2. Determine the domain of the following functions: \( \begin{array}{lll}\text { i) } h(x)=x+\sqrt{3 x-4} & \text { ii) } k(x)=\sqrt{x}+\sqrt{x-1} \\ \text { iii) } f(x)=\sqrt[5]{x+3} & \text { iv) } k(x)=\sqrt[4]{2 x+1} \\ \text { v) } f(x)=x^{2}+\sqrt{x^{2}+4 x-12} & \text { (vi) } f(x)=\sqrt{\frac{x+1}{x-1}}\end{array} \)

To determine the domain of the given functions, we need to analyze the expressions under the square roots and ensure that they remain non-negative (for square roots) or define valid inputs (for other functions). Let's break it down for each function: 1. **For** \( h(x)=x+\sqrt{3x-4} \): The expression \( 3x - 4 \geq 0 \) leads to \( x \geq \frac{4}{3} \). Therefore, the domain is \( [\frac{4}{3}, \infty) \). 2. **For** \( k(x)=\sqrt{x}+\sqrt{x-1} \): The requirements \( x \geq 0 \) (from \( \sqrt{x} \)) and \( x - 1 \geq 0 \) (from \( \sqrt{x-1} \)) imply \( x \geq 1 \). Thus, the domain is \( [1, \infty) \). 3. **For** \( f(x)=\sqrt[5]{x+3} \): The function \( \sqrt[5]{x+3} \) is defined for all real numbers since there's no restriction on odd roots. Hence, the domain is \( (-\infty, \infty) \). 4. **For** \( k(x)=\sqrt[4]{2x+1} \): Since \( \sqrt[4]{2x+1} \) is defined as long as \( 2x + 1 \geq 0 \), we find \( x \geq -\frac{1}{2} \). Thus, the domain is \( [-\frac{1}{2}, \infty) \). 5. **For** \( f(x)=x^2+\sqrt{x^2+4x-12} \): We need the expression inside the square root, \( x^2 + 4x - 12 \geq 0 \). Factoring gives \( (x+6)(x-2) \geq 0 \), leading to the intervals \( (-\infty, -6] \cup [2, \infty) \). Hence, the domain is \( (-\infty, -6] \cup [2, \infty) \). 6. **For** \( f(x)=\sqrt{\frac{x+1}{x-1}} \): To be defined, the fraction must be non-negative and the denominator cannot be zero. Thus, \( x + 1 \geq 0 \) gives \( x \geq -1 \) and \( x - 1 > 0 \) gives \( x > 1 \). The overlapping requirement yields the domain \( (1, \infty) \). So, summarizing, we have: 1. \( [\frac{4}{3}, \infty) \) 2. \( [1, \infty) \) 3. \( (-\infty, \infty) \) 4. \( [-\frac{1}{2}, \infty) \) 5. \( (-\infty, -6] \cup [2, \infty) \) 6. \( (1, \infty) \)