AP Calculus AB and BC

When the Intermediate Value Theorem Fails

The Intermediate Value Theorem rests on one hypothesis: f is continuous on the closed interval. Remove it and a function can step straight over a value without ever taking it. The theorem also only promises values between f(a) and f(b), never outside that range.

The hypotheses, and what each one buys

A theorem is only as strong as its conditions. Below, each hypothesis is dropped on its own while every other one is kept, so you can see precisely what it was holding up.

1. 1

   f is continuous on the closed interval [a, b]

   $$f \in C[a, b]$$

   Continuity is the whole engine. It is what stops the graph from getting from one side of a horizontal line to the other without crossing it. Every proof of the theorem is really a proof that a continuous function cannot skip, whether by bisection or by the least upper bound property.

   Drop it and the theorem fails

   $$f(x) = \frac{x}{|x|} \ \ (x \ne 0), \qquad f(0) = 0$$

   The sign function on [-1, 1], with f(0) = 0

   Here $f(-1) = -1$ and $f(1) = 1$, so $N = 0.5$ lies squarely between them. But this function only ever takes the values $-1$, 0 and $1$. It never takes $0.5$, or any other value between 0 and 1. The single discontinuity at the origin lets it jump the whole gap in one step.

2. 2

   N lies between f(a) and f(b)

   This is not really a condition on the function, it is a description of what is being promised. The theorem hands you values inside the range spanned by the two endpoint values and says nothing whatever about values outside it. There is no counterexample to state here, because a function that misses a value outside that range is not violating anything: $f(x) = x$ on $[0,1]$ never takes the value 2, and no theorem ever said it would.

   No counterexample built from an elementary formula breaks this one on its own, so none is claimed here.

The counterexample above is checked numerically on every build: each function is evaluated and the conclusion is confirmed to fail. A counterexample that stopped working would fail the build rather than sit here misleading you.

Why it is true

1. Assume $f(a) < N < f(b)$; the other order is the same argument with signs flipped.
2. Let $S$ be the set of $x$ in $[a,b]$ with $f(x) < N$. It is nonempty, since $a$ is in it, and bounded above by $b$, so it has a least upper bound $c$.
3. If $f(c) < N$, continuity keeps $f$ below $N$ on a small interval to the right of $c$, so points past $c$ belong to $S$, contradicting that $c$ is an upper bound.
4. If $f(c) > N$, continuity keeps $f$ above $N$ just to the left of $c$, so a smaller number is already an upper bound, contradicting leastness.
5. Both strict inequalities are impossible, so $f(c) = N$. Every step used continuity, which is why the hypothesis cannot be weakened.

What it does not say