AP Calculus AB Exam Question

Let f(x) be the function defined as:

f(x) = (x^2 - 4x + 3)/(x - 3) for x ≠ 3 5 for x = 3

(a) Determine the points at which f(x) is not defined.

(b) Classify the discontinuities, if any, as removable, jump, or infinite.

(c) Determine if f(x) is continuous at x = 3, and justify your answer.

Answer and Explanation

(a) To find the points at which f(x) is not defined, we need to identify the values of x for which the denominator equals zero. In this case, the function is not defined when x - 3 = 0. Solving this equation, we find:

x - 3 = 0 x = 3

Therefore, f(x) is not defined at x = 3.

(b) We need to classify the discontinuities, if any, as removable, jump, or infinite.

For a removable discontinuity, the function can be made continuous by redefining the value at that particular point. A jump discontinuity occurs when there is a jump in the function's value at a specific point. An infinite discontinuity occurs when the function approaches positive or negative infinity at a certain point.

In this case, there is a hole at x = 3, which means the discontinuity is removable. This is because if we simplify the function at x = 3, we get:

f(3) = (3^2 - 4(3) + 3)/(3 - 3) f(3) = (9 - 12 + 3)/(0) f(3) = 0/0

The expression evaluates to 0/0, which is an indeterminate form. We can simplify the function by factoring the numerator:

f(3) = [(x - 1)(x - 3)]/(x - 3)

Now, we can cancel out the common factor:

f(3) = x - 1

By redefining the value of f(3) as x - 1, we can make the function continuous at x = 3. Therefore, the discontinuity at x = 3 is removable.

(c) To determine if f(x) is continuous at x = 3, we need to evaluate the left-hand limit (LHL) and right-hand limit (RHL) of the function at x = 3.

LHL: lim (x->3-) [ (x^2 - 4x + 3)/(x - 3) ] RHL: lim (x->3+) [ (x^2 - 4x + 3)/(x - 3) ]

To evaluate the limits, we can substitute x = 3 into the function:

LHL: lim (x->3-) [ (3^2 - 4(3) + 3)/(3 - 3) ] = lim (x->3-) [ 0/0 ] RHL: lim (x->3+) [ (3^2 - 4(3) + 3)/(3 - 3) ] = lim (x->3+) [ 0/0 ]

As we determined earlier, the discontinuity at x = 3 is removable. This means that the limits will have the same value after the removal of the discontinuity. Therefore:

LHL = RHL = f(3) = 3 - 1 = 2

Since the left-hand and right-hand limits are equal to the value of the function at x = 3, f(x) is continuous at x = 3.

In summary, we have determined that f(x) is not defined at x = 3, the discontinuity at x = 3 is removable, and f(x) is continuous at x = 3.