AP Calculus BC Practice Quiz: Removing Discontinuities

Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026

Test your understanding with short quizzes. This quiz has 9 questions to check your progress.

Question 1 of 9

A function f(x) has a discontinuity at x=c. Under which of the following conditions is it possible to remove the discontinuity?

A)The limit of f(x) as x approaches c exists.B)The function f(c) is undefined.C)The function approaches infinity at x=c.D)The left-hand limit and right-hand limit at x=c are not equal.

All Questions (9)

A function f(x) has a discontinuity at x=c. Under which of the following conditions is it possible to remove the discontinuity?

A) The limit of f(x) as x approaches c exists.

B) The function f(c) is undefined.

C) The function approaches infinity at x=c.

D) The left-hand limit and right-hand limit at x=c are not equal.

Correct Answer: A

According to the provided content, if the limit of a function exists at a discontinuity, it is possible to remove the discontinuity by redefining the function's value at that point to equal the limit. [cite: 1418]

If a function g(x) has a removable discontinuity at x=a and the limit of g(x) as x approaches a is L, how can the discontinuity be removed?

A) By defining g(a) = 0.

B) By ensuring the function is not piecewise.

C) By defining or redefining g(a) = L.

D) It is not possible to remove the discontinuity.

Correct Answer: C

The content states that a discontinuity can be removed by defining or redefining the value of the function at that point, so it equals the value of the limit of the function as x approaches that point. [cite: 1418]

Let f(x) be the function defined by f(x) = { kx + 3, if x ≤ 2; 2x - 1, if x > 2 }. For what value of the constant k is f(x) continuous at x=2?

A) k = -1

B) k = 0

C) k = 1

D) k = 2

Correct Answer: B

For a piecewise function to be continuous at a boundary, the value of the expression on one side must equal the value on the other side. We set kx + 3 = 2x - 1 at x=2. This gives k(2) + 3 = 2(2) - 1, which simplifies to 2k + 3 = 3. Solving for k gives 2k = 0, so k = 0. [cite: 1420, 1415]

The function f(x) = (x² - 9) / (x - 3) is discontinuous at x=3. To make the function continuous at x=3, how should f(3) be defined?

A) f(3) = 0

B) f(3) = 3

C) f(3) = 6

D) The discontinuity cannot be removed.

Correct Answer: C

To remove the discontinuity, we must define f(3) to be equal to the limit of f(x) as x approaches 3. The limit is found by simplifying the expression: lim (x→3) [(x-3)(x+3)]/(x-3) = lim (x→3) [x+3] = 3+3 = 6. Therefore, f(3) should be defined as 6. [cite: 1418]

Let g(x) be the function defined by g(x) = { x² + c, if x < 3; 4x - c, if x ≥ 3 }. For what value of the constant c is g(x) continuous at x=3?

A) c = 1

B) c = 3/2

C) c = 3

D) c = 9

Correct Answer: B

For the function to be continuous at the boundary x=3, the values of the two expressions must be equal at that point. We set x² + c = 4x - c at x=3. This gives (3)² + c = 4(3) - c, which simplifies to 9 + c = 12 - c. Solving for c gives 2c = 3, so c = 3/2. [cite: 1420, 1415]

Let f(x) be the function defined by f(x) = { (x² + x - 6) / (x - 2), if x ≠ 2; k, if x = 2 }. What value of k makes the function f(x) continuous at x=2?

A) k = 1

B) k = 2

C) k = 3

D) k = 5

Correct Answer: D

To make the function continuous at x=2, the value of k must be equal to the limit of the function as x approaches 2. We find the limit: lim (x→2) [(x²+x-6)/(x-2)] = lim (x→2) [(x+3)(x-2)/(x-2)] = lim (x→2) [x+3] = 2+3 = 5. Therefore, k must be 5. [cite: 1418, 1420]

A piecewise function f(x) is defined on two intervals, x < c and x ≥ c. If the limit of f(x) as x approaches c exists, but the function is discontinuous at x=c, which of the following must be true?

A) The value of the expression for x < c is not equal to the value of the expression for x ≥ c at the boundary.

B) The value of f(c) is not equal to the limit of f(x) as x approaches c.

C) The limit from the left does not equal the limit from the right.

D) The discontinuity is non-removable.

Correct Answer: B

The problem states the limit exists. For a piecewise function, this implies the left and right side expressions are equal at the boundary. If the function is still discontinuous, it must be because the defined value of the function at that point, f(c), is not equal to the value of the limit. This is the definition of a removable discontinuity. [cite: 1418, 1420]

Let h(x) be a function defined as h(x) = { x + 1, if x < 1; ax + b, if 1 ≤ x < 4; 7, if x ≥ 4 }. Find the value of 'a' that makes h(x) continuous for all real numbers.

A) a = 1

B) a = 5/3

C) a = 2

D) a = 7/3

Correct Answer: B

For the function to be continuous, the pieces must meet at the boundaries x=1 and x=4. At x=1, we must have x + 1 = ax + b, so 1 + 1 = a(1) + b, which gives 2 = a + b. At x=4, we must have ax + b = 7, so a(4) + b = 7, which gives 4a + b = 7. We solve the system of equations: a + b = 2 and 4a + b = 7. Subtracting the first from the second gives 3a = 5, so a = 5/3. [cite: 1420, 1414]

For a piecewise-defined function to be continuous at a boundary point, which condition is necessary?

A) The expressions defining the function on both sides of the boundary must be linear.

B) The boundary point must be included in the domain of only one of the expressions.

C) The value of the function at the boundary must be zero.

D) The value of the expression on one side of the boundary must equal the value of the expression on the other side of the boundary.

Correct Answer: D

This is a direct application of the principle for continuity in piecewise-defined functions. The value of the expression defining the function on one side of the boundary must equal the value of the expression defining the other side of the boundary, and also equal the function's value at the boundary itself. [cite: 1420]