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Limits & Continuity · Continuity & Theorems

Extreme Value Theorem

f continuous on [a,b]⇒f attains an absolute max and min on [a,b]f \text{ continuous on } [a,b] \Rightarrow f \text{ attains an absolute max and min on } [a,b]

If f is continuous on a closed interval [a, b], then f attains both an absolute maximum and an absolute minimum on that interval.

Conditions. f must be continuous and the interval must be closed and bounded.
  • 2 worked examples
  • 18 in Limits & Continuity

Worked examples

Find the absolute extrema of f(x) = x² on [-1, 3].
  1. f is a polynomial (continuous). Apply EVT: extrema exist.
  2. f'(x) = 2x = 0 → x = 0 (critical point in [-1,3])
  3. Evaluate: f(-1) = 1, f(0) = 0, f(3) = 9

Answer: Absolute min = 0 at x = 0; absolute max = 9 at x = 3.

1 more worked examplePremium

Find the absolute extrema of f(x) = sin(x) on [0, 2π].

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Related formulas

Intermediate Value Theorem

See all 18 Limits & Continuity formulas on one printable sheet - the same statements and conditions as these pages, typeset and laid out for paper.

Common questions

What is the Extreme Value Theorem?

If f is continuous on a closed interval [a, b], then f attains both an absolute maximum and an absolute minimum on that interval. It is one of the limits & continuity formulas in the CalcRef reference.

When does the Extreme Value Theorem apply?

The Extreme Value Theorem holds under this condition: f must be continuous and the interval must be closed and bounded.

How do you use the Extreme Value Theorem?

Worked example. Find the absolute extrema of f(x) = x² on [-1, 3]. f is a polynomial (continuous). Apply EVT: extrema exist. f'(x) = 2x = 0 → x = 0 (critical point in [-1,3]). Evaluate: f(-1) = 1, f(0) = 0, f(3) = 9. Answer: Absolute min = 0 at x = 0; absolute max = 9 at x = 3.

Practice

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