Question

Let `f(x) = (1) / (1-3x)` and `g(x) = (1) / (x^2)` how do you find f(g(x)?

Answer 1

Put `g(x)` in place of `x` in the formula for `f(x)` and simplify to find:

`f(g(x)) = 1+3/(x^2-3)`

Explanation:

`f(g(x))`

`= 1/(1-3g(x))`

`= 1/(1-3(1/(x^2)))`

`= x^2/(x^2-3)`

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

`= 1+3/(x^2-3)`

with restriction `x != 0`

The restriction is necessary because `g(x)` is undefined for `x = 0`, but `1+1/(x^2-3)` is normally defined when `x = 0`.