Questions: The functions f and g are defined as follows. f(x) = x / (x^2 + 36) g(x) = (x + 9) / (x^2 - 81) For each function, find the domain. Write each answer as an interval or union of intervals. Domain of f : Domain of g :

Solution Steps

To find the domain of a function, we need to determine the values of \( x \) for which the function is defined. For rational functions, this means identifying the values of \( x \) that do not make the denominator zero.

1. Domain of \( f(x) \): The function \( f(x) = \frac{x}{x^2 + 36} \) is defined for all real numbers because the denominator \( x^2 + 36 \) is never zero (since \( x^2 + 36 \geq 36 \) for all real \( x \)).

2. Domain of \( g(x) \): The function \( g(x) = \frac{x+9}{x^2 - 81} \) is undefined where the denominator is zero. Solve \( x^2 - 81 = 0 \) to find these points.

The function \( f(x) = \frac{x}{x^2 + 36} \) has a denominator of \( x^2 + 36 \). Since \( x^2 + 36 \) is always positive for all real \( x \) (it is never zero), the domain of \( f \) is all real numbers. Thus, we can express the domain as: \[ \text{Domain of } f: \quad (-\infty, \infty) \]

The function \( g(x) = \frac{x + 9}{x^2 - 81} \) has a denominator of \( x^2 - 81 \). To find the values of \( x \) that make the denominator zero, we solve: \[ x^2 - 81 = 0 \implies x^2 = 81 \implies x = \pm 9 \] Thus, \( g(x) \) is undefined at \( x = -9 \) and \( x = 9 \). Therefore, the domain of \( g \) excludes these points, which can be expressed as: \[ \text{Domain of } g: \quad (-\infty, -9) \cup (-9, 9) \cup (9, \infty) \]

Final Answer