Minimum and Maximum Values

Maximum and minimum values are simply the highest and lowest points a function reaches. ("Extrema" is the umbrella word for both, and "extremum" is one of them.) Finding them is the whole point of optimization: the cheapest design, the strongest beam, the best price. The two theorems look formal, but they say things you'd guess. Fermat's theorem says a peak or valley ($f$ has a relative extremum at $c$) with a slope there ($f'(c)$ exists) must have a flat slope ($f'(c) = 0$). The Extreme Value Theorem says that if the graph has no breaks ($f$ continuous) and you look at a stretch that includes its ends ($[a, b]$), there's always a highest and a lowest point. Bear with the vocabulary; the ideas are friendly.

Definitions. Picture a mountain range. The summit of the tallest mountain is the absolute maximum: $f(c) \ge f(x)$ for every $x$ in the domain. Every other hilltop is a relative maximum: it's higher than everything near it, so $f(c) \ge f(x)$ for all $x$ near $c$, even if there's a taller peak somewhere else. Minimums are the same with $\le$.

Fermat's theorem is what we argued last lesson: at a peak or valley the tangent is flat, or there's no tangent at all. So relative extrema always hide among the critical points. That's why we found critical points first.

A common mistake. Fermat's theorem only goes one way, and it's tempting to run it backward: "$f'(c) = 0$, so there's a max or min at $c$." That's not what it says. $x^3$ has $f'(0) = 0$ but no extremum at all. Flat slope is necessary for a smooth peak, not enough to guarantee one.

The Extreme Value Theorem promises absolute extrema exist, which is good to know before you go looking. But it needs both of its conditions, and you can see why by breaking each one:

• On an open interval, $x^2$ on $(-1, 2)$ gets closer and closer to 4 but never reaches it, since $x = 2$ isn't allowed. There's no largest value.
• With a discontinuity, a function can jump right past where the max would have been.

Before searching for the best design, engineers check that a best one exists: a continuous cost over a closed range of options always has a minimum.