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1.16 Working with the Intermediate Value Theorem (IVT)

Syllabus
2020
Topic
1.16
Level
—

Learning objectives

—FUN-1.A—Explain the behavior of a function on an interval using the Intermediate Value Theorem• FUN-1.A Explain the behavior of a function on an interval using the Intermediate Value Theorem.• FUN-1.A.1 If f is a continuous function on the closed interval [a, b] and d is a number between f(a) and f(b), then the Intermediate Value Theorem guarantees that there is at least one number c between a and b, such that f(c) = d.• Enduring understanding FUN-1: Existence theorems allow us to draw conclusions about a function’s behavior on an interval without precisely locating that behavior.

Using IVT to Prove a Value Exists

The Intermediate Value Theorem (IVT) says that a continuous function cannot skip an output between its endpoint values. If ff is continuous on [a,b][a,b] and dd lies strictly between f(a)f(a) and f(b)f(b), then at least one c∈(a,b)c\in(a,b) satisfies f(c)=df(c)=d.

  1. State why ff is continuous on the entire closed interval [a,b][a,b].
  2. Calculate f(a)f(a) and f(b)f(b).
  3. Show that the target dd lies between those two outputs.
  4. Conclude by IVT that there exists at least one c∈(a,b)c\in(a,b) for which f(c)=df(c)=d.

Let f(x)=x3+xf(x)=x^3+x on [1,2][1,2]. A polynomial is continuous, f(1)=2f(1)=2, and f(2)=10f(2)=10. Because 55 lies between 22 and 1010, IVT guarantees at least one c∈(1,2)c\in(1,2) such that c3+c=5c^3+c=5. The theorem proves that such a solution exists without requiring its exact value.

IVT does not guarantee a unique solution, and endpoint values alone are not enough if the function is discontinuous anywhere on [a,b][a,b]. To prove a zero exists, use the same argument with target d=0d=0 and endpoint outputs of opposite signs.