What is the domain of a division of functions?

Question

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.

Answer 1

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 $0$.

Answer 2

If the domain of $f$ is $A$ and the domain of $g$ is $B$ then the domain of $f/g$ is $$A\cap B\setminus\{x:B|g(x)\ne 0\}.$$ (Of course, we must assume that $A\cap B\setminus\{x:B|g(x)\ne 0\}\ne \emptyset$. In other case $f/g$ doesn't make sense.)

In your example, $f(x)=2$ and $g(x)=1/x.$ We have that

$$\dfrac{f(x)}{g(x)}=2x, \forall x\in\mathbb{R}\setminus\{0\}.$$ Why? Note that $g(0)$ doesn't exist. So we can't consider

$$\dfrac{f(0)}{g(0)}.$$

Answer 3

It is all real numbers without zero i think. Check out this similar example: $$F(x)=\dfrac{(x-2)(x-1)}{(x-2)}$$ You can simplify this fraction to $F(x)=(x-1)$ ONLY when 2 is excluded from the domain So we say that $F$ is defined $$\forall x\in\mathbb{R}\setminus\{2\}.$$ Even though $F(x)=(x-1)$

Answer 4

Usually you define division of functions $f:A\to\mathbb R$ and $g:B\to\mathbb R$ as pointwise division, that is, $$(f/g)(x)=f(x)/g(x),$$ which obviously is only defined where $g(x)\ne 0$. Therefore the domain of the quotient is $D=\{x\in A\cap B\mid x\ne 0\}$.

However while this is the common choice, it is not the only possible choice. Another possible choice for continuous functions might be to define $f/g$ as $$(f/g)(x) = \lim_{y\to x} f(y)/g(y)$$ with the domain being all points in $\mathbb R$ where this limit exists.

With that definition and $f(x)=2$ (with domain $\mathbb R$) and $g(x)=1/x$ (with domain $\mathbb R\setminus\{0\}$), you would indeed find that $(f/g)(x) = 2x$ with domain $\mathbb R$.