Absolute Extrema on a Closed Interval

A highest and a lowest value

A closed interval [a, b] contains both of its ends: every x with $a\le x\le b$a ≤ x ≤ b. The extreme value theorem says that a function continuous on a closed interval reaches a largest value and a smallest value on it, each at some x in the interval. These are the absolute maximum and absolute minimum on [a, b].

The theorem says they exist. It does not say where they are; that takes calculus.

An extreme at an end

Take $f\left(x\right)=x^2-4x+3$f(x) = x² − 4x + 3 on [0, 3]. Its derivative is f'(x) = 2x − 4. At the left end, f'(0) = −4, and at the right end, f'(3) = 6 − 4 = 2. Neither is 0.

Yet f(0) = 3 is the highest value on the interval. The curve is still falling at x = 0; it is highest there only because the interval starts there. Solving f'(x) = 0 cannot find a maximum like this, because the tangent at it is not level.

Every candidate

Inside the interval, a highest or lowest value is a local extreme, so it sits at a critical point, where f' is zero or does not exist. Otherwise it sits at an end. There is nowhere else.

That gives the closed interval method. Find the critical points of f that lie strictly between a and b. Work out f at each of them, and at a and b. The largest of these values is the absolute maximum and the smallest is the absolute minimum.

No test is needed to classify the critical points: comparing the values does that.

Comparing the values

For $x^2-4x+3$x² − 4x + 3 on [0, 3], f'(x) = 2x − 4 = 0 gives x = 2, which lies inside the interval. The candidates are x = 0, x = 2 and x = 3.

f(0) = 3. f(2) = 4 − 8 + 3 = −1. f(3) = 9 − 12 + 3 = 0.

The largest value is 3, so the absolute maximum is 3, at the end x = 0. The smallest is −1, so the absolute minimum is −1, at the critical point x = 2. The value 0 at the other end is neither.

The gold curve $y=x^2-4x+3$y = x² − 4x + 3, drawn from x = 0 to x = 3, with the three candidates marked. The highest is the left end, (0, 3), where the curve is still falling; the lowest is the trough, (2, −1).

The same curve, different intervals

Take $f\left(x\right)=x^3-3x$f(x) = x³ − 3x. Its derivative $3x^2-3$3x² − 3 is zero at x = −1 and x = 1.

On [0, 3], only x = 1 lies inside. The values are f(0) = 0, f(1) = 1 − 3 = −2 and f(3) = 27 − 9 = 18. The absolute minimum is −2 at x = 1, and the absolute maximum is 18 at the end x = 3.

On [−2, 2], both critical points lie inside. The values are f(−2) = −8 + 6 = −2, f(−1) = −1 + 3 = 2, f(1) = −2 and f(2) = 8 − 6 = 2. The absolute maximum is 2, reached twice, at x = −1 and at x = 2; the absolute minimum is −2, reached at x = −2 and at x = 1. An absolute maximum is one value, but it can be taken at more than one x.

The gold curve $y=x^3-3x$y = x³ − 3x on [−2, 2]. The gold line y = 2 touches it at the peak x = −1 and at the end x = 2; the dashed line y = −2 at the end x = −2 and the trough x = 1.

A critical point with no derivative

Take $f\left(x\right)=x^{\frac{2}{3}}$f(x) = x^(2/3) on [−1, 8]. Its derivative $\frac{2}{3\sqrt[3]{x}}$2/(3∛x) is never 0, and does not exist at x = 0, which is inside the interval. So the candidates are x = −1, x = 0 and x = 8.

f(−1) = 1, since the cube root of −1 is −1 and its square is 1. f(0) = 0. $f\left(8\right)=2^2=4$f(8) = 2² = 4. The absolute maximum is 4 at the end x = 8, and the absolute minimum is 0 at the cusp x = 0. Solving f' = 0 alone would have found no candidate inside at all.

When the theorem does not apply

Take $f\left(x\right)=\frac{1}{x}$f(x) = 1/x on (0, 2], the interval with the end 0 left out. Close to 0 the values grow without bound: f(0.01) = 100, f(0.0001) = 10000. There is no largest value, because whatever value is reached, a smaller x gives a larger one. The smallest value, $\frac{1}{2}$1/2 at x = 2, does exist.

Continuity matters as much as the ends. On the closed interval [−1, 1], $\frac{1}{x}$1/x is not continuous, since it has no value at 0, and it has neither a maximum nor a minimum there: it runs off upwards just right of 0 and downwards just left of it.

So check both conditions, closed and continuous, before promising that a maximum exists.

The gold curve $y=\frac{1}{x}$y = 1/x on (0, 2]. It reaches its lowest value, $\frac{1}{2}$1/2, at the end x = 2, which is included; toward the missing end at 0 it climbs past every height.

The usual mistakes

Comparing only the critical points. For $x^2-4x+3$x² − 4x + 3 on [0, 3], the one critical point gives −1, and the maximum, 3, is at an end.

Comparing only the ends. For $x^3-3x$x³ − 3x on [0, 3], the ends give 0 and 18, and the minimum, −2, is inside at x = 1.

Including a critical point outside the interval. For $x^3-3x$x³ − 3x on [0, 3], x = −1 is not in the interval, and its value 2 is not a candidate.

Giving x when the value is asked for. The absolute maximum of $x^2-4x+3$x² − 4x + 3 on [0, 3] is 3; it occurs at x = 0.

A delivery van between two speeds

In the application below, a van must keep between 60 and 120 km/h. The fuel it uses has one stationary point inside that range, and the greatest use is at an end.