Domain Of Composite Functions: M(x) And N(x)

Hey guys, let's dive into the cool world of composite functions and figure out which of our options shares the same domain as $(m n n n)(x)$. We've got our two functions: m(x)=rac{x+5}{x-1} and $n(x)=x-3$. Our mission, should we choose to accept it, is to find the domain of the function that results from plugging $n(x)$ into $m(x)$, and then see which of the answer choices has that exact same domain. It sounds a bit technical, but trust me, it's like solving a fun puzzle once you break it down. We're going to explore what makes a domain restricted and how that restriction carries over when we combine functions. So, grab your calculators, maybe a snack, and let's get this mathematical adventure started!

Understanding Composite Functions and Domains

Alright, let's kick things off by really getting a handle on what $(m n n n)(x)$ means. This notation, $(m n n n)(x)$, is just a fancy way of saying "apply function $n$ first, and then apply function $m$ to the result of $n(x)$." So, we're essentially creating a new function where every $x$ in $m(x)$ is replaced by the entire expression for $n(x)$. To find $(m n n n)(x)$, we substitute $n(x)$ into $m(x)$:

$$(m n n n)(x) = m(n(x)) = m(x-3)$$

Now, we take our $m(x)$ function, which is rac{x+5}{x-1}, and wherever we see an $x$, we pop in $(x-3)$:

(m n n n)(x) = rac{(x-3)+5}{(x-3)-1}

Simplifying the numerator and denominator, we get:

(m n n n)(x) = rac{x+2}{x-4}

Now, here's the crucial part: domain. The domain of a function is simply all the possible input values ($x$-values) for which the function is defined and produces a real number output. For rational functions (those with a fraction where the variable is in the denominator), the biggest restriction comes from the denominator. We can never, ever divide by zero. So, for our composite function (m n n n)(x) = rac{x+2}{x-4}, we need to make sure the denominator, $x-4$, is not equal to zero.

$$x-4 eq 0$$

$$x eq 4$$

This tells us that the composite function $(m n n n)(x)$ is defined for all real numbers except $x=4$. So, the domain of $(m n n n)(x)$ is all real numbers except 4. We can write this in interval notation as $(- n n, 4) n n (4, n n)$.

But wait, there's a little more to it! When we talk about the domain of a composite function, we also need to consider the domain of the inner function, which in this case is $n(x)$. The domain of $n(x) = x-3$ is all real numbers because there are no restrictions (no denominators to be zero, no square roots of negative numbers, etc.). So, the domain of $n(x)$ doesn't add any extra restrictions beyond what we found from the final composite form. Therefore, the domain of $(m n n n)(x)$ is indeed all real numbers except $x=4$. Our goal now is to find which of the given answer choices also has this exact same domain restriction.

Analyzing the Options for Domain Equivalence

Okay, team, we've successfully figured out the domain of our composite function $(m n n n)(x)$ is all real numbers except $x=4$. Now, the real detective work begins: we need to examine each answer choice and determine its domain. Remember, the domain is all about identifying values of $x$ that would cause a problem, like division by zero. Let's go through them one by one:

Option A: h(x) = rac{x+5}{11}

This function, guys, is pretty straightforward. It's a rational function, but the denominator is a constant, 11. Since 11 will never be zero, no matter what value $x$ takes, this function is defined for all real numbers. Its domain is $(- n n, n n)$. This is not the same as the domain of $(m n n n)(x)$, so we can cross this one off the list.

Option B: h(x) = rac{11}{x-1}

Here we have a denominator, $x-1$. To find the restriction, we set the denominator not equal to zero:

$$x-1 eq 0$$

$$x eq 1$$

So, the domain of this function is all real numbers except $x=1$. This is different from our target domain (all real numbers except $x=4$). Onwards!

Option C: h(x) = rac{11}{x-3}

Let's check this one. The denominator is $x-3$. We set it not equal to zero:

$$x-3 eq 0$$

$$x eq 3$$

The domain here is all real numbers except $x=3$. Still not matching our required domain of all real numbers except $x=4$. Keep looking!

Option D: h(x) = rac{11}{x-4}

And finally, we examine this function. The denominator is $x-4$. Let's find the restriction:

$$x-4 eq 0$$

$$x eq 4$$

Bingo! The domain of this function, h(x) = rac{11}{x-4}, is all real numbers except $x=4$. This is exactly the same domain we found for our composite function $(m n n n)(x)$. We've found our match, folks!

The Final Verdict: Matching Domains

So, after all that work, we've definitively found that the composite function $(m n n n)(x)$ simplifies to rac{x+2}{x-4}. The critical restriction for its domain comes from the denominator $x-4$, meaning $x$ cannot be equal to 4. This gives us a domain of all real numbers except 4.

We then meticulously analyzed each of the provided options: h(x)=rac{x+5}{11}, h(x)=rac{11}{x-1}, h(x)=rac{11}{x-3}, and h(x)=rac{11}{x-4}. By examining the denominators of each of these functions, we determined their respective domains. Option A allowed all real numbers. Option B restricted $x$ from being 1. Option C restricted $x$ from being 3.

Option D, h(x)=rac{11}{x-4}, has a denominator of $x-4$. Setting this denominator not equal to zero ($x-4 eq 0$) leads to the restriction $x eq 4$. This is precisely the same domain restriction we found for $(m n n n)(x)$.

Therefore, the function that has the same domain as $(m n n n)(x)$ is Option D: h(x)=rac{11}{x-4}.

It's super important to remember that when composing functions, you need to consider two things:

1. The domain of the inner function ($n(x)$ in this case).
2. The domain of the resulting composite function after substitution and simplification.

In this particular problem, the domain of $n(x)=x-3$ is all real numbers, so it didn't impose any additional restrictions. However, in other problems, it might! Always check both. It’s also a good reminder that even if the simplified form of a composite function looks simple, the domain restrictions can sometimes come from the intermediate steps. Keep practicing these, guys, and you'll be domain masters in no time! It's all about breaking down the problem, understanding the rules of the game (like 'no dividing by zero'), and carefully checking each piece. Great job following along!