4.3 Maxima and Minima

Big idea

Most applications of the derivative ask for a largest or smallest value: the cheapest design, the peak concentration, the highest point of a trajectory. This section sets up the vocabulary and the one structural fact that makes such questions finite, so that later sections can concentrate on deciding what kind of point you have found.

The structural fact is short. At an interior point where the graph turns around, the tangent line has to be horizontal, because a nonzero slope would let you improve the value by stepping one way or the other. So the only interior candidates for an extreme value are points where the derivative is zero, plus the points where the derivative does not exist at all. That turns a search over infinitely many inputs into a search over a short list.

The other half is knowing that a largest value exists in the first place. A continuous function on a closed interval that does not run off to infinity always attains a highest and a lowest value. Drop either hypothesis, continuity or the closed bounded interval, and the guarantee fails. This is why so many statements in this chapter carry the clause “continuous on $[a,b]$.”

Put the two halves together and you get a method that is mechanical on a closed interval: list the critical points inside, add the two endpoints, evaluate, compare. No test for whether a critical point is a peak or a valley is needed, because comparing values decides it. The tests for local behavior come next.

Decoder

If $f$ has a local maximum or a local minimum at an interior point $c$, and $f'(c)$ exists, then $f'(c) = 0$.

Every qualifier in that sentence matters. “Interior” rules out the endpoints of an interval, where the graph can be highest simply because you ran out of room; the derivative there need not vanish. “If $f'(c)$ exists” admits corners and cusps, like the vertex of $|x|$, which is a genuine minimum with no derivative at all.

Read the direction of the implication carefully. It says an extremum forces a zero derivative, not that a zero derivative forces an extremum. The function $f(x) = x^3$ has $f'(0) = 0$ and no extremum at $0$; the graph flattens for an instant and keeps climbing. So the critical points are candidates, not answers. Screening the candidates is the job of the first and second derivative tests in the next section.

The classic mistake is forgetting the endpoints. On $[0, 2]$ the function $f(x) = x$ has no critical point anywhere, yet it certainly has a largest value, attained at $x = 2$. Any search for an absolute extremum over a closed interval that only looks at critical points will miss it.

Definitions and results

Absolute extremum. A function $f$ has an absolute maximum at $c$ on a set $S$ if $f(c) \ge f(x)$ for every $x$ in $S$, and an absolute minimum if $f(c) \le f(x)$ for every $x$ in $S$. The location is $c$; the value is $f(c)$. Say which one you mean.

Local extremum. $f$ has a local maximum at $c$ if $f(c) \ge f(x)$ for all $x$ in some open interval around $c$, and a local minimum with the inequality reversed. Local means you only compare against nearby points.

Relationship between the two. Every absolute extremum attained at an interior point is also a local extremum. The converse fails: a local peak can sit well below the global peak.

The extreme value theorem. If $f$ is continuous on a closed bounded interval $[a,b]$, then $f$ attains an absolute maximum and an absolute minimum on $[a,b]$. The theorem asserts existence only; it gives no method for finding the points.

When the hypotheses fail. On the half open interval $(0,1]$ the continuous function $f(x) = 1/x$ has no maximum, because the interval is not closed. On $[0,2]$ the function equal to $x$ for $x < 1$ and to $0$ for $x \ge 1$ has no maximum, because it is not continuous.

Critical point. A number $c$ in the domain of $f$ is a critical point if $f'(c) = 0$ or $f'(c)$ does not exist. Points outside the domain are never critical points, no matter what the derivative formula does there.

Interior extrema occur at critical points. If $f$ has a local extremum at an interior point $c$, then $c$ is a critical point of $f$. This is the statement in the Decoder, extended to cover the case where $f'(c)$ fails to exist.

The closed interval method. To find the absolute extrema of a continuous $f$ on $[a,b]$: find all critical points in $(a,b)$; evaluate $f$ at each of them and at $a$ and at $b$; the largest of those values is the absolute maximum and the smallest is the absolute minimum.

Worked examples

Listing the critical points

Find the critical points of $f(x) = x^3 - 3x^2 - 9x + 5$.

The function is a polynomial, so it is differentiable everywhere and only the zeros of $f'$ can be critical:

$$ f'(x) = 3x^2 - 6x - 9 = 3(x^2 - 2x - 3) = 3(x - 3)(x + 1) $$

The derivative vanishes at $x = -1$ and $x = 3$, and it exists everywhere, so those two numbers are the complete list. Nothing yet says whether either is a peak or a valley.

Absolute extrema on a closed interval

Find the absolute maximum and minimum of the same $f$ on $[-2, 4]$.

The function is continuous on a closed bounded interval, so both extrema exist. Both critical points $-1$ and $3$ lie inside $(-2,4)$, so the candidate list is $-2$, $-1$, $3$, $4$. Evaluate:

$$ f(-2) = -8 - 12 + 18 + 5 = 3 $$

$$ f(-1) = -1 - 3 + 9 + 5 = 10 $$

$$ f(3) = 27 - 27 - 27 + 5 = -22 $$

$$ f(4) = 64 - 48 - 36 + 5 = -15 $$

The largest value is $10$, so the absolute maximum is $10$, attained at $x = -1$. The smallest is $-22$, so the absolute minimum is $-22$, attained at $x = 3$. Note that the right endpoint $x = 4$ produced $-15$, which beats neither, but on the interval $[-2, 3.5]$ it would have been a different story; the answer depends on the interval, not only on the function.

A critical point with no derivative

Find the absolute extrema of $f(x) = x^{2/3}$ on $[-1, 8]$.

Differentiate:

$$ f'(x) = \frac{2}{3}x^{-1/3} = \frac{2}{3\sqrt[3]{x}} $$

This is never zero, but it is undefined at $x = 0$, and $0$ is in the domain of $f$, so $x = 0$ is a critical point. The graph has a cusp there. Candidates are $-1$, $0$ and $8$:

$$ f(-1) = 1, \qquad f(0) = 0, \qquad f(8) = 4 $$

The absolute maximum is $4$ at $x = 8$, and the absolute minimum is $0$ at $x = 0$. A derivative that fails to exist is not a reason to skip a point; it is a reason to put it on the list.

A flat spot that is not an extremum

Consider $f(x) = x^3$ near $0$, and separately $g(x) = x^3 - 6x^2 + 12x - 7$.

For $f$, the derivative $3x^2$ vanishes at $x = 0$, yet $f$ is negative to the left of $0$ and positive to the right, so $0$ is neither a local maximum nor a local minimum. For $g$,

$$ g'(x) = 3x^2 - 12x + 12 = 3(x - 2)^2 $$

which vanishes only at $x = 2$ and is otherwise positive. Since $g'$ never changes sign, $g$ is increasing throughout and $x = 2$ is again a flat spot with no extremum. Check a value on each side: $g(1) = 1 - 6 + 12 - 7 = 0$ and $g(3) = 27 - 54 + 36 - 7 = 2$, with $g(2) = 8 - 24 + 24 - 7 = 1$ in between, confirming the climb straight through.

These two cases are why a critical point must be screened before it is reported as an extremum.

Practice

These drills cover the whole chain: locate the critical points of a function, decide whether an extremum is local or absolute, and run the closed interval method to produce the largest and smallest values on a given interval.

Quiz

Five items on local and absolute extrema, the hypotheses of the extreme value theorem, critical points, and the closed interval method.