Question

The domain of the function $f(x) = \begin{cases} \frac{x^2 - 9}{x - 3}, & \text{if } x \neq 3 \\ 6, & \text{if } x = 3 \end{cases}$ is:

• A. $(0, 3)$
• B. $(-\infty, 3)$
• C. $(-\infty, \infty)$
• D. $(3, \infty)$
• E. $(-3, 3)$

Hint:

A piecewise definition can remove discontinuities (holes). Here, the undefined point at \( x=3 \) is filled by defining \( f(3)=6 \).

Answer 1

The Correct Option is C

Concept: The domain of a function is the set of all values of \( x \) for which the function is defined. For piecewise functions, domain is the union of domains of all parts.

Step 1: Analyze the first part.
\[ f(x) = \frac{x^2 - 9}{x - 3}, \quad x \neq 3 \] Denominator becomes zero at \( x = 3 \), so this part is defined for: \[ (-\infty, 3) \cup (3, \infty) \]

Step 2: Analyze the second part.
\[ f(x) = 6 \quad \text{at } x = 3 \] So, \( x = 3 \) is included in the domain.

Step 3: Combine both parts.
\[ (-\infty, 3) \cup (3, \infty) \cup \{3\} = (-\infty, \infty) \]

Step 4: Final answer.
\[ \boxed{(-\infty, \infty)} \]