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AP Calculus AB 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

Let $f$ be a continuous function on the closed interval $[0, 4]$. If $f(0) = 5$ and $f(4) = 15$, which of the following values does the Intermediate Value Theorem guarantee for $f(c)$ for some $c$ in the interval $(0, 4)$?

All Questions (7)

Let $f$ be a continuous function on the closed interval $[0, 4]$. If $f(0) = 5$ and $f(4) = 15$, which of the following values does the Intermediate Value Theorem guarantee for $f(c)$ for some $c$ in the interval $(0, 4)$?

A) 0

B) 10

C) 20

D) 4

Correct Answer: B

The Intermediate Value Theorem states that for a continuous function on a closed interval $[a, b]$, the function must take on every value $d$ between $f(a)$ and $f(b)$. In this case, $f(a)=5$ and $f(b)=15$. The value 10 is between 5 and 15, so there must be a $c$ in $(0, 4)$ such that $f(c)=10$. The other values are outside this range.

The function $g$ is continuous on the closed interval $[-2, 6]$. Selected values of $g$ are given in the table below. | x | -2 | 0 | 3 | 6 | |------|----|---|----|---| | g(x) | 5 | 2 | -4 | 1 | On which of the following intervals is there a value $c$ such that $g(c)=0$?

A) On $[-2, 0]$ only

B) On $[0, 3]$ only

C) On $[3, 6]$ only

D) On both $[0, 3]$ and $[3, 6]$

Correct Answer: D

The Intermediate Value Theorem guarantees a root (a value $c$ where $g(c)=0$) exists on an interval $[a, b]$ if $g(a)$ and $g(b)$ have opposite signs, since 0 is between a positive and a negative number. On $[0, 3]$, the function values go from $g(0)=2$ to $g(3)=-4$, so a root is guaranteed. On $[3, 6]$, the function values go from $g(3)=-4$ to $g(6)=1$, so a root is also guaranteed. On $[-2, 0]$, both values are positive, so a root is not guaranteed by the IVT.

Let $f$ be a function such that $f(1) = -2$ and $f(5) = 3$. What additional condition is required to use the Intermediate Value Theorem to conclude that there must be a value $c$ in the interval $(1, 5)$ such that $f(c) = 0$?

A) $f$ is differentiable on $(1, 5)$

B) $f$ is defined for all $x$ in $[1, 5]$

C) $f$ is continuous on $[1, 5]$

D) $f$ is increasing on $[1, 5]$

Correct Answer: C

The primary condition for the Intermediate Value Theorem is that the function must be continuous on the closed interval. Differentiability is a stronger condition than continuity and is not required. Simply being defined is not sufficient, and being increasing is not a requirement of the theorem.

If a function $f$ is continuous on $[a, b]$ and $f(a) \\neq f(b)$, the Intermediate Value Theorem guarantees that for any value $d$ between $f(a)$ and $f(b)$, there is a number $c$ in $(a, b)$ for which $f(c)=d$. What is the key conclusion of this guarantee?

A) There is exactly one number $c$ such that $f(c)=d$.

B) There is at least one number $c$ such that $f(c)=d$.

C) The function must be monotonic on the interval $[a, b]$.

D) The derivative of the function must be zero at some point in $(a, b)$.

Correct Answer: B

The Intermediate Value Theorem is an existence theorem. It guarantees the existence of *at least one* value $c$, but it does not guarantee that this value is unique. There could be multiple values of $c$ for which $f(c)=d$.

Let $h$ be a continuous function on the closed interval $[-1, 3]$. If $h(-1) = -5$ and $h(3) = 7$, which of the following statements must be true?

A) $h(c) = 10$ for some $c$ in $(-1, 3)$.

B) $h(c) = 0$ for some $c$ in $(-1, 3)$.

C) $h(c) = -6$ for some $c$ in $(-1, 3)$.

D) $h(x)$ is always positive for some interval within $[-1, 3]$.

Correct Answer: B

According to the Intermediate Value Theorem, since $h$ is continuous on $[-1, 3]$, it must take on every value between $h(-1)=-5$ and $h(3)=7$. The value 0 is between -5 and 7, so the theorem guarantees its existence. The values 10 and -6 are outside this range and are not guaranteed.

The function $f(x) = \\frac{x^2 - 4}{x-2}$ has $f(1) = 3$ and $f(4) = 6$. A student tries to use the Intermediate Value Theorem to claim there must be a value $c$ in $(1, 4)$ such that $f(c) = 5$. Why is this an incorrect application of the theorem?

A) The value 5 is not between $f(1)$ and $f(4)$.

B) The function $f$ is not continuous on the interval $[1, 4]$.

C) The interval $[1, 4]$ is not a closed interval.

D) The theorem does not apply to rational functions.

Correct Answer: B

The Intermediate Value Theorem requires the function to be continuous on the entire closed interval. The function $f(x) = \\frac{x^2 - 4}{x-2}$ has a removable discontinuity at $x=2$, which is within the interval $[1, 4]$. Because the function is not continuous on the entire interval, the IVT cannot be applied.

A function $f$ is continuous on the closed interval $[a, b]$. If $d$ is a number between $f(a)$ and $f(b)$, the Intermediate Value Theorem is used to explain the behavior of the function. Which behavior is guaranteed by the theorem?

A) The function must cross the x-axis between $a$ and $b$.

B) The function must attain a maximum and minimum value on the interval $[a, b]$.

C) The function's graph is a single, unbroken curve that must pass through the horizontal line $y=d$.

D) The function's rate of change must be constant between $a$ and $b$.

Correct Answer: C

The IVT guarantees that a continuous function takes on all intermediate values. Geometrically, this means that the unbroken curve of the function's graph on $[a, b]$ must cross any horizontal line $y=d$ where $d$ is between the y-values of the endpoints, $f(a)$ and $f(b)$. Option A is only true if $f(a)$ and $f(b)$ have opposite signs. Option B is guaranteed by the Extreme Value Theorem, not the IVT.