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Calculus I · Maxima, Minima, and the Shape of a Function

Where extrema hide: critical points

You cannot search every point of a curve. The first job is to narrow the hunt, and Chapter 6 already did the hard part. Recall Fermat’s lemma: at an interior peak or valley where the derivative exists, the derivative is zero — the curve is momentarily flat at the top of a hill or the bottom of a dip. Promoted to a search strategy, this is:

Fermat’s Theorem. If ff has a local maximum or minimum at an interior point cc, then either f′(c)=0f'(c) = 0 or f′(c)f'(c) does not exist.

A point where f′(c)=0f'(c) = 0 or f′f' fails to exist is called a critical point. Fermat’s theorem says: every interior extremum is a critical point. So the complete list of places an extremum can hide is short — the critical points, plus (on a closed domain) the endpoints. Everything else is ruled out before you start. That is the entire reason “set the derivative to zero” is the first move in every optimization problem: it’s not a ritual, it’s Fermat’s theorem narrowing an infinite search to a finite list.

A warning the theorem makes precise: f′(c)=0f'(c) = 0 is necessary but not sufficient. The function f(x)=x3f(x) = x^3 has f′(0)=0f'(0) = 0, yet 00 is neither a max nor a min — the curve flattens for an instant and keeps climbing. So a critical point is a candidate, not a verdict. The next two tests deliver the verdict.

The first-derivative test: watch the slope change sign

At a critical point, ask what the slope is doing on either side — and Chapter 6’s Consequence 2 (the sign of f′f' controls rising and falling) tells you what that means for the shape:

  • If f′f' goes from positive to negative as xx passes through cc — rising, then falling — then cc is a local maximum. You climbed to a peak and came down.
  • If f′f' goes from negative to positive — falling, then rising — then cc is a local minimum. You descended to a valley floor and climbed out.
  • If f′f' does not change sign, cc is neither (like x3x^3 at 00: positive slope on both sides, just briefly zero).

That’s the whole test, and it rests entirely on monotonicity — which rests on the MVT. (Figure 8.2: a curve annotated with ++ and −- signs for f′f' on each interval, with peaks where +→−+\to- and valleys where −→+-\to+.)

Figure 8.2
Figure 8.2

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