Given: [tex]f(x) = \frac{1}{x-2}[/tex]
[tex]g(x) = \frac{2x+1}{x}[/tex]
A.)Consider
[tex]f(g(x))= f(\frac{2x+1}{x} )[/tex]
[tex]f(\frac{2x+1}{x} )=\frac{1}{(\frac{2x+1}{x})-2}[/tex]
[tex]f(\frac{2x+1}{x} )=\frac{1}{\frac{2x+1-2x}{x}}[/tex]
[tex]f(\frac{2x+1}{x} )=\frac{x}{1}[/tex]
[tex]f(\frac{2x+1}{x} )=1[/tex]
Also,
[tex]g(f(x))= g(\frac{1}{x-2} )[/tex]
[tex]g(\frac{1}{x-2} )= \frac{2(\frac{1}{x-2}) +1 }{\frac{1}{x-2}}[/tex]
[tex]g(\frac{1}{x-2} )= \frac{\frac{2+x-2}{x-2} }{\frac{1}{x-2}}[/tex]
[tex]g(\frac{1}{x-2} )= \frac{x }{1}[/tex]
[tex]g(\frac{1}{x-2} )= x[/tex]
Since, [tex]f(g(x))=g(f(x))=x[/tex]
Therefore, both functions are inverses of each other.
B.
For the Composition function [tex]f(g(x)) = f(\frac{2x+1}{x} )=x[/tex]
Since, the function [tex]f(g(x))[/tex] is not defined for [tex]x=0[/tex].
Therefore, the domain is [tex](-\infty,0)\cup(0,\infty)[/tex]
For the Composition function [tex]g(f(x)) =g(\frac{1}{x-2} )=x[/tex]
Since, the function [tex]g(f(x))[/tex] is not defined for [tex]x=2[/tex].
Therefore, the domain is [tex](-\infty,2)\cup(2,\infty)[/tex]
