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Let $f(x)=x$ and $g(x)=\frac{1}{x}$, Domain$(f)=\mathbb{R}$ and Domain$(g)= \mathbb{R}-\{0\}$. We have to find the domain of $\frac{f(x)}{g(x)}$.

When we solve this expression, as the $x$ of $g(x)$ would go to the numerator, we would get the final term as $x^2$. As $x^2$ is defined for all $\mathbb{R}$, the domain should be $\mathbb{R}$ for $\frac{f(x)}{g(x)}$. Then why do we take the domain as $\mathbb{R}-\{0\}$ instead of $\mathbb{R}$?

N. F. Taussig
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user456
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    You only get $x^2$ after simplifying and when doing so, you'll use operations only allowed for $x \ne 0.$ E.g. the function $f(x) = x^3/x$ is defined only for non-zero $x$ and dividing by $x$, only allowed for $x \ne 0$, gives you $g(x)=x^2$ which is identical to $f(x)$ for all non-zero $x$. – StackTD Apr 04 '16 at 14:05
  • @StackTD Thanks! If $f(x)=\sqrt{x-1}$ and we have to find the domain of $f^2$, what would it be? R or [1,∞)? – user456 Apr 04 '16 at 14:12
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    If you have $(\sqrt{x})^2$, the inner square root is (still) only defined for $x \ge 0$, so... Notice the difference with $\sqrt{x^2}$. – StackTD Apr 04 '16 at 14:14

4 Answers4

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It's because you have to use $g(x)$ to actually get $\dfrac{f(x)}{g(x)}$. If you were just given the function $x^2$ from the start, then yes, the domain would be $\mathbb{R}$. But since you have to actually go through the process of dividing $x$ by $1/x$ to get $x^2$, then you must make sure that both $x$ and $1/x$ are defined. Well, $x$ is defined everywhere, but $1/x$ is not defined at $x=0$. Therefore $x=0$ is not in the domain.

Similar issues arise when solving certain types of equations (for example, logarithmic equations). Solutions must be checked and extraneous solutions must be discarded.

  • Nice explanation! If $f(x)=\sqrt{x-1}$ and we have to find the domain of $f^2$, what would it be? R or [1,∞)? – user456 Apr 04 '16 at 14:10
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    Thanks. I would say $[1,+\infty)$ fore more or less the same reason. A formal explanation would go as follows. In general, the domain of $f(x) \cdot g(x)$ is the domain of $f(x)$ intersected with the domain of $g(x)$. For that example, $f(x)$ and $g(x)$ are both $\sqrt{x-1}$ and the domain of that is $[1,+\infty)$. If you intersect $[1,+\infty)$ with itself you just get $[1,+\infty)$. –  Apr 04 '16 at 14:39
  • for, not fore. Words hard! –  Apr 04 '16 at 15:37
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Good question. There are times when you might want all of $\mathbb{R}$ for the domain, but sometimes not. The answer depends on the context - in particular, whether you may simplify algebraically first.

If you were writing a computer program to compute $f/g$ by finding values for the numerator and denominator and then dividing then $0$ would not be in the domain. This situation isn't just a software construction problem. There are abstract mathematical structures where the rules say "don't simplify".

Ethan Bolker
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$f/g=x^2$ only if x is not 0. Your algebra assumes this to be the case.

Paul
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since the domain of $$g(x)=\frac{1}{x}$$ are all real $x$ without zero,thus the quotient of $$f$$ and $$g$$ has the same domain

Dr. Sonnhard Graubner
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