Question

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

Answer 1

It depends.

Explanation:

Technically, `g circ f` is not a well-defined function. This is because `R_f = [–2,infty)` is not a subset of `D_g = RR` excluding `2`. In order for this function to be well defined, we restrict `D_f` to either `(2, infty)` or `(–infty, –2)`, so that `R_f = (2, infty) sube D_g`

Now that `g circ f` is defined, we start with your statement, that is, `g circ f = g(f(x))`.

Then `g(f(x))=g(x^2-2)=1/((x^2-2)-2)=1/(x^2-4)`.