Question

Find the domain of \(f(x)=\ln \left(x^{2}-25\right)+3\)

Short answer

The domain is \(-\infty, -5)\cup(5, \infty)\.

Step-by-step solution

Understand the function

The given function is \(f(x) = \ln(x^2 - 25) + 3\). We need to find the domain of this function.

Identify restrictions for the logarithmic function

For the natural logarithm function \(\ln(y)\), the argument \(y\) must be positive \(y > 0\). Thus, \(x^2 - 25 > 0\).

Solve the inequality

Solve \(x^2 - 25 > 0\). This can be factored into \((x - 5)(x + 5) > 0\). The solution to this inequality involves finding intervals where this product is positive.

Determine critical points

Solve \(x^2 - 25 = 0\) to find the critical points: \(x = 5\) and \(x = -5\). These points divide the number line into three intervals: \(-\infty, -5\), \(-5, 5\), and \(5, \infty\).

Test intervals

Test a value from each interval to determine if the product \((x - 5)(x + 5)\) is positive in that interval. \(\text{For } x < -5, \ x = -6, (-6-5)(-6+5) = 11 > 0 \; (positive)\). \(\text{For } -5 < x < 5, \ x=0, (0-5)(0+5) = -25 (negative)\). \(\text{For } x > 5, \ x=6, (6-5)(6+5) = 11 (positive)\).

State the domain

The function is defined where \x^2 - 25 > 0\, which corresponds to the union of the intervals where the inequality holds true: \[x\in (-\infty, -5)\cup (5, \infty)\].

Key concepts

natural logarithm

The natural logarithm, denoted as \(\text{ln}\), is a logarithm to the base \(e\), where \(e\) is an irrational and transcendental number approximately equal to 2.71828. It is often used in various scientific and engineering fields because of its relationship with growth processes. If we see a function \(f(x) = \text{ln}(g(x))\), the \(g(x)\) inside the logarithm must always be positive for the function to be defined. So, for \(f(x) = \text{ln}(x^2 - 25) + 3\), we need to ensure that \(x^2 - 25 > 0\). We should remember that one of the key properties of logarithms is that the input, or argument, must be positive. Therefore, the first step in finding the domain always involves ensuring that the argument remains positive.

inequalities

Inequalities are mathematical expressions involving the symbols >, <, ≥, or ≤. They are essential when determining the domain of a logarithmic function. For our problem, the inequality \(x^2 - 25 > 0\) must be solved to find where the natural logarithm function is defined. Inequalities can seem tricky, but they’re just another type of equation that shows a range of possible solutions. By solving \(x^2 - 25 > 0\), we factor it into \((x - 5)(x + 5) > 0\). Here we are determining where this product is positive, leading us to split the inequality into different regions based on critical points.

critical points

Critical points are values of \(x\) where the function changes behavior. For the inequality \(x^2 - 25 > 0\), critical points occur where \(x^2 - 25 = 0\). Solving \(x^2 - 25 = 0\) gives us \(x = 5\) and \(-5\). These are the points where the expression inside the logarithm switches from positive to negative or vice versa. Critical points help us identify the intervals that we need to test for positivity. They also break the number line into regions, which we need to evaluate separately when searching for the domain of a function.

interval testing

Interval testing involves selecting test points within each interval defined by critical points to determine where a function or inequality holds true. For \((x - 5)(x + 5) > 0\), the critical points divide our number line into three intervals: \(-∞, -5\), \(-5, 5\), and \(5, ∞\). We select a test point from each of these intervals to determine if the inequality holds:

• For \(x < -5\), try \(x = -6\), resulting in \(( -6 - 5)( -6 + 5) = 11 > 0\), so it’s positive.
• For \(-5 < x < 5\), try \(x = 0\), resulting in \((0 - 5)(0 + 5) = -25\), so it’s negative.
• For \(x > 5\), try \(x = 6\), resulting in \((6 - 5)(6 + 5) = 11 > 0\), so it’s positive.

Based on interval testing, the initial inequality is satisfied for the intervals: \(-∞, -5) ∪ (5, ∞)\). Hence, the domain of \(f(x) = \text{ln}(x^2 - 25) + 3\) is \(-∞, -5) ∪ (5, ∞)\) because these are the regions where \(x^2 - 25 > 0\).