Since the notation for the composition #(g\circ f)(x)=g(f(x))# means to first apply #f# to #x# to get #f(x)# and then to apply #g# to #f(x)# to get #g(f(x))#, a point #x# will be in the domain of #g\circ f# if and only if #x# is in the domain of #f# and #f(x)# is in the domain of #g#.
For example, let #f(x)=\frac{x+3}{x-4}# and #g(x)=\frac{x-2}{x+5}#. The number #x=4# is not in the domain of #f# and is therefore not in the domain of #g\circ f#. Is there another number that's not in the domain of #g\circ f#? Yes, any value(s) of #x# such that #f(x)=-5# is/are not in the domain of #g\circ f# since #-5# is not in the domain of #g#. In order for #f(x)=-5#, we must have #x+3=-5(x-4)=-5x+20# or #6x=17# or #x=17/6#. The domain of #g\circ f# is therefore #\{x\in R: x!= 4 \mbox{ and } x!=17/6\}#.
If you find a simplified formula for #g\circ f# in such an example, you can be misled in what the correct answer is. In the example above, we can write #(g\circ f)(x)=\frac{f(x)-2}{f(x)+5}=\frac{\frac{x+3}{x-4}-2}{\frac{x+3}{x-4}+5}=\frac{x+3-2(x-4)}{x+3+5(x-4)}=\frac{-x+11}{6x-17}#.
This simplified formula can mislead you into thinking that #17/6# is the only number not in the domain of #g\circ f#, but it's not. The number #4# is also not in the domain as we saw above.
Why does this happen? The reason is that the equality #\frac{\frac{x+3}{x-4}-2}{\frac{x+3}{x-4}+5}=\frac{x+3-2(x-4)}{x+3+5(x-4)}# is not true if #x=4#, because the expression on the left-hand side is undefined there. So, in doing the simplification above, we were implicitly assuming that #x!=4#.
