Four Numbers in a Column, and Two of Them Are Endpoints

You'll learnto build the complete candidate list on a closed interval — every critical point inside it, plus both ends — evaluate f at every entry, and read the answer out of the column of values instead of the column of slopes.

The Extreme Value Theorem promised that a continuous function on a closed, bounded interval actually reaches a highest and a lowest value. Finding them is a three-line procedure: list every critical point inside the interval, add both endpoints, evaluate f at every entry. The largest result is the absolute maximum and the smallest is the absolute minimum. Almost nobody gets that arithmetic wrong. What goes wrong is the list — two entries left off it — or the column, when a table of slopes gets compared instead of a table of heights.

8 STEPS · 6 QUICK CHECKS · EVERY CRITICAL POINT INSIDE, PLUS BOTH ENDS · v1

f(x) = x³ − 3x² − 9x + 5 ON [−2, 6] ⇒ CANDIDATES −2, −1, 3, 6

Before you start

What you're looking at

One cubic, f(x) = x³ − 3x² − 9x + 5, drawn only across the closed interval [−2, 6]. A dashed wall stands at each end of the interval, because the ends are part of the problem rather than the edge of the picture.

The question

Which input on this interval gives the largest value of f, and which gives the smallest? The Extreme Value Theorem already promised both exist. The Candidates Test is how you find them.

Watch for

The only relative maximum on the whole interval is worth 10, and it loses to a right endpoint worth 59 — a point with no local claim to anything at all.

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