Bolzano’s theorem

What you are looking at

The theorem has two conditions and one conclusion, and the picture keeps them apart. The yellow interval is $[\alpha, \beta]$; the chips above the graph say whether $f$ is continuous there and whether $f(\alpha)$ and $f(\beta)$ have opposite signs. When both hold, the conclusion lights up and the roots appear as green points on the axis. It says at least one: with $f(x) = x^3 - 3x + 1$ and a wide enough interval you will find three.

Switch to the discontinuous function and the reason for the first condition becomes visible: $f$ jumps from $-1$ to $1$ at $x = 0$, so it changes sign without ever being zero. Bisect shows the theorem working as a method: each step halves the interval and keeps the half where the sign still changes, so after $n$ steps the root is trapped in an interval of width $\dfrac{\beta - \alpha}{2^n}$.

Try this

1. 1Drag $\beta$ past $x \approx 1.53$. Now $f(\alpha)$ and $f(\beta)$ are both positive and the theorem promises nothing — yet two roots are still there. The converse of Bolzano’s theorem is false.
2. 2Set $\alpha = -2.5$ and $\beta = 2.5$: the signs differ and three roots appear. At least one really means at least one.
3. 3Press Bisect and count the steps: each halves the interval, so after 10 steps it is about a thousandth of its first width.
4. 4Choose Discontinuous: the endpoint values have opposite signs, but the graph jumps over the axis at $x = 0$ and there is no root at all.

Mistakes students make

• Checking only the signs. Without continuity on the whole closed interval the theorem says nothing: $\dfrac{1}{x}$ on $[-1, 1]$ changes sign and has no root.
• Reading it backwards. A root in $(\alpha, \beta)$ does not need $f(\alpha)f(\beta) < 0$: $f(x) = x^2$ on $[-1, 1]$ has a root, yet $f(-1)f(1) = 1 > 0$.
• Claiming exactly one root. The theorem gives existence only; for uniqueness you need more, usually that $f$ is strictly monotonic on the interval.

In the exam

• In an AP free-response answer, name the theorem, state that $f$ is continuous on the closed interval, and give both endpoint values — the justification is marked, not just the conclusion.
• At A-Level this is the change-of-sign method for locating roots, and questions ask when it fails: a discontinuity gives a sign change with no root, and two roots close together give a root with no sign change.

Where it is taught

• AP Calculus AB/BC · Unit 1 (1.16, the Intermediate Value Theorem)
• Edexcel A-Level Pure Year 2 · Numerical methods (locating roots by change of sign)
• IB Maths AA HL · 5.12 (continuity)
• Greek Γ΄ Λυκείου · §1.8