Explain why x = 3 is excluded from the domain of f(x) = (x² − 4x + 3)/(x² − 9), even though x − 3 cancels.

Step-by-step solution

1. 1

   Factor both parts before judging the domain.

   $x^2-4x+3=(x-1)(x-3),\qquad x^2-9=(x-3)(x+3)$

   so

   $f(x)=\frac{(x-1)(x-3)}{(x-3)(x+3)}$

   The factor $x-3$ appears on both sides — that is exactly what makes this case confusing.

2. 2

   The domain is decided by the original denominator, before any cancelling. Set it to zero:

   $x^2-9=0\quad\Longrightarrow\quad x=3 \text{ or } x=-3$

   Substituting $x=3$ into the untouched formula gives

   $f(3)=\frac{9-12+3}{9-9}=\frac{0}{0}$

   which is not a number. Division by zero has no meaning: there is no $y$ with $0\cdot y=$ a prescribed nonzero value, and when the numerator is also $0$ every $y$ would work, so no single value can be assigned.

3. 3

   Cancelling changes the formula but not the function. For every $x\neq 3$,

   $f(x)=\frac{x-1}{x+3}$

   but this simplified expression is a different function unless you keep the restriction $x\neq 3$ attached. Simplifying an expression never adds points to a domain — it only rewrites the rule on the domain you already had.

4. 4

   Classify the two excluded points — they are not the same kind. At $x=-3$ only the denominator vanishes, so $|f(x)|\to\infty$: that is a vertical asymptote. At $x=3$ numerator and denominator vanish together, so the graph has a removable discontinuity — a hole. Its height is the limit of the cancelled form:

   $\lim_{x\to 3}\frac{x-1}{x+3}=\frac{2}{6}=\frac13$

   Numerically, $f(3.00001)\approx 0.333334$, confirming the graph approaches $\tfrac13$ without ever reaching it.

5. 5

   State the domain.

   $\operatorname{dom}(f)=\{x\in\mathbb{R}: x\neq 3,\ x\neq -3\}=(-\infty,-3)\cup(-3,3)\cup(3,\infty)$

   The point $(3,\tfrac13)$ is an open circle on the graph: the curve gets arbitrarily close, but $x=3$ is not in the domain.