Let's analyze the given problem. The functions provided are:

• $f(x) = \frac{1 + x}{1 - x}$
• $g(x) = \frac{x}{1 - x}$

You are asked to find $f(g(x))$ (the composition of $f$ and $g$) and determine its domain.

Step 1: Finding $f(g(x))$

To find $f(g(x))$, we substitute $g(x)$ into $f(x)$. That is: $f(g(x)) = f\left( \frac{x}{1 - x} \right)$ We know $f(x) = \frac{1 + x}{1 - x}$, so: $f(g(x)) = \frac{1 + \frac{x}{1 - x}}{1 - \frac{x}{1 - x}}$ Now, simplify both the numerator and denominator.

Numerator:

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

Denominator:

$1 - \frac{x}{1 - x} = \frac{(1 - x) - x}{1 - x} = \frac{1 - 2x}{1 - x}$

So, we now have: $f(g(x)) = \frac{\frac{1}{1 - x}}{\frac{1 - 2x}{1 - x}} = \frac{1}{1 - 2x}$

Thus, $f(g(x)) = \frac{1}{1 - 2x}$.

Step 2: Finding the Domain of $f(g(x))$

The function $f(g(x)) = \frac{1}{1 - 2x}$ will be undefined when the denominator is zero, i.e., when $1 - 2x = 0$. Solving for $x$:

$1 - 2x = 0 \quad \Rightarrow \quad x = \frac{1}{2}$

Thus, the function $f(g(x))$ is defined for all real numbers except $x = \frac{1}{2}$.

Additionally, since the function $g(x) = \frac{x}{1 - x}$ has its own domain restriction (denominator $1 - x \neq 0$), we must also exclude $x = 1$ from the domain.

Final Answer:

• $f(g(x)) = \frac{1}{1 - 2x}$
• The domain of $f(g(x))$ is $x \in \mathbb{R} \setminus \left\{ 1, \frac{1}{2} \right\}$.

Would you like further details on any step?

Here are some related questions:

1. What is the composition of two arbitrary functions?
2. How do we find the domain of a composite function?
3. Why is division by zero undefined?
4. What are the conditions for a rational function to be defined?
5. How can we graph $f(g(x))$?

Tip: Always check the domain restrictions of each individual function before composing them, as these restrictions will affect the final result.