Intermediate value theorem

What you are looking at

The shaded band runs from $f(\alpha)$ to $f(\beta)$. The theorem says any horizontal line inside it crosses the graph somewhere between $\alpha$ and $\beta$, and the working lists the solutions of $f(x) = \eta$ as you move it. Bolzano’s theorem is the special case $\eta = 0$ — and that is how the usual proof goes: apply Bolzano to $g(x) = f(x) - \eta$.

Outside the band the theorem is silent: the equation may or may not have a solution there. With the discontinuous function the band still exists, but some heights inside it are never reached — the graph jumps over them, and the tool flags it.

Try this

1. 1Press Sweep and watch $\eta$ run from the bottom of the picture to the top: inside the band the green line never misses the graph.
2. 2Stop the line at $\eta = 1$: it meets the graph three times, at $x = -\sqrt{3}$, $0$ and $\sqrt{3}$. The theorem only promised one.
3. 3Lower $\eta$ to $0$, below $f(\alpha)$: the theorem no longer applies, yet the line still crosses the graph twice. Outside the band, anything can happen.
4. 4Choose Discontinuous and set $\eta = 0$: it lies between $f(\alpha)$ and $f(\beta)$, but no $x$ gives $f(x) = 0$.

Mistakes students make

• Using it for values outside $[f(\alpha), f(\beta)]$. The function may well go higher, but the theorem does not say so — that is the extreme value theorem’s job.
• Expecting a unique solution. The theorem gives at least one.
• Forgetting continuity: $\dfrac{1}{x}$ on $[-1, 1]$ has $f(-1) = -1$ and $f(1) = 1$ but never takes the value $0$.

In the exam

• The classic AP item gives a table of values of a continuous function and asks whether $f(c) = k$ must have a solution on an interval: name the theorem, cite continuity, and quote the two values that bracket $k$.
• If you are asked to prove it from Bolzano’s theorem: $g(x) = f(x) - \eta$ is continuous on $[\alpha, \beta]$ with $g(\alpha)\,g(\beta) < 0$, so there is $x_0$ with $g(x_0) = 0$, that is $f(x_0) = \eta$.

Where it is taught

• AP Calculus AB/BC · Unit 1 (1.16)
• IB Maths AA HL · 5.12 (continuity)
• Greek Γ΄ Λυκείου · §1.8