f(x)= x^2+1 g(x)=1/(x-3)
f(g(x))
so (1/(x-3))^2 +1
and from there i am lost on this problem.
thanks
Your working out is correct.
You also don't really need to go any further, unless you are asked to write it down in a different way...
$\displaystyle f(g(x))=\left(\frac{1}{x - 3}\right)^2 + 1$
$\displaystyle = \frac{1}{(x - 3)^2} + 1$
$\displaystyle = \frac{1}{x^2 - 6x + 9} + \frac{x^2 - 6x + 9}{x^2 - 6x + 9}$
$\displaystyle = \frac{x^2 - 6x + 10}{x^2 - 6x + 9}$.
This is about as far as you can go, but like I said, what you had so far is also correct.