AP Calculus AB Exam Question

Consider the function f(x) defined as: [ f(x) = \begin{cases} \frac{x^2 - 4}{x - 2} & x < 2 \ \frac{x^2 - 4}{x + 2} & x > 2 \ \end{cases} ]

1. Determine whether the function f(x) is continuous at x = 2. Justify your answer.

Answer:

To determine the continuity of the function f(x) at x = 2, we need to check if three conditions are satisfied:

1. The function exists at x = 2 (the limit exists).
2. The left-hand limit and the right-hand limit are equal at x = 2.
3. The function value at x = 2 is the same as the limit.

Let's start by finding the left-hand limit (LHL) and right-hand limit (RHL) of the function as x approaches 2:

Left-hand Limit (LHL): To find the LHL of f(x) as x approaches 2, we evaluate the function when x is slightly less than 2: [ LHL = \lim_{{x \to 2^-}} \frac{{x^2 - 4}}{{x - 2}} ]

Substituting x = 2 - h, where h approaches 0, we have: [ LHL = \lim_{{h \to 0^-}} \frac{{(2 - h)^2 - 4}}{{2 - h - 2}} = \lim_{{h \to 0^-}} \frac{{4 - 4h + h^2 - 4}}{{-h}} = \lim_{{h \to 0^-}} (-4 + h) = -4 ]

Right-hand Limit (RHL): To find the RHL of f(x) as x approaches 2, we evaluate the function when x is slightly greater than 2: [ RHL = \lim_{{x \to 2^+}} \frac{{x^2 - 4}}{{x + 2}} ]

Substituting x = 2 + h, where h approaches 0, we have: [ RHL = \lim_{{h \to 0^+}} \frac{{(2 + h)^2 - 4}}{{2 + h + 2}} = \lim_{{h \to 0^+}} \frac{{4 + 4h + h^2 - 4}}{{h}} = \lim_{{h \to 0^+}} (4 + h) = 4 ]

Since the LHL (-4) is not equal to the RHL (4), the limit does not exist at x = 2.

Now, let's evaluate the function value at x = 2:

The function value is undefined at x = 2, as division by zero is not defined.

Therefore, as the limit does not exist and the function value is undefined at x = 2, the function f(x) is discontinuous at x = 2.

This discontinuity at x = 2 is a removable discontinuity since the function can be redefined to be continuous at x = 2 by removing the singularity.