The intermediate value theorem makes precise an intuition you already have: if you draw a continuous curve from one height to another, you cannot get there without passing through every height in between. OpenStax Calculus Volume 1 states it as Theorem 2.11 in Section 2.4. It is one of the first genuine existence theorems in calculus, and it is the usual tool for proving that an equation has a root without solving the equation.

The statement

Let $f$ be continuous on a closed, bounded interval $[a,b]$. If $z$ is any real number between $f(a)$ and $f(b)$, then there is a number $c$ in $[a,b]$ such that $f(c)=z$. Continuity on the closed interval is essential: $f$ must be defined at the endpoints and have no jumps, holes or vertical asymptotes in between. OpenStax first builds that idea with the three-part definition of continuity at a point: $f(a)$ is defined, $\lim_{x\to a}f(x)$ exists, and the two are equal.

Worked example 1: proving a zero exists

OpenStax Example 2.36 asks you to show that $f(x)=x-\cos x$ has at least one zero. The function is continuous everywhere, hence continuous on $[0,\pi/2]$. The endpoint values are $$f(0)=0-\cos 0=-1<0,\qquad f\bigl(\tfrac{\pi}{2}\bigr)=\tfrac{\pi}{2}-\cos\tfrac{\pi}{2}=\tfrac{\pi}{2}>0.$$ The number $0$ lies between $f(0)$ and $f(\pi/2)$, so the theorem supplies a $c$ in $[0,\pi/2]$ with $f(c)=0$. You have not computed $c$. You have proved it exists.

Worked example 2: when the theorem does not apply

OpenStax Example 2.38 considers $f(x)=1/x$ on $[-1,1]$. Here $f(-1)=-1<0$ and $f(1)=1>0$, so the endpoint signs look promising. But $f$ is not continuous on $[-1,1]$: it has an infinite discontinuity at $0$. The hypotheses fail, and in fact there is no zero on that interval. Opposite signs at the endpoints are not enough without continuity.

Worked example 3: you cannot prove a root is absent

OpenStax Example 2.37 asks: if $f$ is continuous on $[0,2]$ and both $f(0)$ and $f(2)$ are positive, can you conclude that $f$ has no zero in the interval? No. The theorem only tells you about values that must occur. It is silent about values that might still occur. The counterexample $f(x)=(x-1)^2$ has $f(0)=1>0$, $f(2)=1>0$, and $f(1)=0$.

Checkpoints

  1. OpenStax Checkpoint 2.26: show that $f(x)=x^3-x^2-3x+1$ has a zero on $[0,1]$.
  2. Does $f(x)=1/x$ have a zero on $[1,2]$? Can you use the theorem to say so?
  3. If $f$ is continuous on $[1,4]$, $f(1)=2$ and $f(4)=8$, must there be a $c$ with $f(c)=5$? Must there be a $c$ with $f(c)=10$?

Common mistakes

  • Dropping continuity. Sign change alone does not guarantee a root.
  • Claiming uniqueness. The theorem says "at least one" $c$, not "exactly one".
  • Claiming a missing value. Same-sign endpoints do not prove there is no root.
  • Confusing it with the mean value theorem. The mean value theorem is about a derivative equalling an average slope. This theorem is about a function hitting an intermediate height.

Related reading

Sources

Last reviewed: 12 November 2026.

Common questions

What does the intermediate value theorem say?

If f is continuous on a closed, bounded interval [a, b] and z is any real number between f(a) and f(b), then there is at least one c in [a, b] with f(c) = z.

Can the intermediate value theorem tell you the value of c?

No. It is an existence theorem. It guarantees that such a c is there; it does not construct it or say there is only one.

Why does continuity matter?

Without continuity on the whole interval the graph can jump over the target value. OpenStax's example f(x) = 1/x on [-1, 1] has opposite signs at the endpoints but no zero, because the function is not continuous on that interval.

If f(a) and f(b) have the same sign, does that mean there is no root?

No. OpenStax points out that f(x) = (x-1)^2 is positive at 0 and at 2 but still has a root at 1. The theorem never lets you conclude that a value is missing.