Respuesta :

Answer:

[tex](-\infty,0)\cup(0, \frac{1}{3})\cup (\frac{1}{3},\infty)[/tex]

Step-by-step explanation:

The given functions are [tex]f(x)=\frac{3x-1}{x-4}[/tex] and [tex]g(x)=\frac{x+1}{x}[/tex] .

We now composed the two functions to find:

[tex](f\circ g)(x)=f(g(x))[/tex]

[tex]\implies (f\circ g)(x)=f(\frac{x+1}{x})[/tex]

[tex]\implies (f\circ g)(x)=\frac{3(\frac{x+1}{x})+1}{\frac{x+1}{x}-4}[/tex]

[tex]\implies (f\circ g)(x)=\frac{4x+3}{1-3x}[/tex]

This function is defined if the denominator is not zero.

[tex]1-3x\ne0[/tex]

[tex]x\ne\frac{1}{3}[/tex]

We write this in interval notation as:

[tex](-\infty,\frac{1}{3})\cup (\frac{1}{3},\infty)[/tex]

We need to be cautious here as x=0 is not in the domain of g(x).

Therefore the domain of [tex](f\circ g)(x)[/tex] is

[tex](-\infty,0)\cup(0, \frac{1}{3})\cup (\frac{1}{3},\infty)[/tex]