Answer:[tex]f(g(x))=\frac{1+x}{(1-x)(2x+1)}[/tex]
The domain is the set of real numbers excluding 1 and -1/2
Explanation:
Given the functions:
[tex]\begin{gathered} f(x)=\frac{1-x}{x+1} \\ \\ g(x)=\frac{x}{1-x} \end{gathered}[/tex]
Then
[tex]\begin{gathered} f(g(x))=\frac{1-\frac{x}{1-x}}{\frac{x}{1+x}+1} \\ \\ =\frac{\frac{1-x-x}{1-x}}{\frac{x+1+x_{}}{1+x}}=\frac{\frac{1}{1-x}}{\frac{2x+1}{1+x}} \\ \\ =\frac{1}{1-x}\times\frac{1+x}{2x+1} \\ \\ =\frac{1+x}{(1-x)(2x+1)} \end{gathered}[/tex]
The domain is the set of real numbers excluding 1 and -1/2
