1.16 - Working with the Intermediate Value Theorem (IVT)

Introduction

The Intermediate Value Theorem (IVT) is a fundamental idea in calculus, and at its core it’s pretty intuitive. The hard part isn’t understanding the concept. It’s learning to properly justify using it by stating three required conditions and writing a formal conclusion.

Understanding the Concept

The Informal Idea

If a continuous function starts at one

y

-value and ends at a different

y

-value, it has to pass through every

y

-value in between.

1.16 - Working with the Intermediate Value Theorem (IVT)

In the graph above, the function is continuous from

x = a

x = b

. Since it starts at height

f(a)

and ends at height

f(b)

, it has to cross every horizontal line

y = k

between those values at least once.

Formal Statement of the IVT

If:

$f$ is continuous on the closed interval $[a, b]$

$f(a) \neq f(b)$ (the endpoints have different $y$ -values)

$k$ is any value between $f(a)$ and $f(b)$

Then:

There exists at least one value

c

in the interval

(a, b)

such that

f(c) = k

The Three Required Conditions

To use the IVT, you have to verify three conditions.

Condition 1: Continuity on the Interval

f \text{ is continuous on } [a, b]

The function has to have no breaks, jumps, or holes anywhere on the interval from

a

b

Condition 2: Different Endpoint Values

f(a) \neq f(b)

The function values at the endpoints have to be different. If they’re the same, the theorem doesn’t apply.

Condition 3: Target Value Between Endpoints

k \text{ is between } f(a) \text{ and } f(b)

The value you’re looking for has to lie between the two endpoint values. You can write this as:

f(a) < k < f(b) \quad \text{or} \quad f(b) < k < f(a)

depending on which endpoint is larger.

Writing the Conclusion

Once all three conditions check out, write the conclusion:

Therefore, by the Intermediate Value Theorem, there exists a value $c$ in $(a, b)$ such that $f(c) = k$ .

Shorthand notation:

\therefore

(three dots in a triangle) means “therefore.”

Example 1: Using IVT with a Polynomial

Problem: Show that

f(x) = x^2 - 2x + 5

takes on the value

20

somewhere on the interval

[-2, 6]

Condition 1: Check continuity

f(x) = x^2 - 2x + 5

is a polynomial, and polynomials are continuous everywhere. So

f

is continuous on

[-2, 6]

. ✓

Condition 2: Check that endpoint values differ

f(-2) = (-2)^2 - 2(-2) + 5 = 4 + 4 + 5 = 13

f(6) = (6)^2 - 2(6) + 5 = 36 - 12 + 5 = 29

Since

f(-2) = 13

and

f(6) = 29

, we know

f(-2) \neq f(6)

. ✓

Condition 3: Check that the target value is between the endpoints

We’re looking for

k = 20

. Is

20

between

13

and

29

13 < 20 < 29

Yes. ✓

Conclusion:

Therefore, by the Intermediate Value Theorem, there exists a value

c

(-2, 6)

such that

f(c) = 20

Example 2: Rational Function with a Restricted Domain

Problem: Determine whether

h(x) = \frac{1}{x - 2}

has to equal

\frac{1}{3}

somewhere on the interval

[4, 8]

Condition 1: Check continuity on $[4, 8]$

The function

h(x) = \frac{1}{x - 2}

has a discontinuity at

x = 2

(a vertical asymptote). But

x = 2

is not in the interval

[4, 8]

, so

h

is continuous on

[4, 8]

. ✓

Condition 2: Check endpoint values

h(4) = \frac{1}{4 - 2} = \frac{1}{2}

h(8) = \frac{1}{8 - 2} = \frac{1}{6}

h(4) \neq h(8)

. ✓

Condition 3: Check the target value

We want

k = \frac{1}{3}

. Is

\frac{1}{3}

between

\frac{1}{6}

and

\frac{1}{2}

\frac{1}{6} < \frac{1}{3} < \frac{1}{2}

Yes. ✓

Conclusion:

Therefore, by IVT, there exists a value

c

(4, 8)

such that

h(c) = \frac{1}{3}

Example 3: A Discontinuous Function Where IVT Does Not Apply

Problem: Does

g(x) = \frac{x^2 - 9}{x - 3}

necessarily equal

7

on the interval

[1, 5]

Condition 1: Check continuity

Factor it:

g(x) = \frac{(x-3)(x+3)}{x-3}

The factor

(x - 3)

cancels, but there’s still a hole at $x = 3$ . Since

3

is inside the interval

[1, 5]

, the function is not continuous on

[1, 5]

. ✗

Conclusion:

The IVT does not apply because Condition 1 fails. We can’t use IVT to conclude whether

g(x) = 7

has a solution on this interval.

