If f is continuous on [a, b] and N lies between f(a) and f(b), then f takes the value N somewhere between a and b.
Step 1: Let's Learn
Read it, or press Listen and follow the words.
The picture
To get from one height to another without lifting your pencil, you must pass through every height in between.
Finding roots
If f(a) is negative and f(b) is positive, the theorem guarantees a root between them. This is how a bisection search works.
It only promises existence
The theorem says a value is attained. It does not say where, nor how many times.
Continuity is not optional
A jump lets a function step over a value entirely. Without continuity the conclusion simply fails.
The interval must be closed
The endpoints have to be included, since their values are what the theorem compares.
A continuous graph cannot skip a value
If f is continuous on [a, b] and N lies between f(a) and f(b), then f takes the value N somewhere in between. It cannot get from one to the other without passing through everything between.
Its main use is locating roots
If f(1) is negative and f(2) positive, a root lies between. That is how numerical root-finding starts, and bisection — repeatedly halving the interval — is the theorem applied over and over.
It proves existence, not location
The theorem guarantees a value exists somewhere in the interval and says nothing about where. Existence theorems of this kind are common in analysis and are often all that is needed.
Continuity is essential
A function with a jump can go from negative to positive without ever equalling zero. Removing the continuity hypothesis makes the theorem false, which is a good demonstration that hypotheses are not decoration.
Step 2: Try It Yourself
Tap and try it out.
Step 3: Watch an Example
One step at a time.
Watch Kofi Guarantee a Root
Kofi must show that x³ − x − 1 = 0 has a solution between 1 and 2.
- Step 1
The function is a polynomial, so it is continuous everywhere, and the theorem applies.
Step 4: Your Turn
Practice makes it stick.
The Endpoints
Problem 1 of 2
f(x) = x³ − x − 1. What is f(1)?
The Other End
Problem 2 of 2
Same function. What is f(2)?
Guarantee It
1 of 8
f(x) = x² − 2. What is f(1)?
2 of 8
Same function. What is f(2)?
3 of 8
Does the theorem guarantee a root between 1 and 2 there? 1 for yes, 0 for no.
4 of 8
f(a) = 3 and f(b) = 9, both positive. Does the theorem guarantee a root? 1 or 0.
5 of 8
f(a) = −4 and f(b) = 4. Does it guarantee a root? 1 or 0.
6 of 8
Does the theorem tell you how many roots there are? 1 for yes, 0 for no.
7 of 8
Put the argument in order.
- 1Evaluate the function at both endpoints.
- 2Check that the target value lies between them.
- 3Conclude the value is attained somewhere inside.
- 4Confirm the function is continuous on the closed interval.
8 of 8
f(x) = x³. What is f(−2)?
Step 5: Quick Check
Show what you know.
Question 1 of 2
f(1) = −3 and f(4) = 6. Does the theorem guarantee a root between them? 1 or 0.
Question 2 of 2
Why does the theorem require continuity?
What You Learned
- A continuous function on a closed interval attains every value between its endpoint values.
- A sign change guarantees a root.
- The theorem promises existence only, never location or count.