Calculus I · Limits and Continuity · lesson

Intermediate Value Theorem

Concept

Learning objectives

Use continuity and endpoint values to prove that a function attains an intermediate output or has a root in an interval; distinguish existence from calculation.

The Intermediate Value Theorem

Concept

A continuous path from below a horizontal line to above that line must cross it somewhere. You may not know exactly where the crossing occurs, but skipping it would require a jump.

Theorem

Intermediate Value Theorem

Suppose $f$ is continuous on $[a,b]$ . If $N$ lies between $f(a)$ and $f(b)$ , then there exists at least one $c\in[a,b]$ such that

\boxed{f(c)=N}.

In particular, if $f(a)$ and $f(b)$ have opposite signs, then there exists $c\in(a,b)$ such that $f(c)=0$ .

Guided walkthrough

Crossing ground level

Suppose a continuous function has

f(1)=-2 \qquad\text{and}\qquad f(4)=5.

The graph starts below the $x$ -axis and ends above it. Because it is continuous, it must cross the axis somewhere between $1$ and $4$ . Thus there is some $c\in(1,4)$ with $f(c)=0$ .

Interactive checkivt-flow-01

Does the IVT guarantee a root of $x^3+x-1$ on $[0,1]$ ?

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Show hint

Check continuity and endpoint signs.

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Worked example

Prove a polynomial has a root

Show that

x^3+x-1=0

has at least one solution in $(0,1)$ .

Show worked solution

Define

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

Polynomials are continuous everywhere, so $f$ is continuous on $[0,1]$ .

Evaluate the endpoints:

f(0)=0+0-1=-1,

f(1)=1+1-1=1.

Because $-1<0<1$ , the Intermediate Value Theorem guarantees some $c\in(0,1)$ such that

f(c)=0.

Therefore, the equation has at least one solution in $(0,1)$ .

Continuous cubic crossing the x-axis between endpoints of opposite sign. — **Read the graph**
Continuity and opposite endpoint signs guarantee at least one root between $0$ and $1$ .

Exam note

The Intermediate Value Theorem does not give the exact root, does not prove the root is unique, and cannot be used unless continuity on the entire closed interval has been established.

After the explanation

Use the section idea

Reading lens

Do the limit, the function value, and the surrounding domain fit together at the point or across the interval?

Mental model

Continuity is a three-part agreement: the value exists, the two-sided limit exists, and those two quantities are equal.

Decision

At a point, test the three conditions in order; on an interval, check the domain and endpoints before invoking any continuity theorem.

Common trap

A sign change supports the Intermediate Value Theorem only when continuity holds on the entire closed interval, and it does not prove uniqueness.

Check yourself

You are ready to continue when you can classify a break, decide whether one value can repair it, and state every IVT hypothesis aloud.

Source & rights

Original instruction with traceable references.

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