Here we need to find fog and gof, and they must be equal to x .
Let's check out fog
fog = f(g(x))
[tex] f(\frac{4x+1}{x}) [/tex]
Substituting the value of g(x) in f(x) for x, we will get
[tex] \frac{1}{\frac{4x+1}{x}-4} =\frac{1}{\frac{4x+1-4x}{x}}=x [/tex]
Domain
Here the input function is g(x), and the denominator should not be 0. So x should not be zero. Therefore, domain is
[tex] (-\infty,0)U(0, \infty) [/tex]
Now let's check gof
gof = g(f(x))
Here we need to insert f(x) in g(x) for x, and on doing that , we will get
[tex] \frac{4(\frac{1}{x-4})+1}{\frac{1}{x-4}} = \frac{4+x-4}{1}=x [/tex]
Domain
Here the input function is f(x), and denominator should not be zero.
SO domain is
[tex] (- \infty,4)U(4, \infty) [/tex]
Since fog = gof =x, so the given function are inverses of each other .