Limits & Continuity · Continuity & Theorems

Intermediate Value Theorem

Q: When does the Intermediate Value Theorem apply?

The Intermediate Value Theorem holds under this condition: f must be continuous on the closed interval [a, b].

Q: How do you use the Intermediate Value Theorem?

Worked example. Show that x³ + x - 1 = 0 has a root in (0, 1). Let f(x) = x³ + x - 1. f is a polynomial, so continuous everywhere. f(0) = 0 + 0 - 1 = -1 0. Since f(0) < 0 < f(1) and f is continuous on [0,1], by IVT there exists c in (0,1) with f(c) = 0. Answer: By IVT, there is at least one root in (0, 1).

f \text{ continuous on } [a,b],\; f(a) < N < f(b) \Rightarrow \exists\, c \in (a,b) : f(c) = N

If f is continuous on [a,b] and N is between f(a) and f(b), then there exists at least one c in (a,b) where f(c) = N. Often used to show a root exists.

Conditions. f must be continuous on the closed interval [a, b].

2 worked examples
18 in Limits & Continuity

Worked examples

Show that x³ + x - 1 = 0 has a root in (0, 1).

Let f(x) = x³ + x - 1. f is a polynomial, so continuous everywhere.
f(0) = 0 + 0 - 1 = -1 < 0
f(1) = 1 + 1 - 1 = 1 > 0
Since f(0) < 0 < f(1) and f is continuous on [0,1], by IVT there exists c in (0,1) with f(c) = 0

Answer: By IVT, there is at least one root in (0, 1).

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Show that cos(x) = x has a solution in (0, π/2).

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Common questions

What is the Intermediate Value Theorem?

If f is continuous on [a,b] and N is between f(a) and f(b), then there exists at least one c in (a,b) where f(c) = N. Often used to show a root exists. It is one of the limits & continuity formulas in the CalcRef reference.

When does the Intermediate Value Theorem apply?

The Intermediate Value Theorem holds under this condition: f must be continuous on the closed interval [a, b].

How do you use the Intermediate Value Theorem?

Worked example. Show that x³ + x - 1 = 0 has a root in (0, 1). Let f(x) = x³ + x - 1. f is a polynomial, so continuous everywhere. f(0) = 0 + 0 - 1 = -1 < 0. f(1) = 1 + 1 - 1 = 1 > 0. Since f(0) < 0 < f(1) and f is continuous on [0,1], by IVT there exists c in (0,1) with f(c) = 0. Answer: By IVT, there is at least one root in (0, 1).

Practice

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