Need help computing  $(f \circ g)(x)$ $$f(x) = \frac{1}{x^2} + 1$$
$$g(x) = \frac{1}{x - 1}$$
Do I just insert the $g(x)$ function into the $f(x)$ function giving me this terrible looking thing?:
$$\frac{1}{\frac{1}{x^2}+1-1}$$
Thanks for any help, I really just need confirmation I'm going in the right direction.

Actually, it would be the other way around. $(f \circ g)(x) = f(g(x))$, and $f(g(x)) = \frac{1}{g(x)^2}+1 = \frac{1}{\frac{1}{(x-1)^2}}+1$. This last expression is almost equal to $(x-1)^2+1$. The difference is that in the former, $ f \circ g$ is undefined at $x = 1$, while in the latter, it is. The correct function would be $(f \circ g )(x) = (x-1)^2 + 1, \forall x \neq 1$.
What you did is actually $(g\circ f)(x) = g(f(x)) = \frac{1}{f(x)-1} = \frac{1}{\frac{1}{x^2}+1-1} = \frac{1}{\frac{1}{x^2}}$. The same note about the function not existing at $x = 0$ applies; therefore, $(g \circ f)(x) = x^2, \forall x \neq 0$.
