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Understanding the domain of a function is crucial as it tells you the values for which a function is defined. For any function, its domain will exclude values that result in mathematical operations, like division by zero, being undefined. Typically, if you have a function like \( f(x) = \frac{1}{x+3} \), it will be undefined when the denominator \( x+3 \) equals zero.
Solving \( x+3 = 0 \) gives \( x = -3 \), which means \( x = -3 \) must be excluded from the domain. Consequently, the domain for such a function includes all real numbers except \(-3\).
When considering the composed function \( f(g(x)) \), it's essential to check the domains of both individual functions and how they operate together. This combined effort ensures you catch all potential undefined values.

Rational functions are expressions that are ratios of two polynomials, for example, \( f(x) = \frac{1}{x+3} \). The core part of understanding rational functions is recognizing that they're generally smooth and defined except in cases where the denominator is zero.
These functions present a unique challenge because specific values that make the denominator zero lead to undefined behaviors, resulting in vertical asymptotes on a graph. For example, \( g(x) = \frac{2}{x-1} \) is undefined at \( x = 1 \).
When working with compositions of rational functions, such as \( f(g(x)) \), you substitute one function into another, and this substitution can introduce additional points of instability, further restricting the domain beyond each individual function.

Undefined values in functions occur predominantly when performing operations like division where the denominator is zero. It’s vital to identify these points to understand the function's behavior fully. For example, functions like \( f(x) = \frac{1}{x+3} \) and \( g(x) = \frac{2}{x-1} \) are undefined where their denominators become zero, at \( x = -3 \) and \( x = 1 \) respectively.
When you compose these functions like \( f(g(x)) \), the potential for undefined values increases as you need to consider instances where either function or their combination is undefined.

• First, ensure \( g(x) \) itself doesn’t render a value for \( f(x) \) undefined.
• Second, verify if the expression created by substituting \( g(x) \) into \( f(x) \), \( \frac{1}{\frac{2}{x-1}+3} \), stays defined.
  This was crucial when we found \( x = \frac{1}{3} \), another point where the function becomes undefined even though it results from the composition, not from the individual functions alone.

Therefore, understanding these undefined points is key to determining the entire domain of the composed function \( f(g(x)) \).