Final answer:
To extend the function f(x) = (x² - 9)/(x + 3) to be continuous at x = -3, we simplify it to f(x) = (x - 3) for x ≠ -3, and define f(-3) = -6 to fill in the gap, ensuring continuity at the point x equals -3.
Step-by-step explanation:
The question involves determining an extended function that is continuous at the point where x equals -3. The original function is f(x) = (x² - 9)/(x + 3). Notice that if we were to plug -3 into this function directly, we would get a division by zero, which is undefined. The function looks discontinuous at x = -3 because of this issue.
To resolve this, we factor the numerator: x² - 9 can be rewritten as (x - 3)(x + 3). This simplifies the function to f(x) = (x - 3) when x is not equal to -3. However, this does not define the function at x = -3. To make the function continuous at x = -3, we can define the value of the function at this point to be the limit of f(x) as x approaches -3. The limit of (x - 3) as x approaches -3 is -6. Therefore, we extend the function by defining f(-3) = -6. Now the function is continuous at x = -3 and is described by:
f(x) = {
(x - 3) if x ≠ -3
-6 if x = -3