This question is about real functions of real variables.
I think that, in general, if the domain of some function $f(x)$ is A, and the domain of another function $g(x)$ is B, then the domain of $(f/g)(x)$ is A$\cap$B and where $g\neq0$.
Now, what happens if I have something like $f(x)=2$, $g(x)=1/x$? In this case, $(f/g)(x)=2x$, which seems to be defined for all real numbers. But my statement above (which I think is correct in general) implies that $x=0$ is not allowed. So I'm conflicted.
Can somebody tell me what the domain of $(f/g)(x)$ is in this case? Is it all real numbers, or all real numbers except $0$?
Thanks.
Best Answer
Your mistake is in thinking that $$\frac2{1/x}\quad\hbox{and}\quad 2x$$ are always equal. They're not. To carefully prove that they are equal, we have $$\frac2{1/x}=\frac2{1/x}\,1=\frac2{1/x}\frac xx=\frac{2x}1=2x\ .$$ But this is not correct when $x=0$, because $\frac00$ is not equal to $1$. So we have to consider $x=0$ separately. In that case we have $2x=0$, but $$\frac2{1/x}=\frac2{1/0}=\frac2{\hbox{nonsense}}=\hbox{nonsense}\ .$$ So in your example, the domain of $f/g$ must exclude $0$.