Intermediate Value Theorem Examples in Math

Start with the recap, study the fully worked examples, then use the practice problems to check your understanding of Intermediate Value Theorem.

This page combines explanation, solved examples, and follow-up practice so you can move from recognition to confident problem-solving in Math.

Concept Recap

If f is continuous on the closed interval [a, b] and N is any value between f(a) and f(b), then there exists at least one c in (a, b) such that f(c) = N.

A continuous function can't skip values. If you start below a line and end above it, you must cross it somewhere. It's like driving from sea level to a mountaintop—you pass through every elevation in between.

Read the full concept explanation →

How to Use These Examples

  • Read the first worked example with the solution open so the structure is clear.
  • Try the practice problems before revealing each solution.
  • Use the related concepts and background knowledge badges if you feel stuck.

What to Focus On

Core idea: Continuity guarantees no gaps in the range. A continuous function on [a, b] hits every value between f(a) and f(b). This is an existence theorem—it tells you a value c exists but doesn't tell you what it is.

Common stuck point: The IVT is an existence theorem: it proves a solution exists but doesn't find it. To approximate the root, combine IVT with bisection—repeatedly halve the interval.

Sense of Study hint: Evaluate f at the endpoints, confirm a sign change, then state that continuity guarantees a root between them.

Worked Examples

Example 1

easy
Show that f(x) = x^3 - x - 1 has a root in the interval (1, 2).

Solution

  1. 1
    f is a polynomial, hence continuous on [1, 2].
  2. 2
    f(1) = 1 - 1 - 1 = -1 < 0.
  3. 3
    f(2) = 8 - 2 - 1 = 5 > 0.
  4. 4
    Since f is continuous, f(1) < 0 < f(2), by the IVT there exists c \in (1,2) with f(c) = 0.

Answer

By IVT, f has at least one root in (1, 2).
The IVT requires continuity and a sign change. Polynomials are continuous everywhere, so verifying the sign change at the endpoints is sufficient.

Example 2

medium
Use the IVT to show that \cos x = x has a solution in (0, 1).

Practice Problems

Try these problems on your own first, then open the solution to compare your method.

Example 1

easy
Show that h(x) = x^2 - 2 has a root in (1, 2).

Example 2

medium
Can IVT be applied to f(x) = \frac{1}{x} on [-1, 1] to conclude it attains the value 0? Explain.
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Background Knowledge

These ideas may be useful before you work through the harder examples.

limitcontinuity types