I had a lesson about operations on funcions. Everything was good until I reach the point of division of function so the lesson was saying that you can divide a function over another function but when it comes to determining the domain I find it weird i.e $$f(x)=\frac{x}{x+1},\qquad g(x)=\frac{x-3}{x+4}$$

The final function is $$\frac{x(x+4)}{(x-1)(x-3)}$$ so the domain becomes $R-[-4,3,1]$

Why do we exclude $-4$ and if we put $f(-4)=0$ not refused. can someone please explain this?

  • $\begingroup$ For some basic information about writing math at this site see e.g. here, here, here and here. $\endgroup$ – AlexR Nov 15 '15 at 16:58

Remember, we are defining our new function $h(x)$ as $f(x)/g(x)$

$h(x) = f(x)/g(x)$

$h(-4) = f(-4)/g(-4)$ where, $g(-4)$ is not defined

Hence, $h(-4)$ is also not defined

If we were to define $h(x)$ independently as $h(x) = x(x+4)/(x+1)(x-3)$, then what you say would be true.


I totally agree with Amit Saxena, I just wanted to add another example for better understanding.

If $f(x):=\frac{x^2}{x}$, then $dom(f) = \mathbb{R \setminus\{0\}}$ although $f(x)$ can be written as $x$ after simplification.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.