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Extreme Value Theorem

Extreme Value Theorem
Min/Max Theorem

A theorem which guarantees the existence of an absolute max and an absolute min for any continuous function over a closed interval.

 

Theorem: If f is continuous over [a,b], then numbers c and d in [a,b] exist where f(c) is absolute min and f(d) is absolute...

Key Formula

If f is continuous on [a,b], then there exist c,d∈[a,b] such that f(c)≤f(x)≤f(d) for all x∈[a,b].\text{If } f \text{ is continuous on } [a, b], \text{ then there exist } c, d \in [a, b] \text{ such that } f(c) \leq f(x) \leq f(d) \text{ for all } x \in [a, b].
Where:
  • ff = A function that is continuous on the closed interval [a, b]
  • [a,b][a, b] = A closed interval with endpoints a and b
  • cc = A point in [a, b] where f attains its absolute minimum
  • dd = A point in [a, b] where f attains its absolute maximum
  • xx = Any point in the interval [a, b]

Worked Example

Problem: Find the absolute maximum and absolute minimum of f(x) = x² − 4x + 3 on the closed interval [0, 5].
Step 1: Verify the hypotheses. The function f(x) = x² − 4x + 3 is a polynomial, so it is continuous everywhere, including on [0, 5]. The interval [0, 5] is closed. Therefore the Extreme Value Theorem guarantees an absolute max and an absolute min exist.
Step 2: Find the critical points by setting the derivative equal to zero.
f′(x)=2x−4=0  ⟹  x=2f'(x) = 2x - 4 = 0 \implies x = 2
Step 3: The critical point x = 2 lies inside [0, 5], so we keep it. Now evaluate f at the critical point and at both endpoints.
f(0)=02−4(0)+3=3f(2)=22−4(2)+3=−1f(5)=52−4(5)+3=8\begin{gathered}f(0) = 0^2 - 4(0) + 3 = 3 \\ f(2) = 2^2 - 4(2) + 3 = -1 \\ f(5) = 5^2 - 4(5) + 3 = 8\end{gathered}
Step 4: Compare the values: f(0) = 3, f(2) = −1, and f(5) = 8. The smallest value is −1 and the largest is 8.
Answer: The absolute minimum is −1, occurring at x = 2. The absolute maximum is 8, occurring at x = 5.

Another Example

This example uses a trigonometric function instead of a polynomial, demonstrating that the theorem applies to any continuous function. It also shows that the absolute extrema can occur at an interior critical point (the max) or at an endpoint (the min).

Problem: Find the absolute maximum and absolute minimum of f(x) = sin(x) + cos(x) on the closed interval [0, π].
Step 1: Verify hypotheses. The function sin(x) + cos(x) is continuous on [0, π], and the interval is closed. The Extreme Value Theorem applies.
Step 2: Find critical points by setting the derivative to zero.
f′(x)=cos⁡(x)−sin⁡(x)=0  ⟹  sin⁡(x)=cos⁡(x)  ⟹  x=π4f'(x) = \cos(x) - \sin(x) = 0 \implies \sin(x) = \cos(x) \implies x = \frac{\pi}{4}
Step 3: The critical point x = π/4 is in [0, π]. Evaluate f at the critical point and both endpoints.
f(0) = \sin(0) + \cos(0) = 0 + 1 = 1$$ $$f\!\left(\frac{\pi}{4}\right) = \sin\!\left(\frac{\pi}{4}\right) + \cos\!\left(\frac{\pi}{4}\right) = \frac{\sqrt{2}}{2} + \frac{\sqrt{2}}{2} = \sqrt{2} \approx 1.414$$ $$f(\pi) = \sin(\pi) + \cos(\pi) = 0 + (-1) = -1
Step 4: Compare: f(0) = 1, f(π/4) = √2, f(π) = −1. The largest value is √2 and the smallest is −1.
Answer: The absolute maximum is √2, occurring at x = π/4. The absolute minimum is −1, occurring at x = π.

Frequently Asked Questions

What happens if the interval is open instead of closed?
The Extreme Value Theorem does not apply on open intervals. For example, f(x) = 1/x on the open interval (0, 1) is continuous there but has no absolute maximum because f(x) increases without bound as x approaches 0 from the right. Both conditions — continuity and a closed interval — are required.
What happens if the function is not continuous on the closed interval?
If the function has a discontinuity on [a, b], the theorem does not guarantee absolute extrema exist. For instance, a function could jump to arbitrarily large values at a point of discontinuity, or it could be undefined in a way that prevents it from attaining a maximum or minimum.
Does the Extreme Value Theorem tell you where the max and min occur?
No. The theorem only guarantees existence — it promises that an absolute maximum and an absolute minimum exist somewhere on [a, b]. To actually find them, you use the Closed Interval Method: compute the derivative, find all critical points in (a, b), evaluate the function at those points and at the endpoints, then compare.

Extreme Value Theorem vs. Intermediate Value Theorem

Extreme Value TheoremIntermediate Value Theorem
What it guaranteesA continuous function on [a, b] attains an absolute maximum and an absolute minimum.A continuous function on [a, b] takes every value between f(a) and f(b).
Hypothesesf is continuous on a closed interval [a, b].f is continuous on a closed interval [a, b].
Conclusion typeExistence of extreme (largest/smallest) values.Existence of intermediate output values (roots, etc.).
Common applicationFinding absolute max/min using the Closed Interval Method.Proving a root or solution exists (e.g., showing f(x) = 0 has a solution).

Why It Matters

The Extreme Value Theorem is essential in optimization problems throughout calculus. When a problem asks you to find the absolute maximum or minimum of a continuous function on a closed interval, this theorem is the reason you know those values exist in the first place. You will encounter it in AP Calculus, college calculus courses, and applied fields like physics, engineering, and economics wherever a quantity must be maximized or minimized over a bounded domain.

Common Mistakes

Mistake: Forgetting to check the endpoints when finding absolute extrema on a closed interval.
Correction: The absolute max or min frequently occurs at an endpoint, not at a critical point. Always evaluate f at both endpoints a and b, in addition to every critical point inside (a, b), then compare all values.
Mistake: Applying the theorem to an open interval or to a function with a discontinuity.
Correction: Both hypotheses are essential: the function must be continuous AND the interval must be closed. If either condition fails, the theorem does not apply and absolute extrema are not guaranteed. Always verify both conditions before invoking the theorem.

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