities, we will assume that the original function was indeed $f(x) = \sqrt{2x - 1}$, so the domain would be $x \ge \frac{1}{2}$.

3. Final Answer

$\{x \in \mathbb{R}: x \ge \frac{1}{2}\}$

(b)