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