Question

How do you find `f@g` which is `f(g(x))` when `f(x)=x^2-2` and `g(x)=1/(x-2)`?

Answer 1

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

Explanation:

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

Hence: `f(g(x))= [g(x)]^2-2`

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

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