Extreme Value Theorem

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Extreme Value Theorem: meaning, method and solved example (AP Calculus AB) – Maths PDF

In short: Extreme value theorem guarantees that a function continuous on a closed, bounded interval attains both an absolute maximum and an absolute minimum somewhere in that interval.

AP Calculus ABCalculusAP Calculus AB Unit 5

What is Extreme Value Theorem?

The extreme value theorem says you cannot trace a connected curve over a closed stretch without reaching a highest point and a lowest point. Both must be actual values the graph hits, not just heights it creeps toward.

Stated precisely: if \(f\) is continuous on \([a, b]\), then there exist numbers \(m\) and \(M\) in \([a, b]\) with \(f(m) \le f(x) \le f(M)\) for every \(x\) in the interval. The extremes occur either at endpoints or at critical points inside.

How to spot it

  • The task asks for the absolute (global) maximum or minimum on a closed interval like [-1, 4].
  • Continuity is given or easy to confirm, and the interval includes both endpoints.
  • You are told to justify that a largest and smallest value exist before looking for them.
  • A contrast is requested: why an open interval or a jump discontinuity can break the guarantee.

The method

  1. Confirm that \(f\) is continuous on the closed interval \([a, b]\); the extreme value theorem then promises both extremes.
  2. Find the critical points inside \((a, b)\) by solving \(f^{\prime}(x) = 0\) or finding where it is undefined.
  3. Evaluate \(f\) at every critical point and at both endpoints.
  4. Compare: the largest number is the absolute maximum, the smallest is the absolute minimum.
min -4min -4max 16x = 4-1
On [-1, 4] the curve y = x^3 – 3x^2 must reach a lowest and a highest value; here they are -4 and 16.

Extreme Value Theorem: solved example

Problem

Find the absolute extreme values of \(f(x) = x^3 – 3x^2\) on \([-1, 4]\).

Solution

The function is a polynomial, hence continuous on the closed interval, so extremes exist. Also \(f^{\prime}(x) = 3x^2 – 6x = 3x(x – 2)\), with critical points at \(x = 0\) and \(x = 2\).

Evaluate: \(f(-1) = -4\), \(f(0) = 0\), \(f(2) = 8 – 12 = -4\), \(f(4) = 64 – 48 = 16\).

The absolute maximum is 16 at \(x = 4\), and the absolute minimum is \(-4\), reached at both \(x = -1\) and \(x = 2\).

Try it

Question 1

Find the absolute maximum of \(g(x) = 12x – x^2\) on \([0, 8]\).

Show the answer

\(g^{\prime}(x) = 12 – 2x = 0\) at \(x = 6\). Values: \(g(0) = 0\), \(g(6) = 36\), \(g(8) = 32\). The maximum is 36.

Question 2

Why does \(h(x) = \frac{1}{x}\) on \([-1, 1]\) fail to have a maximum?

Show the answer

It is not continuous at \(x = 0\), so the hypothesis fails: the values grow without bound near 0 and the theorem gives no guarantee.

Extreme Value Theorem vs. Critical Point

A critical point is a candidate: an input where the derivative is zero or undefined. The theorem is a guarantee that extreme values exist on a closed interval. After the guarantee, you test the candidates and the endpoints to find where the extremes actually occur.

Critical Point

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