How should I think about points of discontinuity of rational functions?

Question

I am re-taking pre-calculus as an adult and I'm finding myself confused by my textbook's section on rational functions. The textbook contains the following passage:

Now, my question is, why do we have to say $x ≠ 3$ after applying these algebraic manipulations? The function $f(x) = x - 2$ is defined for $x = 3$, and by the previous algebraic argument, $x - 2 = \frac{x^2 - 5x + 6}{x - 3}$, so shouldn't the function $f(x) = {x^2 - 5x + 6}{x - 3}$ be defined for $x = 3$ too since these are actually the same function?

Answer 1

To define a function, one has to define a value for each point in its domain. The textbook does not tell what is the domain of $f$, but talks about its "non permissible value". In fact $\frac {x^2-5x+6} {x-3}$ is not defined on $x=3$, because this is $0/0$ which is undefined. So the domain of definition of $f$ is $\mathbb{R} \setminus \{3\}$.

So when simplifying the rational fraction by $x-3$, you can then tell two stories:

• Define $g: \mathbb{R} \to \mathbb{R},\; g(x)=x-2$. Then $f$ and $g$ are different, because their domain of definition is not the same.
• Define $h: \mathbb{R} \setminus \{3\} \to \mathbb{R}, \;h(x)=x-2$. Then $f$ and $h$ are equal.

Note that you may encounter a subtlety that complicates things: when talking about rational fractions, $\frac {X^2-5X+6} {X-3} = X-2$. This is because here people do not talk any more about a function, but about a quotient of two polynomials, that is an algebraic object. It has its own life, so to say, independantly of functions. So simplification in this case is correct.

(And by the way, as commented by @AnotherUser, you should write the text in the picture into some real text; notably because that's better for impaired people).