Intermediate Value Theorem (College Board AP® Calculus AB)

Study Guide

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Written by: Roger B

Reviewed by: Dan Finlay

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 somewhere 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

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 $m^{'} (t) = 4$

Last updated: 5 August 2024

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