What is the domain of the composite function (g@f)(x)?

1 Answer
May 10, 2015

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.