\( g(x)=\frac{2 x+1}{x-2} \). Find the domain of \( g \) and find \( g(-1) \) \( t^{h(x)}=\frac{\sqrt{x^{2}-1}}{x} \). Find the domain of \( h(x) \).

**1. For \( g(x)=\frac{2x+1}{x-2} \):** - **Domain:** The function is undefined when the denominator is zero. Set the denominator equal to zero: \[ x - 2 = 0 \quad \Longrightarrow \quad x = 2. \] Therefore, the domain of \( g(x) \) is all real numbers except \( 2 \): \[ \{x \in \mathbb{R} : x \neq 2\}. \] - **Evaluate \( g(-1) \):** Substitute \( x = -1 \) into the function: \[ g(-1)=\frac{2(-1)+1}{-1-2}=\frac{-2+1}{-3}=\frac{-1}{-3}=\frac{1}{3}. \] --- **2. For \( t^{h(x)}=\frac{\sqrt{x^{2}-1}}{x} \):** We need to determine the values of \( x \) for which the equation makes sense. Note that since the left side \( t^{h(x)} \) (with \( t>0 \), \( t\neq1 \)) is always positive, the right side must also be positive. - **Step 1: Domain for the square root.** The expression under the square root must be nonnegative: \[ x^2 - 1 \geq 0 \quad \Longrightarrow \quad x^2 \geq 1, \] which implies \[ |x| \geq 1. \] - **Step 2: The quotient \( \frac{\sqrt{x^{2}-1}}{x} \) must be positive.** - For \( x > 0 \): Since \( |x| \geq 1 \) then \( x \geq 1 \). When \( x > 1 \), we have \(\sqrt{x^{2}-1}>0\) and \( x>0 \), hence \[ \frac{\sqrt{x^{2}-1}}{x} > 0. \] At \( x=1 \), we get: \[ \sqrt{1^{2}-1} = \sqrt{0} = 0, \quad \text{so} \quad \frac{0}{1} = 0, \] but \( 0 \) is not positive, so \( x=1 \) must be excluded. - For \( x < 0 \): Here, \( |x| \geq 1 \) implies \( x \leq -1 \). At \( x=-1 \), again \[ \sqrt{(-1)^{2}-1} = 0, \] so the quotient equals \( 0 \) which is not allowed. For \( x < -1 \), the numerator is positive while the denominator is negative, making the quotient negative. This does not satisfy the requirement of being positive. Thus, the only interval that works is: \[ x > 1. \] - **Domain of \( h(x) \):** \[ (1, \infty). \] --- **Final Answers:** - The domain of \( g \) is \( \{x \in \mathbb{R} : x \neq 2\} \) and \( g(-1)=\frac{1}{3} \). - The domain of \( h(x) \) is \( (1, \infty) \).