Extrema

Learn local and absolute minima, closed-interval extrema.

Lesson 2 of 32 parts Practice Flashcards

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Local and Absolute Minima

Objective: Explain local and absolute minima, apply the stated method, and check its conditions.

A derivative changing from negative to positive at a critical input gives a local minimum. An absolute minimum requires comparison over the full domain or a supporting global argument. An upward parabola’s vertex is its absolute minimum on the real line. A local minimum is not automatically the absolute minimum.

Formulas

First derivative test: − to + gives a local minimum

Worked examples

Try itFind the minimum value of f(x)=(x−7)²+7 over all real x.Show answer

Step 1: f′(x)=2(x−7) gives critical input 7. Step 2: The derivative changes from negative to positive. Step 3: f(7)=7, the global minimum.

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Local and Absolute Minima

Objective: Explain local and absolute minima, apply the stated method, and check its conditions.

Formulas

First derivative test: − to + gives a local minimum

Worked examples

Try itFind the minimum value of f(x)=(x−7)²+7 over all real x.Show answer

Step 1: f′(x)=2(x−7) gives critical input 7. Step 2: The derivative changes from negative to positive. Step 3: f(7)=7, the global minimum.

Closed-Interval Extrema

Objective: Explain closed-interval extrema, apply the stated method, and check its conditions.

A continuous function on a closed bounded interval attains absolute maximum and minimum values. Evaluate interval endpoints and interior critical points, then compare outputs. Endpoint candidates matter even when the derivative does not vanish there. Never omit endpoints in a closed-interval extrema problem.

Formulas

Candidates: endpoints and interior critical points

Worked examples

Try itFind the maximum of f(x)=x² on the closed interval [0,7].Show answer

Step 1: Check endpoints and any interior critical inputs. Step 2: f(0)=0 and f(7)=49; f′(x)=2x is positive inside. Step 3: The maximum is 49.

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