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Is the function $f(x) = \frac{x^2}{x}$ defined for every $x \in \mathbb R$, or only defined on $\mathbb R \setminus \{0 \}$?

Background:

Say we are given $P(x) = x^2 - 4x + 3$ and $Q(x) = (x - 1)(2x + 3) - (2x - 2)(3x + 5)$. The question asks to factorize both expressions and then consider the function: $F(x) = \frac{P(x)}{Q(x)}$. Then, the question asks to find the domain of definition of $F(x)$.

I teach my students that $F$ is defined for $Q(x) \neq 0$.

However, shouldn't one try to simplify $F(x)$ before one finds the domain of definition?

I can write for instance:

$$F(x) = \frac{P(x)(x - 10)}{Q(x)(x - 10)}$$

Should this writing change the domain of definition of $F$?

Then, I suppose that we should try to simplify $F$ before determining its domain of definition.

Can someone correct/validate my supposition?

Thank you.

2 Answers2

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The function $f(x)$ that we take to be the evaluation of the expression $x^2/x$ has a maximum domain $\mathbb{R}\setminus \{0\}$ in the reals.

It is clear to see that everywhere but at $x=0$ we have $f(x) = x$, and there is a hole left at $x=0$. In this circumstance, we can fill in the hole to achieve a continuous function by defining a new function: $$F(x) = \left\{ \begin{array}{cc} f(x) & x\neq 0\\ 0 & x=0 \end{array}\right.$$

This function agrees with $f$ over $f$'s domain, but it also fills in the gap left by $f$ in such a way that it is a continuous function.

Note that $F(x)$ is the only continuous function on $\mathbb{R}$ that agrees with $f$ on $\mathbb{R}\setminus \{ 0\}$. Thus it is often convention to write $f(x) = F(x)$, though this is an abuse of notation since they technically have different domains.

Joel
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No, definitely no. First you define a function, then you manipulate its formula. Of course you can sometimes extend a function across a removable singularity like $f(x)=\frac{x^2}{x}$ at $x=0$. However a function is defined by giving a formula and a domain of definition.

For a rational function you can clearly try to cancel common factors, but this will give you another, different function.

Siminore
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