Answer:
x ∈ R, x ≠ 1
Step-by-step explanation:
f(g(x))
= f( [tex]\frac{4}{x-2}[/tex] )
= [tex]\frac{1}{\frac{4}{x-2}+4 }[/tex]
= [tex]\frac{1}{\frac{4+4(x-2)}{x-2} }[/tex]
= [tex]\frac{1}{\frac{4+4x-8}{x-2} }[/tex]
= [tex]\frac{x-2}{4x-4}[/tex]
The denominator of f(g(x)) cannot be zero as this would make f(g(x)) undefined. Equating the denominator to zero and solving gives the value that x cannot be, that is
4x - 4 = 0 ⇒ 4x = 4 ⇒ x = 1 ← excluded value
Domain : x ∈ R, x ≠ 1, that is
(- ∞, 1) ∪ (1, ∞ ) ← in interval notation
