We are given the functions $f(g(x)) = \frac{|x|}{2(x^2 - 1)}$ and $g(x) = 4x^2 + 1$, and the task is to find the domain of $f(x)$.

Step 1: Analyze $g(x)$

The function $g(x) = 4x^2 + 1$ is a quadratic function. Since $4x^2$ is non-negative for all real $x$, and adding 1 makes the function strictly positive, $g(x)$ is always positive. Therefore, the domain of $g(x)$ is all real numbers $\mathbb{R}$, as there are no restrictions.

Step 2: Analyze $f(g(x))$

The function $f(g(x)) = \frac{|x|}{2(x^2 - 1)}$ depends on two factors:

1. The denominator $2(x^2 - 1)$ cannot be zero because division by zero is undefined.
2. The absolute value $|x|$ in the numerator does not impose any restrictions on the domain.

For $2(x^2 - 1) = 0$, the denominator becomes zero when: $x^2 - 1 = 0 \quad \Rightarrow \quad x^2 = 1 \quad \Rightarrow \quad x = \pm 1$ Thus, $f(g(x))$ is undefined at $x = 1$ and $x = -1$.

Step 3: Determine the domain of $f(x)$

The function $f(g(x))$ is defined for all real numbers except $x = 1$ and $x = -1$, as these values make the denominator zero.

Therefore, the domain of $f(x)$ is: $\boxed{(-\infty, -1) \cup (-1, 1) \cup (1, \infty)}$

Would you like more details, or do you have any further questions?

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Here are 5 related questions to deepen understanding:

1. What is the range of the function $g(x) = 4x^2 + 1$?
2. How does the absolute value in $f(g(x))$ affect its behavior for negative values of $x$?
3. How would the domain change if $g(x) = x^2 - 4$ instead of $4x^2 + 1$?
4. Can you describe how to find the inverse of $f(x)$ if it exists?
5. How would the domain of $f(x)$ be affected if the denominator included a square root?

Tip: When working with rational functions, always check for points where the denominator becomes zero, as these are restrictions on the domain.