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The functions \( f \) and \( g \) are defined as follows. \[ f(x)=\frac{x+8}{x^{2}-64} \] \[ g(x)=\frac{x}{x^{2}+1} \] For each function, find the domain. Write each answer as an interval or union of intervals. \begin{tabular}{|l|l|l|l|}\hline Domain of \( f: \square \) & {\( [\square, \square) \)} & \( \square \) & \( \square \cup \square \) \\ \hline & \( (\square, \square) \) & {\( [\square, \square] \)} & \( (\square, \square] \) \\ \hline\end{tabular}

Ask by Reid Stephens. in the United States
Feb 04,2025

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The domain of \( f(x) \) is all real numbers except \( x = -8 \) and \( x = 8 \), which can be written as \( (-\infty, -8) \cup (-8, 8) \cup (8, \infty) \). The domain of \( g(x) \) is all real numbers, expressed as \( (-\infty, \infty) \).

Solución

To find the domain of the functions \( f(x) \) and \( g(x) \), we need to identify the values of \( x \) for which each function is defined. This typically involves finding values that would make the denominator zero, as these values are not included in the domain. ### Step 1: Find the domain of \( f(x) \) The function \( f(x) \) is given by: \[ f(x) = \frac{x + 8}{x^2 - 64} \] The denominator \( x^2 - 64 \) can be factored as: \[ x^2 - 64 = (x - 8)(x + 8) \] Setting the denominator equal to zero to find the restrictions: \[ x^2 - 64 = 0 \implies x - 8 = 0 \quad \text{or} \quad x + 8 = 0 \] This gives us: \[ x = 8 \quad \text{and} \quad x = -8 \] Thus, the function \( f(x) \) is undefined at \( x = 8 \) and \( x = -8 \). Therefore, the domain of \( f(x) \) is all real numbers except these two points: \[ \text{Domain of } f: (-\infty, -8) \cup (-8, 8) \cup (8, \infty) \] ### Step 2: Find the domain of \( g(x) \) The function \( g(x) \) is given by: \[ g(x) = \frac{x}{x^2 + 1} \] The denominator \( x^2 + 1 \) is always positive for all real numbers \( x \) because \( x^2 \) is non-negative and adding 1 ensures it is never zero. Therefore, there are no restrictions on \( x \) for this function. Thus, the domain of \( g(x) \) is all real numbers: \[ \text{Domain of } g: (-\infty, \infty) \] ### Final Answers Now we can fill in the table with the domains we found: \[ \begin{tabular}{|l|l|l|l|}\hline Domain of \( f: & (-\infty, -8) & (-8, 8) & (8, \infty) \\ \hline & & & \\ \hline Domain of \( g: & (-\infty, \infty) & & \\ \hline \end{tabular} \] So, the final answers for the domains are: - Domain of \( f: (-\infty, -8) \cup (-8, 8) \cup (8, \infty) \) - Domain of \( g: (-\infty, \infty) \)

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To find the domain of the functions \( f(x) \) and \( g(x) \), we need to identify any values of \( x \) that would make the functions undefined. For \( f(x) = \frac{x+8}{x^2-64} \): 1. The denominator \( x^2 - 64 \) must not be zero. Solving \( x^2 - 64 = 0 \) gives \( x^2 = 64 \) or \( x = \pm 8 \). 2. Thus, the domain excludes \( x = 8 \) and \( x = -8\). The domain of \( f \) in interval notation is: \[ (-\infty, -8) \cup (-8, 8) \cup (8, \infty) \] For \( g(x) = \frac{x}{x^2+1} \): 1. The denominator \( x^2 + 1 \) is always positive since \( x^2 \) is non-negative for all \( x \), and thus \( x^2 + 1 \) is never zero. So the domain of \( g \) is all real numbers: \[ (-\infty, \infty) \] Putting the answers in the specified format, we have: Domain of \( f: (-\infty, -8) \cup (-8, 8) \cup (8, \infty) \) & \( [\text{Interval: } -8, -8] \) & \( \infty \) & \( (-\infty, \infty) \) \\ \hline \( \) & \( (-\infty, -8) \) & \( [\infty] \) & \( \infty \) \\ \hline

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