Example 4: Using a Table of Values

Problem: The function

f

is continuous. Use the table to determine whether there has to be a value where

f(x) = 5

on the interval

[1, 4]

$x$	1	2	3	4
$f(x)$	3	7	8	9

Condition 1: Continuity

We’re told

f

is continuous. ✓

Condition 2: Endpoint values differ

f(1) = 3, \quad f(4) = 9 \quad \Rightarrow \quad f(1) \neq f(4)

✓

Condition 3: Target value between endpoints

We want

k = 5

. Is

3 < 5 < 9

? Yes. ✓

Conclusion:

Therefore, by IVT, there exists a value

c

(1, 4)

such that

f(c) = 5

Example 5: Counting Minimum Crossings

Problem: Given the table for the continuous function

g

, what is the minimum number of times

g(x) = 4

on the interval

[0, 10]

$x$	0	2	5	7	10
$g(x)$	2	6	1	5	3

We need to find the intervals where

g(x)

crosses

y = 4

Interval $[0, 2]$ :

g(0) = 2

and

g(2) = 6

. Since

2 < 4 < 6

, the function has to equal 4 at least once. ✓

Interval $[2, 5]$ :

g(2) = 6

and

g(5) = 1

. Since

1 < 4 < 6

, the function has to equal 4 at least once. ✓

Interval $[5, 7]$ :

g(5) = 1

and

g(7) = 5

. Since

1 < 4 < 5

, the function has to equal 4 at least once. ✓

Interval $[7, 10]$ :

g(7) = 5

and

g(10) = 3

. Since

3 < 4 < 5

, the function has to equal 4 at least once. ✓

Answer: The function has to equal 4 at least 4 times on

[0, 10]

Common Mistakes to Avoid

Mistake 1: Forgetting to check continuity

❌ Wrong: “Since

f(2) = 3

and

f(5) = 8

, and

6

is between them, there must be a solution.”

✓ Correct: First verify that

f

is continuous on

[2, 5]

Mistake 2: Checking the wrong interval

When you’re given a table with lots of values, make sure you only check the specified interval, not the entire table.

Mistake 3: Not checking whether $k$ is strictly between

k = f(a)

k = f(b)

, the value is already sitting at an endpoint. Technically the IVT still applies (the value exists at the endpoint), but it isn’t interesting.

Mistake 4: Forgetting to state all three conditions

On AP Calculus exams, you have to explicitly state all three conditions to get full credit.

Writing a Complete IVT Justification

Template for Full Credit

1. State continuity:

“

f

is continuous on

[a, b]

” with a brief justification if needed, like “because

f

is a polynomial.”

2. Calculate and compare endpoints:

“

f(a) = \underline{\hspace{1cm}}

f(b) = \underline{\hspace{1cm}}

, therefore

f(a) \neq f(b)

”

3. Check the target value:

“

k = \underline{\hspace{1cm}}

is between

f(a)

and

f(b)

”

You can write it as

f(a) < k < f(b)

f(b) < k < f(a)

4. Write the conclusion:

“Therefore, by the Intermediate Value Theorem, there exists a value

c

(a, b)

such that

f(c) = k

.”

Practice Problems

Problem 1

Show that

f(x) = x^3 - x - 1

has a zero on

[1, 2]

Answer:

$f$ is continuous on $[1, 2]$ since it’s a polynomial

$f(1) = -1$ and $f(2) = 5$ , so $f(1) \neq f(2)$

$k = 0$ is between $-1$ and $5$

Therefore, by IVT, there exists a value

c

(1, 2)

such that

f(c) = 0

Problem 2

Can IVT be used to show

g(x) = \frac{x}{x^2 - 4}

equals

1

[-3, 3]

Answer:

No.

g(x)

has discontinuities at

x = -2

and

x = 2

, and both are inside

[-3, 3]

. Condition 1 fails, so IVT does not apply.

Problem 3

Given the continuous function

h

$x$	0	3	6
$h(x)$	5	2	7

Must

h(x) = 4

somewhere on

[0, 6]

Answer:

Check

[0, 3]

h(0) = 5

and

h(3) = 2

. Is

2 < 4 < 5

? Yes.

So yes, by IVT there exists a value

c

(0, 3)

where

h(c) = 4

Key Takeaways

The IVT guarantees that continuous functions hit all intermediate values

Three conditions have to be verified: continuity, different endpoints, and target between endpoints

Always state all three conditions explicitly for full credit

The conclusion is always: “there exists a value $c$ in $(a, b)$ such that $f(c) = k$ ”

Discontinuities inside the interval mean IVT does not apply

With tables, check specific intervals, not just the endpoints of the entire table

IVT tells you a solution exists, but not where it is or how many there are

Use IVT to establish existence, not to find exact values

Polynomial, exponential, and trigonometric functions are continuous everywhere (for trig, on their domains)

Rational functions are continuous except where the denominator equals zero

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