AP Calculus BC Practice Quiz: Working with the Intermediate Value Theorem (IVT)
Written by AP Content Team, Verified for 2026 AP Exams, Last updated: May 2026
Test your understanding with short quizzes. This quiz has 7 questions to check your progress.
Question 1 of 7
All Questions (7)
A) f(c) = 5
B) f(c) = 25
C) f(c) = 35
D) f(c) = 10
Correct Answer: B
The Intermediate Value Theorem guarantees that for any value d between f(a) and f(b), there is a c such that f(c) = d. Here, f(2) = 10 and f(8) = 30. The only option between 10 and 30 is 25.
A) The function f must be increasing on [a, b].
B) The function f must be differentiable on [a, b].
C) The function f must be continuous on [a, b].
D) The function f must have a positive value at f(a) or f(b).
Correct Answer: C
The provided text explicitly states that the Intermediate Value Theorem applies 'If f is a continuous function on the closed interval [a, b]'. Continuity is a required condition.
A) The function has a minimum value at x=a.
B) There is exactly one number c between a and b such that f(c) = 0.
C) The function is always increasing on the interval [a, b].
D) There is at least one number c between a and b such that f(c) = 0.
Correct Answer: D
Since f is continuous and 0 is a number between f(a) (a negative value) and f(b) (a positive value), the IVT guarantees there is at least one number c between a and b such that f(c) = 0. This explains the behavior of the function by guaranteeing it crosses the x-axis.
A) [-1, 5]
B) (-∞, ∞)
C) [12, -4]
D) [-4, 12]
Correct Answer: D
The Intermediate Value Theorem applies to any value d that is between f(a) and f(b). In this case, f(a) = -4 and f(b) = 12. The interval containing all numbers between -4 and 12 is [-4, 12].
A) There is exactly one such c.
B) There is at least one such c.
C) There are at most two such c values.
D) There are no such c values.
Correct Answer: B
The theorem states that 'there is at least one number c'. It does not specify an upper limit; there could be one, two, or more such values, but it only guarantees a minimum of one.
A) There exists a c in (0, 10) such that f(c) = 10.
B) There exists a c in (0, 10) such that f(c) = 0.
C) There exists a c in (0, 10) such that f(c) = 20.
D) The function's behavior can be partially explained by showing it must take on all values between -5 and 15.
Correct Answer: C
The Intermediate Value Theorem only guarantees the existence of f(c) = d for values of d *between* f(a) and f(b). Since f(0) = -5 and f(10) = 15, the theorem applies to any d in [-5, 15]. The value 20 is outside this range, so its existence is not guaranteed by the IVT.
A) [2, 3]
B) [3, 4]
C) [1, 2]
D) The theorem does not guarantee a value of 0 on any of these subintervals.
Correct Answer: C
To guarantee h(c) = 0, we need to find an interval [a, b] where 0 is between h(a) and h(b). On the interval [1, 2], the function values go from h(1) = 10 to h(2) = -2. Since 0 is between 10 and -2, the IVT guarantees a c in (1, 2) such that h(c) = 0. On [2, 3], values go from -2 to 1, so a root is also guaranteed there. However, [1, 2] is the first option that satisfies the condition.