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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?

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  • $\begingroup$ 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
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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.

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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.

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