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Function composition is a fundamental concept in mathematics where one function is applied to the results of another function. If you have two functions, say \(f(x)\) and \(g(x)\), composing them means creating a new function, such as \((f \circ g)(x)\) or \(f(g(x))\). For instance, given \(f(x) = \frac{1}{x-2}\) and \(g(x) = \frac{x+2}{x}\), to find \((f \circ g)(x)\), we substitute \(g(x)\) into \(f(x)\), which gives us \(f(g(x)) = \frac{1}{\frac{x+2}{x} - 2}\). It's critical to simplify the expression correctly to find the proper composed function.

The domain of a function refers to all the possible values that can be input into the function. Different functions have various restrictions. For example, rational functions like \(f(x) = \frac{1}{x-2}\) are undefined when the denominator equals zero, meaning \(x = 2\) is not in the domain of \(f(x)\). When dealing with composite functions, you need to consider the domains of each individual function and where they might overlap to ensure the overall function remains defined. For instance, with \((f \circ g)(x) = \frac{x}{2-x}\), the domain must exclude values that make either the inner function \(g(x)\) or the final expression undefined, which in this case are \(x = 0\) and \(x = 2\).

Rational functions are functions that can be expressed as the ratio of two polynomials. They're denoted as \( \frac{P(x)}{Q(x)} \), where both \(P(x)\) and \(Q(x)\) are polynomials. For example, \(f(x) = \frac{1}{x-2}\) is a rational function where the numerator is a constant polynomial (1), and the denominator is a linear polynomial \(x-2\). The domain of rational functions typically includes all real numbers except where the denominator is zero. Simplifying rational functions during composition, as we did when finding \((f \circ g)(x)\) and \((g \circ f)(x)\), requires careful algebraic manipulation to maintain the correct domains.

Precalculus problems often involve various functions and their compositions, aiming to prepare students for calculus. These problems might require finding composite functions, determining domains, and simplifying expressions. For example, in the given exercise, we found \((f \circ g)(x)\) and \((g \circ f)(x)\) and their respective domains. Precalculus bridges the gap to more advanced topics and solidifies foundational mathematical concepts. When solving these problems, it's beneficial to follow step-by-step solutions to understand the proper technique, such as ensuring the composed function is simplified correctly and identifying all restrictions for the domain.