Given \( g \) and \( h \) as defined below, determine \( g+h, h-g, g h \), and \( \frac{h}{g} \). Using interval notation, report the domain of each result. \( g(x)=\frac{x+2}{x^{2}-36} \) and \( h(x)=\frac{-5 x+7}{x^{2}-36} \) (a) \( (g+h)(x)=\square \) Domain of \( (g+h)(x): \) (b) \( (h-g)(x)=\square \) Domain of \( (h-g)(x): \) (c) \( (g h)(x)=\square \) Domain of \( (g h)(x): \square \) (d) \( (x)=\square \) Domain of \( \left(\frac{h}{g}\right)(x): \)

To solve the problem, we first need to analyze the functions \( g(x) \) and \( h(x) \): \[ g(x) = \frac{x+2}{x^2 - 36} \] \[ h(x) = \frac{-5x + 7}{x^2 - 36} \] The denominator \( x^2 - 36 \) can be factored as \( (x - 6)(x + 6) \). This means that both functions \( g(x) \) and \( h(x) \) are undefined at \( x = 6 \) and \( x = -6 \). Therefore, the domain of both functions is: \[ \text{Domain of } g(x) \text{ and } h(x): (-\infty, -6) \cup (-6, 6) \cup (6, \infty) \] Now, we will compute \( g+h \), \( h-g \), \( gh \), and \( \frac{h}{g} \). ### (a) \( (g+h)(x) \) \[ (g+h)(x) = g(x) + h(x) = \frac{x+2}{x^2 - 36} + \frac{-5x + 7}{x^2 - 36} \] Combining the fractions: \[ (g+h)(x) = \frac{(x + 2) + (-5x + 7)}{x^2 - 36} = \frac{-4x + 9}{x^2 - 36} \] **Domain of \( (g+h)(x) \):** The domain is the same as the individual functions, excluding the points where the denominator is zero: \[ \text{Domain of } (g+h)(x): (-\infty, -6) \cup (-6, 6) \cup (6, \infty) \] ### (b) \( (h-g)(x) \) \[ (h-g)(x) = h(x) - g(x) = \frac{-5x + 7}{x^2 - 36} - \frac{x + 2}{x^2 - 36} \] Combining the fractions: \[ (h-g)(x) = \frac{(-5x + 7) - (x + 2)}{x^2 - 36} = \frac{-6x + 5}{x^2 - 36} \] **Domain of \( (h-g)(x) \):** The domain is the same as the individual functions: \[ \text{Domain of } (h-g)(x): (-\infty, -6) \cup (-6, 6) \cup (6, \infty) \] ### (c) \( (gh)(x) \) \[ (gh)(x) = g(x) \cdot h(x) = \left(\frac{x+2}{x^2 - 36}\right) \cdot \left(\frac{-5x + 7}{x^2 - 36}\right) \] Multiplying the fractions: \[ (gh)(x) = \frac{(x + 2)(-5x + 7)}{(x^2 - 36)^2} \] **Domain of \( (gh)(x) \):** The domain is the same as the individual functions: \[ \text{Domain of } (gh)(x): (-\infty, -6) \cup (-6, 6) \cup (6, \infty) \] ### (d) \( \left(\frac{h}{g}\right)(x) \) \[ \left(\frac{h}{g}\right)(x) = \frac{h(x)}{g(x)} = \frac{\frac{-5x + 7}{x^2 - 36}}{\frac{x + 2}{x^2 - 36}} = \frac{-5x + 7}{x + 2} \] **Domain of \( \left(\frac{h}{g}\right)(x) \):** The function is undefined where \( g(x) = 0 \) (i.e., where \( x + 2 = 0 \) or \( x = -2 \)) and where the original denominators are zero: \[ \text{Domain of } \left(\frac{h}{g}\right)(x): (-\infty, -6) \cup (-6, -2) \cup (-2, 6) \cup (6, \infty) \] ### Summary of Results (a) \( (g+h