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5.2 Extreme Value Theorem, Global Versus Local Extrema, and Critical Points

Syllabus
2020
Topic
5.2
Level
—

Learning objectives

—FUN-1.C—Justify conclusions about functions by applying the Extreme Value Theorem• FUN-1.C Justify conclusions about functions by applying the Extreme Value Theorem.• FUN-1.C.1 If a function f is continuous over the interval [a, b], then the Extreme Value Theorem guarantees that f has at least one minimum value and at least one maximum value on [a, b].• FUN-1.C.2 A point on a function where the first derivative equals zero or fails to exist is a critical point of the function.• FUN-1.C.3 All local (relative) extrema occur at critical points of a function, though not all critical points are local extrema.• Enduring understanding FUN-1: Existence theorems allow us to draw conclusions about a function’s behavior on an interval without precisely locating that behavior.

Guarantee Extrema, Then Check the Candidates

The Extreme Value Theorem (EVT) is an existence guarantee. If ff is continuous on the closed interval [a,b][a,b], then ff attains at least one absolute (global) minimum value and at least one absolute maximum value somewhere on [a,b][a,b]. The theorem guarantees that these values exist; it does not locate them.

Idea Meaning Key point
Global extremum Greatest or least value on the entire stated interval May occur at an endpoint or an interior point
Local extremum Greatest or least value compared with nearby values Must occur at a critical point
Critical point A point on the function where f′(x)=0f'(x)=0 or f′(x)f'(x) does not exist It is only a candidate; it need not be an extremum

To find global extrema on [a,b][a,b]: first verify continuity on the whole closed interval. Next find every interior point where f′(x)=0f'(x)=0 or f′(x)f'(x) does not exist. Evaluate ff at those critical points and at both endpoints. The largest output is the global maximum value, and the smallest is the global minimum value.

For f(x)=x2−2xf(x)=x^2-2x on [0,3][0,3], continuity guarantees both global extrema. Since f′(x)=2x−2f'(x)=2x-2, the only interior critical point is x=1x=1. Compare f(0)=0f(0)=0, f(1)=−1f(1)=-1, and f(3)=3f(3)=3. Thus the global minimum is −1-1 at x=1x=1, while the global maximum is 33 at the endpoint x=3x=3.

Do not reverse the critical-point statement. Every local extremum is a critical point, but a critical point may be neither a maximum nor a minimum: for f(x)=x3f(x)=x^3, f′(0)=0f'(0)=0, yet the function keeps increasing through x=0x=0.