Given f(x) = 4x/(x-1), find the composite function (f o f)(x) and its domain.

The composite function $(f \circ f)(x)$ means applying the function $f$ to $f(x)$.
So, $(f \circ f)(x) = f(f(x))$.
To calculate this, we substitute the entire expression for $f(x)$ into $f(x)$ wherever $x$ appears.

$$(f \circ f)(x) = f\left(\frac{4x}{x-1}\right)$$

Substitute $\frac{4x}{x-1}$ for $x$ in the definition of $f(x)$:

$$(f \circ f)(x) = \frac{4\left(\frac{4x}{x-1}\right)}{\left(\frac{4x}{x-1}\right)-1}$$

First, multiply the numerator:

$$\frac{4\left(\frac{4x}{x-1}\right)}{\left(\frac{4x}{x-1}\right)-1} = \frac{\frac{16x}{x-1}}{\frac{4x}{x-1}-1}$$

Next, simplify the denominator by finding a common denominator:

$$\frac{4x}{x-1}-1 = \frac{4x}{x-1} - \frac{x-1}{x-1} = \frac{4x - (x-1)}{x-1} = \frac{4x - x + 1}{x-1} = \frac{3x+1}{x-1}$$

Now, substitute this back into the complex fraction:

$$(f \circ f)(x) = \frac{\frac{16x}{x-1}}{\frac{3x+1}{x-1}}$$

To simplify a complex fraction, multiply the numerator by the reciprocal of the denominator:

$$(f \circ f)(x) = \frac{16x}{x-1} \cdot \frac{x-1}{3x+1}$$

Cancel out the common term $(x-1)$:

$$(f \circ f)(x) = \boxed{\frac{16x}{3x+1}}$$

For a composite function $(f \circ f)(x)$ to be defined, two conditions must be met:

1. The input $x$ must be in the domain of the inner function, $f(x)$. The function $f(x) = \frac{4x}{x-1}$ is undefined when its denominator is zero, so $x-1 \neq 0$, which means $x \neq 1$.

2. The output of the inner function, $f(x)$, must be in the domain of the outer function, $f$. This means $f(x)$ cannot take the value that makes $f(x)$ undefined, which is 1. So, $f(x) \neq 1$. Substitute the expression for $f(x)$:

   $$\frac{4x}{x-1} \neq 1$$

   Multiply both sides by $(x-1)$ (assuming $x \neq 1$):

   $$4x \neq 1(x-1)$$ $$4x \neq x-1$$

   Subtract $x$ from both sides:

   $$3x \neq -1$$

   Divide by 3:

   $$x \neq -\frac{1}{3}$$

Additionally, the denominator of the simplified composite function $(f \circ f)(x) = \frac{16x}{3x+1}$ cannot be zero.

$$3x+1 \neq 0$$ $$3x \neq -1$$ $$x \neq -\frac{1}{3}$$

This condition is already covered by the second point above.

Combining these conditions, the domain of $(f \circ f)(x)$ is all real numbers $x$ such that $x \neq 1$ and $x \neq -\frac{1}{3}$.