Question

Need help with this question on composition of two rational functions. I have a screenshot of the question where i need help

Answer 1

using ( f ○ g )(x) = f(g(x))

substitute x =
`(5)/(x-1)` into f(x)

f(g(x)) = 5 / (
`(5)/(x-1)` + 2 ) ×
`(x-1)/(x-1)`

=
`(5(x-1))/(5+2(x-1))`

=
`(5x-5)/(5+2x-2)` =
`(5x-5)/(2x+3)`

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 = -
`(3)/(2)` ← excluded value

domain is x ∈ ( - ∞, -
`(3)/(2)`) ∪ ( -
`(3)/(2)`, + ∞ )