← Back to course

The Intermediate Value Theorem

You’ll be able to

The IVT: no gaps means no skipped values

The Intermediate Value Theorem (IVT) says: if f is continuous on a closed interval [a, b], then f takes on every value between f(a) and f(b) somewhere on that interval. In symbols, for any target N strictly between f(a) and f(b), there is at least one c in (a, b) with f(c) = N. Because a continuous graph has no breaks, it cannot jump over an intermediate height — it must pass through it.

The classic use: guaranteeing a root

The most common application sets the target to N = 0. If f is continuous on [a, b] and f(a) and f(b) have opposite signs, then 0 lies between them, so the IVT guarantees a c in (a, b) with f(c) = 0 — a root. Note what the theorem does and does not give: it guarantees at least one solution exists, but it does not tell you how many there are or where exactly they are.

Intermediate Value Theorem
f continuous on [a, b] and N between f(a) and f(b) ⟹ ∃ c in (a, b) with f(c) = N
An existence theorem: it promises a c exists but gives no formula for it.

Continuity is not optional

The IVT’s conclusion fails the moment continuity is dropped. A function with a jump can leap right over an intermediate value without ever attaining it. This is why every correct IVT justification must first assert (or verify) that f is continuous on the closed interval. On the AP exam, a solution that skips the words "f is continuous" typically loses the point even when the numerical setup is right.

Worked example

Show that f(x) = x³ + x − 1 has a root in the interval [0, 1].

  1. 1.f is a polynomial, so it is continuous on [0, 1] — the IVT hypothesis is satisfied.
  2. 2.Evaluate the endpoints: f(0) = 0 + 0 − 1 = −1 (negative).
  3. 3.Evaluate the other endpoint: f(1) = 1 + 1 − 1 = 1 (positive).
  4. 4.Since f(0) < 0 < f(1), the target N = 0 lies between the endpoint values, so by the IVT there is a c in (0, 1) with f(c) = 0.
Answer: Because f is continuous and changes sign from −1 to 1 across [0, 1], the IVT guarantees at least one root in (0, 1).
Tip

For a "show a solution exists" prompt, follow a three-line ritual: (1) state f is continuous on the closed interval, (2) compute the two endpoint values, (3) note the target lies between them and invoke the IVT by name. That structure is exactly what earns full credit.

Checkpoint

A function f is continuous on [1, 5] with f(1) = −2 and f(5) = 6. Which value is the IVT guaranteed to produce on (1, 5)?

Checkpoint

Which condition is essential for applying the Intermediate Value Theorem on [a, b]?

On the exam

The IVT is an existence theorem — it never locates or counts solutions. If a prompt asks "how many" or "find the value," the IVT alone is insufficient; you would need additional tools like a sign analysis or the Mean Value Theorem.

Answer the 2 checkpoints as you read.

Sign in to save your progress