Intermediate Value Theorem

The central idea

If $f$ is continuous on the closed interval $[a,b]$ and $N$ lies between $f(a)$ and $f(b)$, then there exists at least one $c\in[a,b]$ such that $f(c)=N$. For root existence, take $N=0$: opposite signs at the endpoints guarantee a zero inside, provided continuity holds.

Definitions, hypotheses, and notation

Continuity on the entire closed interval is essential. A discontinuous function can have opposite endpoint signs while jumping over zero, so endpoint arithmetic alone is not enough. The target value may equal an endpoint output, in which case the theorem permits $c=a$ or $c=b$; strict between-ness places a solution in the open interval.

Bisection operationalizes the theorem. After finding an interval with opposite signs, evaluate the midpoint and retain the half whose endpoint values still bracket zero. Continuity preserves existence in every retained interval, while shrinking width controls location error. This algorithm still does not establish uniqueness; monotonicity or another argument must supply that.

Conceptual meaning

A continuous graph cannot move from one endpoint height to another without passing through every intermediate height. The theorem guarantees existence but generally gives neither a formula for $c$ nor uniqueness.

A dependable method and decision rule

1. Identify a closed interval and the target output $N$.
2. Justify continuity on the entire interval, including any domain restrictions.
3. Evaluate the two endpoint values.
4. Show explicitly that $N$ lies between those values.
5. Conclude that at least one appropriate $c$ exists; do not claim uniqueness without more information.

Fully worked example

Graphical or geometric meaning

Plotting the endpoint points, any unbroken path joining them must cross the $x$-axis. Bisection repeatedly applies the same sign-change logic to smaller intervals, converting the existence theorem into a numerical approximation method.

Common mistakes and why they fail

Verification and reasonableness checks

• Verify the interval does not cross a denominator zero or other domain break.
• State endpoint values and the between-ness inequality.
• Use monotonicity separately if uniqueness is needed.

Existence is not a formula

On $[a,b]$, establish continuity throughout and show the target $L$ lies between $f(a)$ and $f(b)$. The theorem then guarantees some $c$ with $f(c)=L$, but does not locate it or guarantee uniqueness. For roots, take $L=0$ and bracket it with opposite signs. A sign change without continuity is insufficient: $1/x$ changes sign across zero and has no root. After proving existence, bisection can narrow the location by repeatedly keeping a subinterval whose endpoint outputs bracket the target. Each step preserves continuity on the smaller interval, which is why the numerical procedure rests on the theorem rather than on guesswork.

Practice

1. Why does $x^2-2$ have a root in $(1,2)$?
2. Does equal endpoint sign prove there is no root?
3. What extra fact can prove uniqueness?

Connections and next steps

• Continuity
• Limits
• Increasing and Decreasing Functions