using ( f ○ g )(x) = f(g(x))
substitute x = [tex]\frac{5}{x-1}[/tex] into f(x)
f(g(x)) = 5 / ( [tex]\frac{5}{x-1}[/tex] + 2 ) × [tex]\frac{x-1}{x-1}[/tex]
= [tex]\frac{5(x-1)}{5+2(x-1)}[/tex]
= [tex]\frac{5x-5}{5+2x-2}[/tex] = [tex]\frac{5x-5}{2x+3}[/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
solve 2x + 3 = 0 ⇒ x = - [tex]\frac{3}{2}[/tex] ← excluded value
domain is x ∈ ( - ∞, - [tex]\frac{3}{2}[/tex]) ∪ ( - [tex]\frac{3}{2}[/tex], + ∞ )
