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AP Calculus AB Flashcards: Working with the Intermediate Value Theorem (IVT)

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

Review key ideas with interactive flashcards. This set includes 10 cards to help you master important concepts.

A function $f$ is continuous on $[0, 5]$, with $f(0) = -2$ and $f(5) = 4$. Can the IVT be used to show that $f(c) = 6$ for some $c$ in $(0, 5)$?
No, the IVT cannot be used for this case. The value $d=6$ is not between the function's endpoint values of $f(0)=-2$ and $f(5)=4$.
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A function $f$ is continuous on $[0, 5]$, with $f(0) = -2$ and $f(5) = 4$. Can the IVT be used to show that $f(c) = 6$ for some $c$ in $(0, 5)$?
No, the IVT cannot be used for this case. The value $d=6$ is not between the function's endpoint values of $f(0)=-2$ and $f(5)=4$.
What is the primary use of the Intermediate Value Theorem in analyzing a function?
The IVT is used to explain the behavior of a continuous function on an interval by guaranteeing it takes on every intermediate value between its starting and ending points.
What does the Intermediate Value Theorem guarantee about the number $c$?
The IVT guarantees the *existence* of at least one number $c$ between $a$ and $b$ where the function's value is $d$. It does not provide a method for finding the exact value of $c$.
If the IVT guarantees a value $c$ such that $f(c)=d$, where must the value $c$ be located?
The theorem guarantees that the number $c$ is located between $a$ and $b$, meaning it is within the open interval $(a, b)$.
A function $g$ has values $g(1)=-3$ and $g(4)=3$. Why can't we use the IVT to guarantee a root (where $g(c)=0$) exists between 1 and 4?
We cannot apply the IVT because we do not know if the function $g$ is continuous on the closed interval $[1, 4]$, which is a mandatory condition for the theorem.
What are the two essential conditions that must be met to apply the Intermediate Value Theorem?
First, the function $f$ must be continuous on the closed interval $[a, b]$. Second, the value $d$ must be a number between the function's endpoint values, $f(a)$ and $f(b)$.
Does the Intermediate Value Theorem guarantee that there is *only one* number $c$ for which $f(c)=d$?
No, the theorem does not guarantee uniqueness. It only guarantees that there is *at least one* such number $c$ on the interval.
What is the Intermediate Value Theorem (IVT)?
If a function $f$ is continuous on a closed interval $[a, b]$ and $d$ is a number between $f(a)$ and $f(b)$, then the IVT guarantees there is at least one number $c$ between $a$ and $b$ such that $f(c)=d$.
A function $f$ is continuous on $[2, 8]$, with $f(2) = 10$ and $f(8) = 20$. Does the IVT guarantee a value $c$ such that $f(c) = 15$?
Yes, it does. Since the function is continuous on the closed interval and $d=15$ is between $f(2)=10$ and $f(8)=20$, the IVT guarantees at least one $c$ between 2 and 8 exists.
In the statement of the IVT, what is the required property of the function $f$ on the interval $[a, b]$?
The function $f$ must be a continuous function on the closed interval $[a, b]$.