Unlocking The Domain Of Composite Functions A Step-by-Step Analysis

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In the realm of mathematics, understanding function domains is crucial for comprehending their behavior and applications. When dealing with composite functions, determining the domain becomes even more intricate. This article delves into the intricacies of finding the domain of a composite function, using the example of $(m \circ n)(x)$ where $m(x)=\frac{x+5}{x-1}$ and $n(x)=x-3$. We will explore the steps involved in identifying the domain and then compare it to the domains of other functions to find a match.

Deconstructing Composite Functions and Their Domains

To truly grasp the concept, let's first understand what a composite function is. A composite function, denoted as $(m \circ n)(x)$, is formed by applying one function to the result of another. In simpler terms, it means plugging the function $n(x)$ into the function $m(x)$. Mathematically, $(m \circ n)(x) = m(n(x))$.

Finding the domain of a composite function involves a two-step process. First, we need to determine the domain of the inner function, $n(x)$, because the input to the composite function must first be a valid input for $n(x)$. Second, we need to consider the domain of the outer function, $m(x)$, but not in its original form. Instead, we need to ensure that the output of $n(x)$, which becomes the input for $m(x)$, is within the domain of $m(x)$. This is where the complexities arise, as we are essentially dealing with a restricted input for the outer function.

Let's apply this understanding to our given functions, $m(x)=\frac{x+5}{x-1}$ and $n(x)=x-3$. The function $n(x) = x - 3$ is a linear function, and linear functions are defined for all real numbers. Therefore, the domain of $n(x)$ is $(-\infty, \infty)$. However, the function $m(x) = \frac{x+5}{x-1}$ is a rational function, which means it has a denominator. Rational functions are undefined when the denominator equals zero. In this case, the denominator is $x-1$, so $m(x)$ is undefined when $x-1 = 0$, which means $x = 1$. Thus, the domain of $m(x)$ is all real numbers except 1, which can be written as $(-\infty, 1) \cup (1, \infty)$.

Unraveling the Domain of $(m \circ n)(x)$

Now, let's find the composite function $(m \circ n)(x)$. We substitute $n(x)$ into $m(x)$: $(m \circ n)(x) = m(n(x)) = m(x-3) = \frac{(x-3)+5}{(x-3)-1} = \frac{x+2}{x-4}$. At first glance, it might seem that the domain of $(m \circ n)(x)$ is all real numbers except 4, since the denominator $x-4$ becomes zero when $x=4$. However, we must remember the crucial second step: we need to consider the domain of the inner function, $n(x)$, and ensure that the output of $n(x)$ does not violate the domain of $m(x)$. We know that the domain of $m(x)$ is all real numbers except 1. Therefore, we need to find the values of $x$ for which $n(x) = 1$, because if $n(x) = 1$, then $m(n(x)) = m(1)$, which is undefined. Setting $n(x) = 1$, we get $x-3 = 1$, which means $x = 4$. This confirms our initial observation that $x=4$ is not in the domain. However, we also need to consider if there are any other restrictions imposed by the original domain of $m(x)$. Since the domain of $n(x)$ is all real numbers, the only restriction comes from the fact that $n(x)$ cannot be equal to 1. Therefore, the domain of $(m \circ n)(x)$ is all real numbers except 4. In interval notation, this is $(-\infty, 4) \cup (4, \infty)$. This critical step ensures we're not just looking at the simplified composite function but also considering the original domain restrictions.

Identifying Functions with Matching Domains

Having determined the domain of $(m \circ n)(x)$, which is all real numbers except 4, we can now compare it to the domains of the given options:

  • A. $h(x)=\frac{x+5}{11}$: This is a rational function, but the denominator is a constant (11), which never equals zero. Therefore, the domain of $h(x)$ is all real numbers, or $(-\infty, \infty)$. This does not match the domain of $(m \circ n)(x)$.
  • B. $h(x)=\frac{11}{x-1}$: This is a rational function with a denominator of $x-1$. The denominator is zero when $x=1$. Therefore, the domain of $h(x)$ is all real numbers except 1, or $(-\infty, 1) \cup (1, \infty)$. This does not match the domain of $(m \circ n)(x)$.
  • C. $h(x)=\frac{11}{x-4}$: This is a rational function with a denominator of $x-4$. The denominator is zero when $x=4$. Therefore, the domain of $h(x)$ is all real numbers except 4, or $(-\infty, 4) \cup (4, \infty)$. This matches the domain of $(m \circ n)(x)$.

Therefore, the function that has the same domain as $(m \circ n)(x)$ is option C, $h(x)=\frac{11}{x-4}$.

Key Takeaways and Domain Determination Strategies

Mastering the art of finding the domain of composite functions is essential for success in mathematics, particularly in calculus and analysis. The process involves carefully considering the domains of both the inner and outer functions and any restrictions they impose on the composite function. Remember these key takeaways:

  • Always start with the inner function: Determine its domain first, as this sets the initial restrictions for the composite function.
  • Consider the outer function's domain: But apply it to the output of the inner function. This means you need to find the values that make the inner function's output fall outside the outer function's domain.
  • Look for potential restrictions: These often arise from denominators in rational functions, radicands in square roots (or other even-indexed roots), and arguments of logarithms.
  • Combine the restrictions: The final domain is the intersection of all the individual restrictions found.
  • Simplify the composite function: While this can help in some cases, always remember to go back and consider the original domains of the individual functions.

By meticulously following these steps, you can confidently navigate the complexities of composite function domains and avoid common pitfalls. This thorough approach ensures you're not just finding the simplified domain but the true domain, which reflects all the underlying restrictions.

In conclusion, understanding the domains of functions, especially composite functions, is a fundamental skill in mathematics. By carefully considering the restrictions imposed by each function involved and following a systematic approach, you can accurately determine the domain of any composite function. The example discussed in this article provides a solid foundation for tackling more complex problems in the future. Remember, practice makes perfect, so continue to explore various examples and solidify your understanding of this crucial concept.