Determining The Domain Of P(x) = Ln(√((x+1)/(x-1)))

Determining the domain of a function is a fundamental concept in mathematics. It involves identifying the set of all possible input values (x-values) for which the function produces a valid output. In this article, we embark on a journey to unravel the intricacies of the function $p(x) = \ln\left(\sqrt{\frac{x+1}{x-1}}\right)$ and meticulously determine its domain. This exploration will not only enhance your understanding of domain determination but also provide valuable insights into the behavior of logarithmic and radical functions.

Demystifying the Function: A Step-by-Step Approach

To effectively determine the domain of $p(x)$, we need to break it down into its constituent parts and analyze the restrictions imposed by each. The function comprises two key components: a natural logarithm and a square root. Each of these operations has specific requirements for its input, which will ultimately shape the overall domain of $p(x)$.

The Logarithmic Constraint: Unveiling the Positive Argument Requirement

The natural logarithm, denoted as $\ln(x)$, is defined only for positive arguments. In simpler terms, the input to the logarithm must be strictly greater than zero. This stems from the fundamental definition of logarithms as the inverse of exponential functions. Since exponential functions always produce positive outputs, their inverses (logarithms) are only defined for positive inputs. Applying this principle to our function, we encounter the first crucial restriction: $\sqrt{\frac{x+1}{x-1}} > 0$. This inequality forms the cornerstone of our domain determination, dictating that the expression inside the square root must be positive.

The Radical Constraint: Ensuring Non-Negativity and Avoiding Division by Zero

The square root function, symbolized as $\sqrt{x}$, introduces another layer of complexity. While it is defined for non-negative inputs (i.e., $x \geq 0$), our expression involves a fraction inside the square root. This introduces an additional constraint: the denominator of the fraction cannot be zero. Division by zero is an undefined operation in mathematics, and we must carefully exclude any x-values that would lead to this scenario. Therefore, we have two conditions to satisfy: $\frac{x+1}{x-1} \geq 0$ (to ensure a non-negative radicand) and $x-1 \neq 0$ (to avoid division by zero). These conditions, combined with the logarithmic constraint, will guide us towards the final domain of $p(x)$.

Untangling the Inequalities: A Journey into Critical Values and Interval Analysis

Now that we have identified the key constraints, the next step involves solving the inequalities to determine the specific x-values that satisfy them. To achieve this, we will employ a powerful technique called critical value analysis. This method involves identifying the values of x where the expressions in the inequalities change sign. These critical values act as dividing points on the number line, separating intervals where the expressions maintain a consistent sign.

Identifying Critical Values: The Signposts of Inequality Solutions

To find the critical values, we need to consider both the numerator and the denominator of the fraction inside the square root. The numerator, $x+1$, becomes zero when $x=-1$. This is our first critical value. The denominator, $x-1$, becomes zero when $x=1$. This is our second critical value. These two values, -1 and 1, are the signposts that will guide us through the interval analysis. They mark the points where the expression $\frac{x+1}{x-1}$ can potentially change from positive to negative or vice versa.

Constructing the Sign Table: A Visual Map of Inequality Solutions

With the critical values in hand, we can now construct a sign table. This table is a visual tool that helps us organize and analyze the sign of the expression $\frac{x+1}{x-1}$ in different intervals. We divide the number line into intervals based on the critical values: $(-\infty, -1)$, $(-1, 1)$, and $(1, \infty)$. Within each interval, we choose a test value and evaluate the expression. The sign of the result tells us the sign of the expression throughout the entire interval. Let's walk through the construction of the sign table step-by-step:

• Interval $(-\infty, -1)$: Choose a test value, say $x=-2$. Then $\frac{x+1}{x-1} = \frac{-2+1}{-2-1} = \frac{-1}{-3} = \frac{1}{3} > 0$. The expression is positive in this interval.
• Interval $(-1, 1)$: Choose a test value, say $x=0$. Then $\frac{x+1}{x-1} = \frac{0+1}{0-1} = \frac{1}{-1} = -1 < 0$. The expression is negative in this interval.
• Interval $(1, \infty)$: Choose a test value, say $x=2$. Then $\frac{x+1}{x-1} = \frac{2+1}{2-1} = \frac{3}{1} = 3 > 0$. The expression is positive in this interval.

The sign table visually summarizes these findings, clearly indicating the intervals where the expression is positive, negative, or zero. This table is instrumental in determining the solution to our inequalities and, ultimately, the domain of $p(x)$.

Weaving Together the Constraints: Defining the Domain

Having analyzed the sign of $\frac{x+1}{x-1}$, we can now revisit our original constraints. We need to satisfy both $\sqrt{\frac{x+1}{x-1}} > 0$ (from the logarithmic constraint) and $\frac{x+1}{x-1} \geq 0$ with $x-1 \neq 0$ (from the radical constraint). The logarithmic constraint requires the expression inside the square root to be strictly positive, meaning we need to consider only the intervals where $\frac{x+1}{x-1} > 0$. From our sign table, these intervals are $(-\infty, -1)$ and $(1, \infty)$. The radical constraint adds the condition that $\frac{x+1}{x-1}$ must be non-negative, which includes the intervals where the expression is positive and where it is equal to zero. However, we must also exclude $x=1$ because it makes the denominator zero.

The Final Verdict: Articulating the Domain in Interval Notation

Considering all the constraints, we arrive at the final domain of $p(x)$. The function is defined for all x-values less than -1 and greater than 1. In interval notation, we express this as $(-\infty, -1) \cup (1, \infty)$. This notation signifies the union of two intervals, representing all the x-values that satisfy the conditions for the function to be valid. The domain $(-\infty, -1) \cup (1, \infty)$ encapsulates the essence of the function's permissible inputs, providing a comprehensive understanding of its behavior.

Visualizing the Domain: A Graphical Representation

To further solidify our understanding, let's visualize the domain graphically. We can represent the domain on a number line, highlighting the intervals where the function is defined. A number line with open circles at -1 and 1, and shading extending to the left of -1 and to the right of 1, provides a clear visual representation of the domain $(-\infty, -1) \cup (1, \infty)$. This graphical representation complements the analytical solution, offering a holistic perspective on the function's domain.

Conclusion: A Journey Completed, Understanding Gained

In this comprehensive exploration, we have successfully navigated the intricacies of the function $p(x) = \ln\left(\sqrt{\frac{x+1}{x-1}}\right)$ and meticulously determined its domain. By carefully analyzing the constraints imposed by the logarithmic and radical components, employing critical value analysis and sign tables, and expressing the solution in interval notation, we have gained a deep understanding of the function's behavior. The domain $(-\infty, -1) \cup (1, \infty)$ represents the set of all permissible inputs for $p(x)$, providing a foundation for further analysis and applications of this fascinating function. This journey into domain determination not only enhances our mathematical skills but also provides valuable insights into the interconnectedness of different mathematical concepts.