AP®MathsCollege BoardCalculus ABStudy GuidesUnit 1: Limits & ContinuityContinuityIntermediate Value Theorem

Intermediate Value Theorem (College Board AP® Calculus AB): Study Guide

Written by: Roger B

Reviewed by: Dan Finlay

Updated on 4 May 2026

Intermediate value theorem

What is the intermediate value theorem?

The intermediate value theorem says that:
- If $f$ is a continuous function on the closed interval $[a, b]$
- and if $d$ is a number between $f (a)$ and $f (b)$
- then there is at least one number $c$ between $a$ and $b$ such that $f (c) = d$
In practical terms this means that
- If a function continuous on an interval $[a, b]$ starts with value $f (a)$ and ends with value $f (b)$
- Then between $a$ and $b$ , the function takes on every value between $f (a)$ and $f (b)$
This seems really obvious if you think about the 'a function whose graph I can sketch without taking my pencil off the paper' way of describing continuity
- But it is an incredibly important result in mathematics
For example, consider the continuous function $f$ such that $f (1) = 5$ and $f (3) = 7$
- The IVT tells you that there is a value $c$ in the interval $1 < c < 3$ such that $f (c) = 6$
  - This means the equation $f (x) = 6$ has at least one solution

Graph of y = f(x) on [a, b]; point c between a and b has f(c) = d, marked by dashed vertical and horizontal lines intersecting the curve. — **An illustration of the intermediate value theorem**

What does the intermediate value theorem tell me about zeros of a function?

If the value of a continuous function is positive for one $x$ -value and negative for another $x$ -value, then the function has at least one zero between the two $x$ -values
- This follows from the intermediate value theorem by letting $d = 0$

What does the intermediate value theorem not tell me?

The IVT does not tell you where the function takes the value
- It tells you there is a value in the interval $a < x < b$
- But it does not tell you the actual value
The IVT does not tell you how many times the function takes the value
- It tells you there is at least one value that satisfies $f (x) = d$
- But there could be multiple values
The IVT does not tell you anything about the maximum or minimum values
- If the value $d$ is outside the interval between $f (a)$ and $f (b)$
- Then you cannot use the IVT to be certain that there is a value $a < x < b$ such that $f (x) = d$
For example, consider the continuous function $f$ such that $f (1) = 5$ and $f (3) = 7$
- The IVT does not help you find the actual value of a solution to $f (x) = 6$
- The IVT does not tell you the number of solutions to the equation $f (x) = 6$
- The IVT does not tell you whether there is a solution to $f (x) = 9$

Examiner Tips and Tricks

On the exam it's important to justify any use of the intermediate value theorem.

Often this means explaining how you know the function in question is continuous
Remember that if a function is differentiable, then it is continuous
- And if a function is twice-differentiable then both the function and its derivative are continuous

Worked Example

A social sciences researcher is using a function $m$ to model the total mass of all the garden gnomes appearing on lawns in a particular neighborhood at time $t$ . The function $m$ is twice-differentiable, with $m (t)$ measured in kilograms and $t$ measured in days.

The table below gives selected values of $m^{'} (t)$ , the rate of change of the mass, over the time interval $0 \leq t \leq 12$ .

$t$

(days)

$m^{'} (t)$

(kilograms per day)

2.6

4.8

12.2

0.7

-1.3

Is there a time $t$ , $0 \leq t \leq 3$ , for which $m^{'} (t) = 4$ ? Justify your answer.

Answer:

This is a job for the intermediate value theorem, but first you have to justify why $m^{'} (t)$ is continuous

$m$ is twice-differentiable, which means $m$ and $m^{'}$ are both continuous

$m (t)$ is twice-differentiable, which means $m^{'} (t)$ is differentiable, which means $m^{'} (t)$ is continuous

$m^{'} (0) = 2.6 < 4 < 4.8 = m^{'} (3)$

Therefore by the intermediate value theorem there is a time $t$ , $0 \leq t \leq 3$ , for which $m^{'} (t) = 4$

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