Given \( g \) and \( h \) as defined below, determine \( g+h, h-g, g h \), and \( \frac{h}{g} \). Using interval notation, repa the domain of each result. \( g(x)=\frac{-x-11}{x^{2}-100} \) and \( h(x)=\frac{-7 x-2}{x^{2}-100} \) (a) \( (g+h)(x)=\square \) Domain of \( (g+h)(x): \square \) (b) \( (h-g)(x)=\square \) Domain of \( (h-g)(x): \square \) (c) \( (g h)(x)=\square \) Domain of \( (g h)(x): \square \) (d) \( \left(\frac{h}{g}\right)(x)=\square \) Domain of \( \left(\frac{h}{g}\right)(x): \square \)

To solve the problem, we first need to analyze the functions \( g(x) \) and \( h(x) \): \[ g(x) = \frac{-x-11}{x^2-100} \] \[ h(x) = \frac{-7x-2}{x^2-100} \] ### Step 1: Determine the domain of \( g(x) \) and \( h(x) \) The denominators of both functions are \( x^2 - 100 \). We need to find where this expression is equal to zero: \[ x^2 - 100 = 0 \implies x^2 = 100 \implies x = 10 \text{ or } x = -10 \] Thus, the domain of both \( g(x) \) and \( h(x) \) is all real numbers except \( x = 10 \) and \( x = -10 \): \[ \text{Domain of } g(x) \text{ and } h(x): (-\infty, -10) \cup (-10, 10) \cup (10, \infty) \] ### (a) Calculate \( (g+h)(x) \) \[ (g+h)(x) = g(x) + h(x) = \frac{-x-11}{x^2-100} + \frac{-7x-2}{x^2-100} \] Combining the fractions: \[ (g+h)(x) = \frac{(-x-11) + (-7x-2)}{x^2-100} = \frac{-8x - 13}{x^2-100} \] **Domain of \( (g+h)(x) \)**: The domain remains the same as \( g(x) \) and \( h(x) \): \[ \text{Domain of } (g+h)(x): (-\infty, -10) \cup (-10, 10) \cup (10, \infty) \] ### (b) Calculate \( (h-g)(x) \) \[ (h-g)(x) = h(x) - g(x) = \frac{-7x-2}{x^2-100} - \frac{-x-11}{x^2-100} \] Combining the fractions: \[ (h-g)(x) = \frac{(-7x-2) + (x+11)}{x^2-100} = \frac{-6x + 9}{x^2-100} \] **Domain of \( (h-g)(x) \)**: The domain remains the same: \[ \text{Domain of } (h-g)(x): (-\infty, -10) \cup (-10, 10) \cup (10, \infty) \] ### (c) Calculate \( (gh)(x) \) \[ (gh)(x) = g(x) \cdot h(x) = \left(\frac{-x-11}{x^2-100}\right) \cdot \left(\frac{-7x-2}{x^2-100}\right) \] Multiplying the fractions: \[ (gh)(x) = \frac{(-x-11)(-7x-2)}{(x^2-100)^2} = \frac{(x+11)(7x+2)}{(x^2-100)^2} \] **Domain of \( (gh)(x) \)**: The domain remains the same: \[ \text{Domain of } (gh)(x): (-\infty, -10) \cup (-10, 10) \cup (10, \infty) \] ### (d) Calculate \( \left(\frac{h}{g}\right)(x) \) \[ \left(\frac{h}{g}\right)(x) = \frac{h(x)}{g(x)} = \frac{\frac{-7x-2}{x^2-100}}{\frac{-x-11}{x^2-100}} = \frac{-7x-2}{-x-11} \] This simplifies to: \[ \left(\frac{h}{g}\right)(x) = \frac{7x + 2}{x + 11} \] **Domain of \( \left(\frac{h}{g}\right)(x) \)**: The domain excludes the points where \( g(x) = 0 \) (which occurs when \( -x-11 = 0 \) or \( x = -11 \)) and where \( g(x) \) is undefined (i.e., \( x = 10 \)