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Maxima, minima, critical numbers

Optimization starts with a fact from geometry: at a smooth peak, the curve is momentarily flat, so the derivative is zero.

Extrema defined

A function has a local maximum at c if every nearby point gives a value no larger than f(c), and a local minimum if no smaller. A value that is largest or smallest over the entire domain is an absolute maximum or minimum. On a closed interval the Extreme Value Theorem guarantees both an absolute max and an absolute min exist for any continuous function.

f(c) \geq f(x) \text{ for all } x \text{ near } c

Critical numbers

A critical number of f is an interior point c where f'(c) = 0 or where f' does not exist. Fermat's theorem says every local extremum of a differentiable function happens at a critical number, so these points are the only candidates. You must still test them, since a critical number can also be a plateau like x = 0 for x^3.

f'(c) = 0 \quad \text{or} \quad f'(c) \text{ undefined}

Worked example

Find the absolute extrema of f(x) = x^3 - 3x on the closed interval from -2 to 3. The derivative 3x^2 - 3 is zero at x = -1 and x = 1. Comparing f(-2) = -2, f(-1) = 2, f(1) = -2, and f(3) = 18, the absolute maximum is 18 at x = 3 and the absolute minimum is -2 at both endpoints and x = 1.

f'(x) = 3x^2 - 3 = 0 \implies x = \pm 1

Flashcard

f'(c) = 0

Stationary point, candidate for local extremum

Where a differentiable function has a local max or min, its derivative must be zero by Fermat's theorem. The converse fails: a zero derivative can also sit on a level stretch like x cubed at zero.

Check your understanding

For f(x) = x^2 - 4x + 5, where is the local minimum and what is its value?

Take the unit quiz

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