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Calculus / Calculus 1 · Limits

Intermediate Value Theorem

The intermediate value theorem says something that sounds obvious: a continuous function that starts below a value and ends above it must hit that value somewhere in between. It's the reason you can prove an equation has a solution without ever solving it.

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Chapter 2 · Lesson 11 of 13Calculus 1 · Limits
  1. 1.Introduction to Limits
  2. 2.Finding Limits From a Graph
  3. 3.One-Sided Limits
  4. 4.Limit Laws
  5. 5.Evaluating Limits
  6. 6.Squeeze Theorem
  7. 7.Infinite Limits
  8. 8.Limits at Infinity
  9. 9.Continuity
  10. 10.Types of Discontinuity
  11. 11.Intermediate Value Theorem
  12. 12.Epsilon-Delta Definition of a Limit
  13. 13.L'Hôpital's Rule
See all calculus chapters →

If you were 1 metre tall at age three and 1.6 metres at age fifteen, there was a moment when you were exactly 1.3 metres tall. You can’t grow from one height to another without passing through every height in between. The intermediate value theorem (IVT) is the precise version of that idea for functions, and it’s the first of several big theorems that depend on continuity.

Intermediate value theorem: if f is continuous on [a, b] and N is between f(a) and f(b), then f(c) = N for some c in (a, b)
The intermediate value theorem

What Is the Intermediate Value Theorem?

Suppose is continuous on the closed interval , and is any number between and . Then there is at least one number strictly between and with .

A continuous curve from the point (a, f(a)) to (b, f(b)). A dashed horizontal line at height N, between f(a) and f(b), crosses the curve at x = c
The curve has to cross the line y = N

On a graph: draw any horizontal line between the heights of the two endpoints. Because the curve is unbroken and runs from one side of the line to the other, it has to cross the line at least once. The theorem guarantees that a crossing exists; it doesn’t say where it is, and there may be more than one.

Why Continuity Is Essential

Two graphs with a dashed line at height N: a continuous line crosses it; a function with a jump goes from below the line to above it without ever touching it
A jump can leap over the value N

Take away continuity and the theorem fails. equals at and 1 at , yet it never equals 0. It jumps from to at , an infinite discontinuity, so it isn’t continuous on , and the theorem doesn’t apply. Whenever you use the IVT, say why the function is continuous on the whole closed interval.

Example 1: Showing an Equation Has a Root

Show that has a solution between 0 and 1.

Let . It’s a polynomial, so it’s continuous on . Now check the endpoints:

Since 0 lies between and 1, the intermediate value theorem guarantees some in with . That’s the whole proof: a sign change plus continuity. Notice that we never found the root, only proved it exists.

Example 2: An Equation You Can’t Solve by Algebra

Show that has a solution in .

No algebra will isolate here, so move everything to one side: , which is continuous everywhere. Then and . By the IVT, for some in between, so . (Numerically, .) This “move everything to one side, then look for a sign change” step is how most IVT problems start.

Example 3: Hitting a Value Other Than Zero

Show that takes the value 5 somewhere in .

is continuous, and , and 5 is between 4 and 9, so for some in . Of course, , which is between 2 and 3 as promised. The theorem works for any target value, not just zero.

Approximating a Root: the Bisection Method

The IVT doesn’t locate the root, but you can use it over and over to trap the root in smaller and smaller intervals. Halve the interval, check the sign at the midpoint, and keep the half where the sign still changes.

Bisection for x³ + x − 1: the interval [0, 1] shrinks to [0.5, 1], [0.5, 0.75], [0.625, 0.75] and [0.625, 0.6875], each with a sign change, closing in on the root near 0.6823
Each step halves the interval that contains the root

For : , so the root is in . , so it’s in . , so . , so . After a few more steps the root settles at about 0.6823. This is the bisection method, and it’s how many calculators and computer programs find roots.

The Idea Behind the Proof

Why is the theorem true? The full proof rests on a property of the real numbers called completeness, which says the number line has no gaps. Suppose . Consider every in where , and let be the smallest number that is greater than or equal to all of them. If were less than , continuity would keep below a little to the right of , contradicting how was chosen. If were greater than , continuity would keep above a little to the left of , again a contradiction. The only option left is .

What the Theorem Does Not Say

  • It doesn’t say where is, only that it exists.
  • It doesn’t say how many values of there are; there could be several.
  • The converse is false. A function can hit every value between and and still be discontinuous.
  • No sign change doesn’t mean no root. on is positive at both ends but has two roots in between. The IVT simply can’t detect them.

Common Mistakes to Avoid

  • Skipping the continuity check. Always state why is continuous on .
  • Using an interval that contains a discontinuity, like for .
  • Claiming you found the root. The IVT proves existence; finding the value is a separate step.
  • Forgetting to move everything to one side before looking for a sign change.

Practice Problems

Work each one before opening the answer.

Problem 1. Does have a root in ?

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is a continuous polynomial, and : a sign change.

Answer: yes, by the IVT

Problem 2. Can the IVT show that has a root in ?

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and : no sign change. (In fact there’s no real root at all.)

Answer: no

Problem 3. Show that has a solution in .

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is continuous, and .

Answer: a solution exists in ; it’s

Problem 4. Why doesn’t the IVT apply to on ?

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has an infinite discontinuity at , inside the interval.

Answer: is not continuous on

Problem 5. Use one bisection step to say which half of contains the root of .

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, , . The sign changes between 1.5 and 2.

Answer: (the root is )

Next, the epsilon-delta definition of a limit: the precise language that turns “gets close to” into mathematics.

Frequently Asked Questions

What is the intermediate value theorem?+

If is continuous on the closed interval and is any number between and , then there is at least one number in with . A continuous function can't skip any value between its endpoint values.

How do you use the intermediate value theorem to show a root exists?+

Check that the function is continuous on , then show and have opposite signs. Since 0 lies between them, the theorem guarantees some in with . For on , and , so there's a root between 0 and 1.

Why does the intermediate value theorem need continuity?+

A function with a jump can leap over a value. For example, is at and 1 at , but it never equals 0, because it isn't continuous on .

Does the intermediate value theorem tell you where the root is?+

No. It only guarantees that at least one exists. It doesn't say where, or how many there are. To locate the root, you can repeatedly halve the interval (the bisection method) or solve the equation directly.

How is the intermediate value theorem proved?+

The proof uses the completeness of the real numbers. Look at all the points in where and take to be the smallest number that is at least as big as all of them. Continuity then rules out both and , so .

Related lessons

ContinuityTypes of DiscontinuitySolving Equations and Inequalities: Every Lesson in OrderFinding the Zeros of a Polynomial