Verified Answer :

  • The logarithmic function \(\log_{b}(f(x))\) requires \(f(x) > 0\).
  • Set up the inequality \(\frac{x+9}{x} > 0\).
  • Identify critical points at \(x = -9\) and \(x = 0\) to test the intervals \((-\infty, -9)\), \((-9, 0)\), and \((0, \infty)\).
  • Determine that the inequality holds for \((-\infty, -9) \cup (0, \infty)\).
  • The final domain is \(\boxed{(-\infty, -9) \cup (0, \infty)}\).

Explanation

  1. Problem Analysis

The function provided is \(h(x) = \log_{6}\left(\frac{x+9}{x}\right)\). For a logarithmic function \(\log_{b}(f(x))\) to be well-defined in the set of real numbers, the base \(b\) must be positive and not equal to 1, and the argument \(f(x)\) must be strictly positive. Here, the base is \(6\), which is valid. Therefore, the function \(h(x)\) is well-defined if and only if the argument \(\frac{x+9}{x} > 0\).

  1. Finding Critical Points and Intervals

To solve the inequality \(\frac{x+9}{x} > 0\), we first identify the critical points, which are where the numerator or denominator equals zero. The numerator \(x+9 = 0\) implies \(x = -9\), and the denominator \(x = 0\) implies \(x = 0\). These critical points divide the real number line into three intervals: \((-\infty, -9)\), \((-9, 0)\), and \((0, \infty)\). We will test each interval to see if the expression is positive.

  1. Evaluating Intervals

We examine the sign of \(f(x) = \frac{x+9}{x}\) on each interval:

  1. For \(x < -9\) (e.g., \(x = -10\)): \(f(-10) = \frac{-10+9}{-10} = \frac{-1}{-10} = \frac{1}{10} > 0\). Thus, the inequality holds for \(x \in (-\infty, -9)\).
  2. For \(-9 < x < 0\) (e.g., \(x = -1\)): \(f(-1) = \frac{-1+9}{-1} = \frac{8}{-1} = -8 < 0\). Thus, the inequality does not hold for \(x \in (-9, 0)\).
  3. For \(x > 0\) (e.g., \(x = 1\)): \(f(1) = \frac{1+9}{1} = \frac{10}{1} = 10 > 0\). Thus, the inequality holds for \(x \in (0, \infty)\).

Combining these results, the argument is positive for \(x \in (-\infty, -9) \cup (0, \infty)\).

  1. Conclusion and Final Answer

Based on our analysis, the function \(h(x) = \log_{6}\left(\frac{x+9}{x}\right)\) is defined when \(\frac{x+9}{x} > 0\). This occurs when \(x\) is in the union of the intervals \((-\infty, -9)\) and \((0, \infty)\). Therefore, the domain of the function is \(\boxed{(-\infty, -9) \cup (0, \infty)}\).

Examples

The concept of a domain is essential when calculating functions in real-world scenarios. Imagine you are designing a budget model for a project where the cost-to-benefit ratio is represented by a logarithmic function, such as \(\log(\frac{x+9}{x})\). Here, \(x\) represents your initial investment. The domain ensures that your investment \(x\) remains in a range where the ratio \(\frac{x+9}{x}\) is positive, keeping your model mathematically valid and preventing impossible results like taking the logarithm of a negative number. This is crucial for planning and financial forecasting, as it helps determine valid ranges for inputs to keep operations functioning correctly.

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