Problem 8 - Brief Applied Calculus

Question

Find functions \(f\) and \(g\) such that the given function is the composition \(f(g(x))\). $$ \sqrt{\frac{x-1}{x+1}} $$

Step-by-Step Solution

Answer

The functions are \(g(x) = \frac{x-1}{x+1}\) and \(f(x) = \sqrt{x}\).

Step 1: Understand the Composition

We need to find two functions, \(f(x)\) and \(g(x)\), such that their composition results in the given function \(f(g(x)) = \sqrt{\frac{x-1}{x+1}}\). In a composition \(f(g(x))\), \(g(x)\) is applied first and then \(f(x)\) is applied to the result of \(g(x)\).

Step 2: Identify the Inner Function g(x)

For the composition \(f(g(x)) = \sqrt{\frac{x-1}{x+1}}\), we start by setting \(g(x)\) to be the innermost part of the expression under the square root. Here, set \(g(x) = \frac{x-1}{x+1}\).

Step 3: Identify the Outer Function f(x)

Given that \(g(x) = \frac{x-1}{x+1}\), the overall function \(f(g(x)) = \sqrt{\frac{x-1}{x+1}}\) suggests that \(f(x)\) should simply take the square root of its argument. Therefore, set \(f(x) = \sqrt{x}\).

Step 4: Compose the Functions to Verify

Verify that \(f(g(x)) = f\left(\frac{x-1}{x+1}\right)\) indeed simplifies to \(\sqrt{\frac{x-1}{x+1}}\). Substitute \(g(x)\) into \(f(x)\), resulting in \(\sqrt{\frac{x-1}{x+1}}\), confirming our solution.

Key Concepts

Function Decomposition

Decomposing a function involves breaking down a complex function into simpler components. This method is helpful for understanding the structure of composite functions, like the one given: \(\sqrt{\frac{x-1}{x+1}}\). The goal of function decomposition is to identify two functions, \(f\) and \(g\), such that when \(g(x)\) is applied first, and then \(f(x)\) is applied to the result, it yields the original function.

• In the context of our problem, think of \(g(x)\) as the inside function, responsible for the transformation inside the square root.
• It often simplifies the expression substantially.
• Then, \(f(x)\) is the outer function that operates on the result of \(g(x)\).

Identifying these components correctly can greatly simplify the process of working with complex expressions. Here, we set \(g(x) = \frac{x-1}{x+1}\) and \(f(x) = \sqrt{x}\), making
\(f(g(x)) = \sqrt{g(x)} = \sqrt{\frac{x-1}{x+1}}\). This decomposition perfectly reconstructs the original function.

Square Root Function

The square root function is one of the most common functions studied in elementary algebra. It maps a non-negative number to its principal square root. In mathematical notation, it is written as \(f(x) = \sqrt{x}\).

• The square root function is defined only for non-negative values because no real number squared yields a negative result.
• When working with compositions, it often acts as an outer function, taking the output of another function—like \(\frac{x-1}{x+1}\) in our case—and computing the square root.

This process transforms the input in a way that is easy to conceptualize: whatever is inside the square root simply becomes non-negative.
This makes interpreting compound functions more intuitive as in our expression \(\sqrt{\frac{x-1}{x+1}}\). It's an excellent practice to identify the range and possible values of \(x\) before applying the square root to avoid imaginary outputs.

Rational Expressions

Rational expressions are fractions where both the numerator and the denominator are polynomials. They can be represented as \(\frac{p(x)}{q(x)}\) where \(p(x)\) and \(q(x)\) are polynomials, and \(q(x) eq 0\).

• In our function, \(\frac{x-1}{x+1}\) is a classic example of a rational expression.
• The denominator must not be zero, as this would make the entire expression undefined.

Understanding the behavior of rational expressions is critical:

• They often flip sign depending on the input values, especially across the roots of the denominator.
• This affects the validity and the domain of the entire function composition.

In our problem, the rational expression determines much of the output behavior of the entire function: ensuring its value remains within limits desired by any subsequent functions, such as the square root. Recognizing these details can help manage compound functions effectively, avoiding undefined regions and complex results.