Weber Romero
04/06/2024 · Elementary School
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 \( \left(\frac{h}{g}\right)(x): \)
UpStudy ThothAI Solution
Tutor-Verified Answer
Step-by-step Solution
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 original 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 original 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 original 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 functions are undefined:
\[
\text{Domain of } \left(\frac{h}{g}\right)(x): (-\infty, -6) \cup (-6, -2) \cup (-2, 6) \cup (6, \infty)
\]
### Summary of Results
(a) \( (g+h)(x) = \frac
Quick Answer
(a) \( (g+h)(x) = \frac{-4x + 9}{x^2 - 36} \)
Domain of \( (g+h)(x): (-\infty, -6) \cup (-6, 6) \cup (6, \infty) \)
(b) \( (h-g)(x) = \frac{-6x + 5}{x^2 - 36} \)
Domain of \( (h-g)(x): (-\infty, -6) \cup (-6, 6) \cup (6, \infty) \)
(c) \( (gh)(x) = \frac{(x+2)(-5x + 7)}{(x^2 - 36)^2} \)
Domain of \( (gh)(x): (-\infty, -6) \cup (-6, 6) \cup (6, \infty) \)
(d) \( \left(\frac{h}{g}\right)(x) = \frac{-5x + 7}{x + 2} \)
Domain of \( \left(\frac{h}{g}\right)(x): (-\infty, -6) \cup (-6, -2) \cup (-2, 6) \cup (6, \infty) \)
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