Limits & Continuity · Continuity & Theorems

Intermediate Value Theorem

f continuous on [a,b],  f(a)<N<f(b)c(a,b):f(c)=Nf \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].

Worked examples

Show that x³ + x - 1 = 0 has a root in (0, 1).
  1. Let f(x) = x³ + x - 1. f is a polynomial, so continuous everywhere.
  2. f(0) = 0 + 0 - 1 = -1 < 0
  3. f(1) = 1 + 1 - 1 = 1 > 0
  4. 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).

Show that cos(x) = x has a solution in (0, π/2).
  1. Let f(x) = cos(x) - x, continuous on [0, π/2]
  2. f(0) = 1 - 0 = 1 > 0
  3. f(π/2) = 0 - π/2 ≈ -1.57 < 0
  4. By IVT, there exists c in (0, π/2) with f(c) = 0, i.e., cos(c) = c

Answer: By IVT, cos(x) = x has a solution in (0, π/2).

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