AP Calculus AB Exam Question

Consider the function $f(x) = \frac{x^2 - 4}{x-2}$.

a) Determine the value of $f(2)$.

b) Determine if the function $f(x)$ is continuous at $x = 2$. Justify your answer.

c) Identify and classify any discontinuities, if they exist, of the function $f(x)$ over its domain.

Answer

a) To determine the value of $f(2)$, we substitute $x = 2$ into the function:

$f(2) = \frac{2^2 - 4}{2-2}$

Simplifying, we obtain:

$f(2) = \frac{4 - 4}{0}$

Since division by zero is undefined, we cannot evaluate $f(2)$ using direct substitution.

b) To determine if the function $f(x)$ is continuous at $x = 2$, we need to check if the limit of $f(x)$ as $x$ approaches 2 exists and is finite.

First, let's evaluate the left-hand limit as $x$ approaches 2:

$\lim_{{x \to 2^-}} f(x) = \lim_{{x \to 2^-}} \frac{x^2 - 4}{x-2}$

Using direct substitution, we find:

$\lim_{{x \to 2^-}} f(x) = \frac{2^2 - 4}{2-2} = \frac{0}{0}$

Again, we have an indeterminate form, so we need to simplify further.

Factoring the numerator, we get:

$\lim_{{x \to 2^-}} f(x) = \lim_{{x \to 2^-}} \frac{(x-2)(x+2)}{x-2}$

Simplifying, we find:

$\lim_{{x \to 2^-}} f(x) = \lim_{{x \to 2^-}} (x+2) = 2 + 2 = 4$

Now, let's evaluate the right-hand limit as $x$ approaches 2:

$\lim_{{x \to 2^+}} f(x) = \lim_{{x \to 2^+}} \frac{x^2 - 4}{x-2}$

Using direct substitution, we find:

$\lim_{{x \to 2^+}} f(x) = \frac{2^2 - 4}{2-2} = \frac{0}{0}$

Once again, we have an indeterminate form, so we need to simplify further.

Factoring the numerator again, we get:

$\lim_{{x \to 2^+}} f(x) = \lim_{{x \to 2^+}} \frac{(x-2)(x+2)}{x-2}$

Simplifying, we find:

$\lim_{{x \to 2^+}} f(x) = \lim_{{x \to 2^+}} (x+2) = 2 + 2 = 4$

Since both the left-hand and right-hand limits of $f(x)$ as $x$ approaches 2 are equal to 4, we can conclude that the limit of $f(x)$ as $x$ approaches 2 exists and is finite. Therefore, the function $f(x)$ is continuous at $x = 2$.

c) The function $f(x)$ does not have any discontinuities over its domain, which is all real numbers except $x = 2$. The limit of $f(x)$ as $x$ approaches any value other than 2 will exist and be finite, as shown in the previous steps.

Hence, the function $f(x)$ has no discontinuities.