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The functions $f$ and $g$ are defined as follows. $\newline$ \begin{align} &f(x) = \frac{x-2}{x^2 + 4x - 12},\ &g(x) = \frac{x}{x^2 + 64} \end{align} $\newline$ For each function, find the domain. $\newline$ Write each answer as an interval or union of intervals.

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Solve higher-degree inequalities

Full solution

Q. The functions

f

and

g

are defined as follows.

\newline

\begin{align*} &f(x) = \frac{x-2}{x^2 + 4x - 12},\ &g(x) = \frac{x}{x^2 + 64} \end{align*}

\newline

For each function, find the domain.

\newline

Write each answer as an interval or union of intervals.

Identify Domain of $f(x)$ : To find the domain of $f(x)$ , we need to determine the values of $x$ for which the function is defined. The function $f(x)$ has a denominator of $x^2 + 4x - 12$ , which cannot be equal to zero because division by zero is undefined. Therefore, we need to find the values of $x$ that make the denominator zero and exclude them from the domain.
Factor Denominator of $f(x)$ : We factor the denominator of $f(x)$ to find the values of $x$ that make it zero: $x^2 + 4x - 12 = (x + 6)(x - 2)$ . Setting each factor equal to zero gives us $x = -6$ and $x = 2$ .
Determine Domain Exclusions: The values $x = -6$ and $x = 2$ are the points where the function $f(x)$ is undefined. Therefore, the domain of $f(x)$ is all real numbers except $x = -6$ and $x = 2$ . In interval notation, this is written as $(-\infty, -6) \cup (-6, 2) \cup (2, \infty)$ .
Find Domain of $g(x)$ : Now, we find the domain of $g(x)$ . The function $g(x)$ has a denominator of $x^2 + 64$ , which is always positive since $x^2$ is always non-negative and $64$ is positive. Therefore, the denominator of $g(x)$ can never be zero, and $g(x)$ is defined for all real numbers.
Determine Domain of $g(x)$ : Since $g(x)$ is defined for all real numbers, the domain of $g(x)$ is $(-\infty, \infty)$ .

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