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 \( g \circ f \).

To find \( f \circ g \), also known as \( f(g(x)) \), we substitute \( g(x) \) into \( f(x) \): \[ f(g(x)) = f\left(\frac{1}{x}\right) = \frac{2}{\frac{1}{x} - 1} = \frac{2x}{1 - x} \] For \( g \circ f \), which is \( g(f(x)) \), we substitute \( f(x) \) into \( g(x) \): \[ g(f(x)) = g\left(\frac{2}{x-1}\right) = \frac{1}{\frac{2}{x-1}} = \frac{x-1}{2} \] Now, let's look at the domain of each function. The function \( f(x) \) has a domain of all real numbers except \( x=1 \) (to avoid division by zero), while \( g(x) \) has a domain of all real numbers except \( x=0 \). For \( f \circ g \), we need \( g(x) \) to be in the domain of \( f \), which excludes \( x=0 \) and \( x=1 \). So the domain of \( f \circ g \) is all real numbers except \( x=0 \) and \( 1 \). In the case of \( g \circ f \), we need \( f(x) \) to be in the domain of \( g \), which also excludes \( x=1 \). Hence, the domain of \( g \circ f \) is all real numbers except \( x=1 \). In summary: - Domain of \( f \): \( x \in \mathbb{R}, x \neq 1 \) - Domain of \( g \): \( x \in \mathbb{R}, x \neq 0 \) - Domain of \( f \circ g \): \( x \in \mathbb{R}, x \neq 0, 1 \) - Domain of \( g \circ f \): \( x \in \mathbb{R}, x \neq 1 \)