#Intermediate Value Theorem

#Table of Contents

What is the Intermediate Value Theorem?
Practical Understanding
Exam Tip
Worked Example
Practice Questions
Glossary
Summary and Key Takeaways

#What is the Intermediate Value Theorem?

Key Concept

The Intermediate Value Theorem states that if $f$ is a continuous function on the closed interval $[a, b]$ and $d$ is any number between $f(a)$ and $f(b)$ , then there exists at least one number $c$ in the interval $(a, b)$ such that $f(c) = d$ .

#Mathematical Definition

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 exists at least one number $c$ in the interval $(a, b)$ such that $f(c) = d$

#Practical Understanding

In practical terms: - Imagine a function

f

is continuous on the interval

[a, b]

, starting at

f(a)

and ending at

f(b)

- Somewhere between

f(a)

and

f(b)

, the function must take on every value between

f(a)

and

f(b)

This can be visualized as sketching a graph of the function without lifting the pencil off the paper.

Note: Although this might seem intuitive, it is a fundamental result in mathematics.

#Exam Tip

Exam Tip

When using the Intermediate Value Theorem in exams, always justify why the function in question is continuous. Remember that:

If a function is differentiable, it is also continuous.
If a function is twice-differentiable, both the function and its first derivative are continuous.

#Worked Example

A social sciences researcher is using a function

m(t)

to model the total mass of all the garden gnomes appearing on lawns in a particular neighborhood over time. The function

m(t)

is twice-differentiable, with

t

measured in days.

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

t(days)	0	3	7	10	12
$m'(t)$ (kg/day)	2.6	4.8	12.2	0.7	-1.3

Question: Is there a time $t$ , $0 \le t \le 3$ , for which $m'(t) = 4$ ? Justify your answer.

Answer:

First, justify why $m'(t)$ is continuous:
- Since $m(t)$ is twice-differentiable, $m(t)$ and $m'(t)$ are both continuous.
Check the values at the endpoints:
- $m'(0) = 2.6$ and $m'(3) = 4.8$
Since $2.6 < 4 < 4.8$ , by the Intermediate Value Theorem, there exists a time $t$ in the interval $0 \le t \le 3$ such that $m'(t) = 4$ .

#Practice Questions

Practice Question

1. Given a continuous function $f$ on the interval $[1, 4]$ where $f(1) = 3$ and $f(4) = 7$ , is there a value $c$ in $(1, 4)$ for which $f(c) = 5$ ? Justify your answer.

Practice Question

2. A function $g$ is continuous on $[2, 5]$ and differentiable on $(2, 5)$ . If $g(2) = -1$ and $g(5) = 4$ , show that there is a number $c$ in $(2, 5)$ such that $g(c) = 0$ .

#Glossary

Continuous Function: A function without any breaks, jumps, or holes in its domain.
Differentiable: A function that has a derivative at every point in its domain.
Twice-Differentiable: A function that has a second derivative at every point in its domain.

#Summary and Key Takeaways

#Summary

The Intermediate Value Theorem (IVT) asserts that for any continuous function on a closed interval, any value between the function's values at the endpoints of the interval must be taken by the function at some point within the interval.

#Key Takeaways

The IVT is applicable only to continuous functions.
The theorem guarantees the existence of at least one $c$ in the interval where the function takes on a specified value.
Differentiability implies continuity, which is crucial when applying the IVT.

By understanding and applying the Intermediate Value Theorem, you can solve problems involving continuous functions and their values over specific intervals.

This comprehensive set of notes and practice questions should help solidify your understanding of the Intermediate Value Theorem.

Continuity

Emily Davis

5 min read

Next Topic - Average Rate of Change

Listen to this study note

Study Guide Overview

This study guide covers the Intermediate Value Theorem (IVT), including its formal definition, practical applications, and exam tips. It explains the concept of continuous functions and their relevance to the IVT. Worked examples and practice questions demonstrate how to apply the theorem, along with a glossary of key terms like differentiable and twice-differentiable.

#Intermediate Value Theorem

#What is the Intermediate Value Theorem?

Key Concept

#Mathematical Definition

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 exists at least one number $c$ in the interval $(a, b)$ such that $f(c) = d$

#Practical Understanding

In practical terms: - Imagine a function

f

is continuous on the interval

[a, b]

, starting at

f(a)

and ending at

f(b)

- Somewhere between

f(a)

and

f(b)

, the function must take on every value between

f(a)

and

f(b)

This can be visualized as sketching a graph of the function without lifting the pencil off the paper.

Note: Although this might seem intuitive, it is a fundamental result in mathematics.

#Exam Tip

Exam Tip

When using the Intermediate Value Theorem in exams, always justify why the function in question is continuous. Remember that:

If a function is differentiable, it is also continuous.
If a function is twice-differentiable, both the function and its first derivative are continuous.

#Worked Example

A social sciences researcher is using a function

m(t)

to model the total mass of all the garden gnomes appearing on lawns in a particular neighborhood over time. The function

m(t)

is twice-differentiable, with

t

measured in days.

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

t(days)	0	3	7	10	12
$m'(t)$ (kg/day)	2.6	4.8	12.2	0.7	-1.3

Question: Is there a time $t$ , $0 \le t \le 3$ , for which $m'(t) = 4$ ? Justify your answer.

Answer:

First, justify why $m'(t)$ is continuous:
- Since $m(t)$ is twice-differentiable, $m(t)$ and $m'(t)$ are both continuous.
Check the values at the endpoints:
- $m'(0) = 2.6$ and $m'(3) = 4.8$
Since $2.6 < 4 < 4.8$ , by the Intermediate Value Theorem, there exists a time $t$ in the interval $0 \le t \le 3$ such that $m'(t) = 4$ .

#Practice Questions

Practice Question

1. Given a continuous function $f$ on the interval $[1, 4]$ where $f(1) = 3$ and $f(4) = 7$ , is there a value $c$ in $(1, 4)$ for which $f(c) = 5$ ? Justify your answer.

Practice Question

2. A function $g$ is continuous on $[2, 5]$ and differentiable on $(2, 5)$ . If $g(2) = -1$ and $g(5) = 4$ , show that there is a number $c$ in $(2, 5)$ such that $g(c) = 0$ .

#Glossary

Continuous Function: A function without any breaks, jumps, or holes in its domain.
Differentiable: A function that has a derivative at every point in its domain.
Twice-Differentiable: A function that has a second derivative at every point in its domain.

#Summary and Key Takeaways

#Summary

#Key Takeaways

The IVT is applicable only to continuous functions.
The theorem guarantees the existence of at least one $c$ in the interval where the function takes on a specified value.
Differentiability implies continuity, which is crucial when applying the IVT.

By understanding and applying the Intermediate Value Theorem, you can solve problems involving continuous functions and their values over specific intervals.

This comprehensive set of notes and practice questions should help solidify your understanding of the Intermediate Value Theorem.

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Question 1 of 7

If a function $f$ is continuous on the closed interval $[a, b]$ , and $d$ is a value between $f(a)$ and $f(b)$ , what does the Intermediate Value Theorem guarantee? 🤔

There exists exactly one number $c$ in $(a, b)$ such that $f(c) = d$

There exists at least one number $c$ in $(a, b)$ such that $f(c) = d$

There exists a number $c$ in $[a, b]$ such that $f(c) = d$

The function $f$ is differentiable on $(a, b)$