Question: How would you determine the domain of g(f(x)) with g(x) = 3/x and f(x) = 6/(1-4x)?
Attempt: If you composite the function, the final answer would be (1-4x)/2. However, the domain would not include 1/4. Why is that? I learned that the domains of the original functions carry over to the composition, but why is 1/4 excluded and 0 is still included?
On
The composition $g(f(x))=3/f(x)=3/6/(1-4x)$$=1/2(1-4x)$. The domain of the function is intuitively all values of $x$ that can be plugged into the function that gives us a real number as the output. The only restriction for this function is $1/4$, since if we plug $1/4$ in for $x$, then we would be dividing by $0$, which means the function is undefined at $x=1/4$. Hence, the domain of $g(f(x))=\mathbb{R}\setminus \{1/4\}$ (all real numbers except $1/4$).
*Notice that zero is included in the domain, since if you plug in zero for x, we have $1/2(1-4(0))=1/2.$
To answer your specific question, $1/4$ is excluded because $f(x)$ is not defined there. Walk through the calculation of $g(f(x))$ step by step (in other words, first calculate $f(x)$, and then calculate $g(f(x))$, and you'll see the problem.
Your intuition is misleading you for a natural reason: there is a continuous extension of $g(f(x))$ that is defined at $x=1/4$. If you graph the composite function, you'll find that it's a nice straight line except for a "hole" at $x=1/4$. Your intuition wants to fill in that hole. When you learn about limits, you'll learn how to do that rigorously.
For now, the general answer to your question is that $\text{dom}(g(f(x)) = \{x~\vert~x\in \text{dom}(f)~ \land ~ f(x) \in \text{dom}(g)\}$. So in your specific question, $1/4 \notin \text{dom}(g(f(x))$ because $1/4 \notin \text{dom}(f)$.