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C05 · Unit C1 Limits and continuity

Intermediate Value Theorem and existence proofs

CoreReview editionNot yet human-reviewed
Time
About 45 minutes, untimed
Course framework reference
AP 1.16
Builds on
Part of

Progress on this lesson counts in every pathway listed.

Understand the idea

If f is continuous on [a,b], every value between f(a) and f(b) is attained at least once. The IVT is an existence theorem: it does not identify the input or guarantee uniqueness. For a root, find endpoint values of opposite signs and explicitly state continuity. Additional monotonicity information may establish uniqueness later.

Worked example

For f(x) equals x cubed +x minus 1, f(0) equals minus 1 and f(1) equals 1. Polynomial continuity on [0,1] and IVT guarantee c in (0,1) with f(c) equals 0. Since f minutes (x) equals 3x squared +1>0 everywhere, calculus can additionally establish uniqueness.

Explore the mathematics

Intermediate Value Theorem and existence proofs

Check continuity and opposite endpoint signs before claiming a root exists.

Setting 3 of 5

Scroll sideways to see the full diagram.

Predict whether the curve must cross the horizontal axis.xy-2-11234f(x)=x²−4, continuous on [0,3]

Predict before continuing

Predict whether the curve must cross the horizontal axis.

Read the complete explanation
  1. Predict whether the curve must cross the horizontal axis.
  2. Endpoint values have opposite signs: f(0)<0<f(3).
  3. Continuity and the Intermediate Value Theorem guarantee at least one zero in (0,3).
  4. This example has positive root √4≈2. IVT guarantees existence, not uniqueness in general.
See the idea: Intermediate Value Theorem and existence proofs
See the idea: Intermediate Value Theorem and existence proofsThe continuous polynomial x³+x−1 starts below zero and ends above it on [0,1]. IVT guarantees an intercept between them. Its strictly positive derivative additionally proves uniqueness.xy0-10.25-0.50.500.750.511

The continuous polynomial x³+x−1 starts below zero and ends above it on [0,1]. IVT guarantees an intercept between them. Its strictly positive derivative additionally proves uniqueness.

Discuss: explain what each label represents, then draw a related example with different values.

A common misconception

Opposite endpoint signs alone are insufficient when the function has a discontinuity between them, as 1 over x on [ minus 1,1] shows.

Investigate and explain

Apply three bisection steps to x cubed minus 2. Keep a sign table and distinguish a guaranteed interval from an exact root.
Identify one step in the worked example, name the mathematical fact that permits it, and state its domain conditions.