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

Technically, $g \circ f$ is not a well-defined function. This is because R_f = [–2,infty) is not a subset of ${D}_{g} = \mathbb{R}$ excluding $2$. In order for this function to be well defined, we restrict ${D}_{f}$ to either $\left(2 , \infty\right)$ or (–infty, –2), so that ${R}_{f} = \left(2 , \infty\right) \subseteq {D}_{g}$
Now that $g \circ f$ is defined, we start with your statement, that is, $g \circ f = g \left(f \left(x\right)\right)$.
Then $g \left(f \left(x\right)\right) = g \left({x}^{2} - 2\right) = \frac{1}{\left({x}^{2} - 2\right) - 2} = \frac{1}{{x}^{2} - 4}$.