Calculus I · Limits and Continuity · lesson

Intermediate Value Theorem

Learn Intermediate Value Theorem with plain-language explanations, guided examples, worked homework methods, interactive checks, and exam-style practice.

Course progress40 of 47 core pages40 of 47

Where this chapter fits

Chapter 5: Continuity

Connect limits to function values, classify discontinuities, repair piecewise definitions, and use the Intermediate Value Theorem.

Reading lens: Do the limit, the function value, and the surrounding domain fit together at the point or across the interval? Keep that question in view while reading Intermediate Value Theorem; the worked mathematics is evidence for the idea, not a substitute for it.

This page connects Two-Parameter Continuity Problems to Bisection Method After the IVT. Read the explanation first, predict each example’s next move, and only then compare the written solution.

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 $[0,1]$ ?

Your work stays on this device. No account or AI grader is used.

Show hint

Check continuity and endpoint signs.

Attempt once to unlock the solution

Submit an answer first. The hint is available now.

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 $(0,1)$ .

Graph reading guide

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.

Source & rights

Original instruction with traceable references.

The exposition is original. No Active Calculus exercise is reproduced verbatim. Public-domain examples were modernized and recomposed when used as inspiration.

The verified handoff declares original composition and requires owner provenance review. BetterGrades-original material remains separate from public-domain references; no source textbook PDF is published here.

Chapter 5: Continuity

Connect limits to function values, classify discontinuities, repair piecewise definitions, and use the Intermediate Value Theorem.

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 $[0,1]$ ?

Your work stays on this device. No account or AI grader is used.

Show hint

Check continuity and endpoint signs.

Attempt once to unlock the solution

Submit an answer first. The hint is available now.

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 $(0,1)$ .

Graph reading guide

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.

Source & rights

Original instruction with traceable references.

The exposition is original. No Active Calculus exercise is reproduced verbatim. Public-domain examples were modernized and recomposed when used as inspiration.

Chapter 5: Continuity

The Intermediate Value Theorem

Related resources

Original instruction with traceable references.

Chapter 5: Continuity

The Intermediate Value Theorem

Related resources

Original instruction with traceable references.