Intermediate Values and Root Bracketing

A guarantee of crossing

If $f$ is continuous on $[a,b]$ , it reaches every value between $f(a)$ and $f(b)$ . This is the intermediate value theorem. It formalizes the idea that a continuous path cannot move from below a height to above it without passing through that height.

The theorem guarantees existence, not a formula for the input and not uniqueness. For a root, a convenient condition is $f(a)f(b)<0$ : the endpoint outputs have opposite signs, so some $c\in(a,b)$ satisfies $f(c)=0$ . An endpoint that already equals zero is a root without further argument.

Visual guide

VISUAL GUIDEA continuous graph cannot skip the level

f(x) = x³ + x
Target level y = 1

The continuous curve x³ + x runs from −2 to 2 on [−1, 1]. It must cross the horizontal level y = 1 somewhere in between. Continuity is what rules out jumping over that level.

Worked example: a root without factoring

Let $f(x)=x^3+x-1$ . It is a polynomial, so it is continuous on $[0,1]$ . Since $f(0)=-1$ and $f(1)=1$ , there is a root between zero and one.

To narrow it, evaluate the midpoint: $f(0.5)=-0.375$ . The sign change now lies on $[0.5,1]$ . Next, $f(0.75)=0.171875$ , so the new bracket is $[0.5,0.75]$ . Each bisection halves the bracket length. After $n$ halvings, its length is $(b-a)/2^n$ , and its midpoint has error at most half that length.

This bound assumes the bracket continues to contain the root. Preserve the opposite-sign endpoints and verify that the function is continuous throughout the interval.

Worked example: why the condition matters

For $g(x)=1/x$ on $[-1,1]$ , endpoint values have opposite signs, but there is no root. The missing hypothesis is continuity across zero. A sign change across a vertical asymptote does not bracket a zero.

Conversely, $h(x)=x^2$ has a root at zero but no sign change across it. Opposite endpoint signs are sufficient, not necessary. A graph touching the axis can hide an even-multiplicity root from a sign-change search.

Existence versus uniqueness

For $x^3+x-1$ , the derivative $3x^2+1$ is positive everywhere, so the function is strictly increasing. That additional fact makes the root unique. Intermediate values alone cannot do this: a continuous oscillating function may cross the same height many times.

A mathematical bracket is also different from a rounded numerical report. If displayed endpoint values round to zero, calculate with enough precision before deciding which half contains the sign change.

Practice

Prove that $x^3-2$ has a root in $(1,2)$ .
After ten bisections of $[1,2]$ , bound the error of the final midpoint.
Explain why equal endpoint signs do not prove the absence of roots.

Show worked solutions

The polynomial is continuous and has values $-1$ and $6$ at the endpoints, so it crosses zero.
The interval width is $1/1024$ ; midpoint error is at most $1/2048$ .
$x^2$ on $[-1,1]$ has positive endpoint values and a zero inside. The sign-change criterion only gives a sufficient condition.