2) Let $$ f(x)=\frac{2}{x-1} $$ and $$ g(x)=\frac{1}{x} $$ i) Find $$ f \circ g $$ and $$ g \circ f $$ i) The domain of $$ f, g, f \circ g $$ and $$ g \circ f $$.

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Alright, I need to find the compositions of the functions $$ f(x) = \frac{2}{x-1} $$ and $$ g(x) = \frac{1}{x} $$. Let's start by recalling what function composition means. The composition $$ f \circ g $$ means applying $$ g $$ first and then $$ f $$ to the result, and similarly for $$ g \circ f $$.

First, let's find $$ f \circ g $$. That would be $$ f(g(x)) $$. So, substitute $$ g(x) $$ into $$ f(x) $$:
$$
f(g(x)) = f\left(\frac{1}{x}\right) = \frac{2}{\frac{1}{x} - 1}
$$
Simplifying the denominator:
$$
\frac{1}{x} - 1 = \frac{1 - x}{x}
$$
So,
$$
f(g(x)) = \frac{2}{\frac{1 - x}{x}} = \frac{2x}{1 - x}
$$
Therefore, $$ f \circ g(x) = \frac{2x}{1 - x} $$.

Next, let's find $$ g \circ f $$. That would be $$ g(f(x)) $$. So, substitute $$ f(x) $$ into $$ g(x) $$:
$$
g(f(x)) = g\left(\frac{2}{x-1}\right) = \frac{1}{\frac{2}{x-1}} = \frac{x - 1}{2}
$$
So, $$ g \circ f(x) = \frac{x - 1}{2} $$.

Now, let's determine the domains of these functions. The domain of a function consists of all real numbers $$ x $$ for which the function is defined.

For $$ f(x) = \frac{2}{x-1} $$, the denominator cannot be zero, so $$ x - 1 \neq 0 $$, which means $$ x \neq 1 $$. Therefore, the domain of $$ f $$ is all real numbers except $$ x = 1 $$.

For $$ g(x) = \frac{1}{x} $$, the denominator cannot be zero, so $$ x \neq 0 $$. Therefore, the domain of $$ g $$ is all real numbers except $$ x = 0 $$.

For $$ f \circ g(x) = \frac{2x}{1 - x} $$, the denominator $$ 1 - x $$ cannot be zero, so $$ 1 - x \neq 0 $$, which means $$ x \neq 1 $$. Additionally, since $$ g(x) $$ is involved, $$ x $$ must also not be zero because $$ g(0) $$ is undefined. Therefore, the domain of $$ f \circ g $$ is all real numbers except $$ x = 0 $$ and $$ x = 1 $$.

For $$ g \circ f(x) = \frac{x - 1}{2} $$, there are no restrictions on $$ x $$ because the denominator is a constant. Therefore, the domain of $$ g \circ f $$ is all real numbers.

Wait a second, let me double-check that. For $$ g \circ f(x) $$, since $$ g(f(x)) = \frac{x - 1}{2} $$, and $$ f(x) = \frac{2}{x-1} $$, $$ f(x) $$ is defined for all $$ x \neq 1 $$. However, $$ g(f(x)) $$ would require $$ f(x) \neq 0 $$ because $$ g(0) $$ is undefined. So, $$ \frac{2}{x-1} \neq 0 $$, which implies $$ x - 1 \neq 0 $$, so $$ x \neq 1 $$. But earlier, I thought $$ g \circ f $$ has no restrictions, which seems contradictory. Let me clarify.

Actually, $$ g(f(x)) = \frac{x - 1}{2} $$ is a linear function and is defined for all real numbers except where $$ f(x) $$ is undefined. Since $$ f(x) = \frac{2}{x-1} $$ is undefined at $$ x = 1 $$, $$ g(f(x)) $$ is also undefined at $$ x = 1 $$. Therefore, the domain of $$ g \circ f $$ is all real numbers except $$ x = 1 $$.

So, summarizing:
- Domain of $$ f $$: All real numbers except $$ x = 1 $$.
- Domain of $$ g $$: All real numbers except $$ x = 0 $$.
- Domain of $$ f \circ g $$: All real numbers except $$ x = 0 $$ and $$ x = 1 $$.
- Domain of $$ g \circ f $$: All real numbers except $$ x = 1 $$.

But wait, earlier I thought $$ g \circ f $$ has no restrictions, but upon re-evaluating, it seems that $$ g \circ f $$ is undefined at $$ x = 1 $$ because $$ f(1) $$ is undefined. Therefore, the domain of $$ g \circ f $$ is all real numbers except $$ x = 1 $$.

i) - $$ f \circ g(x) = \frac{2x}{1 - x} $$ - $$ g \circ f(x) = \frac{x - 1}{2} $$ ii) - Domain of $$ f $$: All real numbers except $$ x = 1 $$. - Domain of $$ g $$: All real numbers except $$ x = 0 $$. - Domain of $$ f \circ g $$: All real numbers except $$ x = 0 $$ and $$ x = 1 $$. - Domain of $$ g \circ f $$: All real numbers except $$ x = 1 $$.

2) Let \( f(x)=\frac{2}{x-1} \) and \( g(x)=\frac{1}{x} \) i) Find \( f \circ g \) and \( g \circ f \) i) The domain of \( f, g, f \circ g \) and \( g \circ f \).

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