The Intermediate Value Theorem is a key idea in calculus that connects the shape of a graph to the values a function must take. If a function is continuous on a closed interval, it cannot jump over any y-value between its endpoint values. This matters because it gives a reliable way to prove that solutions exist, even when we cannot find them exactly.
It is especially useful for showing that an equation has a root on an interval.
Key Facts
- If f is continuous on [a, b] and N is between f(a) and f(b), then there is some c in [a, b] such that f(c) = N.
- Root version: if f is continuous on [a, b] and f(a)f(b) < 0, then there is some c in (a, b) such that f(c) = 0.
- The theorem requires continuity on the entire closed interval [a, b].
- A sign change from f(a) < 0 to f(b) > 0 guarantees at least one x-intercept between a and b.
- The Intermediate Value Theorem proves existence, but it does not usually give the exact value of c.
- The theorem does not say the solution is unique, since a continuous function may cross the same value many times.
Vocabulary
- Continuous function
- A function is continuous on an interval if its graph can be drawn there without breaks, holes, or jumps.
- Closed interval
- A closed interval [a, b] includes both endpoints a and b.
- Intermediate value
- An intermediate value is any y-value between f(a) and f(b).
- Root
- A root of a function is an input c where f(c) = 0.
- Existence theorem
- An existence theorem proves that at least one object or solution exists without necessarily finding it exactly.
Common Mistakes to Avoid
- Ignoring continuity, because the theorem only works if the function is continuous on the whole interval. A jump or hole can allow the graph to skip an intermediate value.
- Thinking a sign change is required for every use of the theorem, because the theorem applies to any value between f(a) and f(b), not only 0. A sign change is just the special case used to guarantee a root.
- Claiming the theorem finds the exact root, because it only guarantees that a root exists somewhere in the interval. Extra methods such as bisection or Newton's method are needed to approximate it.
- Assuming there is only one solution, because a continuous curve can cross the same horizontal level more than once. The theorem guarantees at least one solution, not exactly one.
Practice Questions
- 1 Let f(x) = x^3 - x - 2. Compute f(1) and f(2), then use the Intermediate Value Theorem to explain why f(x) = 0 has a solution between 1 and 2.
- 2 Let g(x) = x^2 - 5. Use the Intermediate Value Theorem on the interval [2, 3] to show that there is a number c with g(c) = 0. What familiar number is c?
- 3 A function is continuous on [0, 4], with f(0) = -3 and f(4) = 7. Explain why the graph must cross the x-axis, and state whether the Intermediate Value Theorem guarantees exactly one crossing.