Calculus I · Unit 2B · lesson

The Closed Interval Method

Concept

Learning objectives

Apply the Extreme Value Theorem and closed interval method.

Absolute Extrema on a Closed Interval

Explanation

Before the formulas

The theorem in The Closed Interval Method connects global information on an interval with local derivative behavior inside it. Read the hypotheses and conclusion separately. The theorem guarantees existence of at least one point; it may not identify the point, make it unique, or place it at the midpoint.

Use a diagram to understand the claim, then return to algebra to find candidate values when the problem asks for them. A correct theorem citation should name the interval and explain why each hypothesis is satisfied.

Explanation

Absolute extrema are found by comparing a finite candidate list

On a closed interval, a continuous function must attain both an absolute maximum and an absolute minimum. Candidates occur at endpoints and critical numbers inside the interval. The method is therefore simple: find candidates, evaluate the original function, and compare outputs.

Do not compare derivative values. The derivative finds candidates; the original function decides which candidate is highest or lowest.

On a closed interval, absolute extrema may occur at interior critical numbers or at endpoints. The closed-interval method is therefore a finite comparison procedure: find every candidate, evaluate the original function, and compare the outputs.

This method is reliable because continuity on a closed bounded interval guarantees that absolute extrema exist. Without those hypotheses, a function may approach a best value without ever attaining it.

Theorem

Extreme Value Theorem

If $f$ is continuous on a closed interval $[a,b]$ , then $f$ attains an absolute maximum and an absolute minimum on $[a,b]$ .

Method

Closed interval method

• Find critical numbers inside $(a,b)$ . • Evaluate $f$ at every interior critical number. • Evaluate $f(a)$ and $f(b)$ . • Compare the function values. Largest is absolute maximum; smallest is absolute minimum.

Guided walkthrough

Absolute extrema of a cubic

Find the absolute extrema of

f(x)=x^3-3x

on $[-2,3]$ .

Answer reveal

Worked solution

Your setup or answer

Write a real attempt before opening the supplied answer.

Common mistake

Endpoints do not need to be critical numbers to contain absolute extrema. Omitting endpoints from a closed-interval problem is a reliable way to calculate the wrong answer with impressive algebra.

After the explanation

Use the section idea

Reading lens

Turn derivative signs and theorem hypotheses into a defensible account of extrema, monotonicity, concavity, and global shape.

Mental model

Critical numbers divide the domain into testable intervals; endpoints and discontinuities keep local evidence from becoming an unjustified global claim.

Decision

List the domain and candidates, test derivative signs, compare endpoint values, and verify each theorem's hypotheses explicitly.

Common trap

Calling every point with f-prime zero an extremum or every point with f-double-prime zero an inflection point.

Check yourself

Can every turn, bend, endpoint result, and asymptote in your sketch be traced to algebraic evidence?

Source & rights

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