Question #N482
If the function $f(x) = \frac{x^2 - 9}{x - 3}$ is defined for all real numbers except $x = 3$, what is the value of $f(4)$?
A. -1
B. 3
C. 7
D. undefined
Correct Answer is: C
First, we can factor the numerator: $f(x) = \frac{(x+3)(x-3)}{x-3}$. Since $x \neq 3$, we can cancel out the common factor of $(x-3)$ in the numerator and denominator, leaving us with $f(x) = x + 3$. Now, substituting $x = 4$, we get $f(4) = 4 + 3 = 7$.