The extreme value theorem
Before hunting for the largest and smallest value of a function on an interval, can we be sure there is a largest and a smallest value to find?
Where you are 5 in Continuity
Why this exists
Many questions ask for the largest or smallest value of a quantity: the greatest area, the least cost, the highest point of a path. Each one assumes there is a largest or smallest value to find. That assumption can fail. On the interval (0, 1] the function f(x) = \dfrac{1}{x} has no largest value: f(0.1) = 10, f(0.001) = 1000, and every value is beaten by a value closer to 0.
So before any search for a highest or lowest value, there is an earlier question: does one exist? The extreme value theorem answers it for a large and common class of functions, the functions continuous on a closed interval. It is the second existence theorem about continuous functions, beside the intermediate value theorem of The intermediate value theorem.
Definition
Let f be defined on an interval I.
- f has an absolute maximum on I at c if c is in I and f(c) \ge f(x) for every x in I. The number f(c) is the absolute maximum value.
- f has an absolute minimum on I at d if d is in I and f(d) \le f(x) for every x in I. The number f(d) is the absolute minimum value.
Together these are the absolute extreme values of f on I (see Maxima, minima and critical points for the whole family). The value is unique, but it may be reached at more than one input: \cos x on [0, 4\pi] reaches its maximum 1 at 0, 2\pi and 4\pi.
Two words must be kept apart. f is bounded above on I if some number B has f(x) \le B for every x in I. The smallest such B is the least upper bound, written \sup_{x \in I} f(x). An absolute maximum exists exactly when the least upper bound is itself a value f(c). A function can be bounded without reaching its bound: x^2 on [0, 1) stays below 1 and comes as close to 1 as we like, yet no input gives 1. Bounded below, the greatest lower bound and the absolute minimum work the same way.
The extreme value theorem. If f is continuous on the closed interval [a, b], then f has an absolute maximum and an absolute minimum on [a, b]: there are numbers c and d in [a, b] with
Intuition
Draw the graph of a continuous f from the point above a to the point above b without lifting the pencil. The pencil starts at a definite place, ends at a definite place, and never leaves the page in between. Somewhere along that unbroken stroke it is highest, and somewhere it is lowest.
Each hypothesis closes one escape route: a missing end, a jump, and an interval with no end each let the values approach a height that is never reached. Figure 2 shows one of each.
Results
Extreme value theorem. A function continuous on [a, b] reaches an absolute maximum and an absolute minimum on [a, b].
Bounded. In particular, a function continuous on [a, b] is bounded there: f(d) \le f(x) \le f(c).
Each hypothesis is needed. Drop any one and the conclusion can fail.
| Hypothesis dropped | Function | What goes wrong |
|---|---|---|
| closed (an end left out) | \dfrac{1}{x} on (0, 1] | values grow without bound near 0: no maximum |
| continuous | x^2 on [0, 1), x - 2 on [1, 2] | values approach 1 from below: no maximum |
| bounded interval | e^x on [0, \infty) | values grow without bound: no maximum |
The converse is false. A function may reach both extremes without meeting the hypotheses. The theorem gives a guarantee when its hypotheses hold and says nothing when they fail.
Existence only. The theorem says that c and d exist. It does not say where they are, how many there are, or what the extreme values equal.
With the intermediate value theorem. If f is continuous on [a, b] with minimum value m and maximum value M, then the set of values of f on [a, b] is exactly the closed interval [m, M] (Exercise 19).
Why it is true
The theorem is stated here without proof. Like the intermediate value theorem, its proof rests on the completeness of the real numbers: every nonempty set of reals with an upper bound has a least upper bound. The full proof belongs to a course in analysis. The sketch below shows where completeness enters.
Step 1: f is bounded. Suppose f were not bounded above on [a, b]. Then it is not bounded above on at least one of the two halves; keep that half and halve again, exactly as in the bisection method. The nested intervals close in on a single number p in [a, b] (completeness). Since f is continuous at p (one-sidedly if p is an end), f(x) < f(p) + 1 for every x of [a, b] close enough to p. But the kept intervals eventually lie that close to p, and f is unbounded on each of them. This is a contradiction.
Step 2: the bound is reached. By Step 1 and completeness, M = \sup_{x \in [a, b]} f(x) exists. Suppose no input gave M. Then M - f(x) > 0 on all of [a, b], so g(x) = \dfrac{1}{M - f(x)} is continuous there, and by Step 1 some K > 0 has g(x) \le K. Then M - f(x) \ge \dfrac{1}{K}, so f(x) \le M - \dfrac{1}{K} for every x: a smaller upper bound than the least one. This is a contradiction, so M = f(c) for some c. The minimum follows by applying the same argument to -f.
Worked examples
Ganni checks the hypotheses
Ganni wants to know whether each function is guaranteed an absolute maximum and an absolute minimum on the interval given. (a) f(x) = \dfrac{x + 1}{x - 4} on [0, 3]. (b) The same f on [2, 5]. (c) \ln x on (0, e].
Solution(a) f is a rational function, continuous everywhere its denominator is not zero. The denominator x - 4 is zero only at x = 4, which is not in [0, 3], so f is continuous on the closed interval [0, 3]. The theorem applies: both extremes exist.
(b) Now 4 lies in [2, 5] and f(4) is undefined, so f is not continuous on [2, 5] and the theorem says nothing. In fact both extremes fail. As x \to 4^+ the numerator approaches 5 and the denominator is a small positive number, so \lim_{x \to 4^+} f(x) = \infty: there is no maximum. Likewise \lim_{x \to 4^-} f(x) = -\infty: there is no minimum.
(c) \ln x is continuous on (0, \infty), so it is continuous on (0, e], but the interval is not closed and the theorem says nothing. Because \ln x is increasing, its maximum on (0, e] is \ln e = 1, at x = e. There is no minimum, since \lim_{x \to 0^+} \ln x = -\infty.
Pawlu’s function stays below 2
Pawlu looks at h(x) = \dfrac{2x^2}{x^2 + 3} on [0, \infty). Show that h is bounded above by 2, that no smaller number is an upper bound, and that h has no absolute maximum. Does it have an absolute minimum?
SolutionRewrite h by dividing:
The fraction \dfrac{6}{x^2 + 3} is positive for every x, so h(x) < 2: the number 2 is an upper bound, and it is never reached.
No smaller number L < 2 is an upper bound, because h(x) > L as soon as \dfrac{6}{x^2 + 3} < 2 - L, which holds for all large x. For instance, with L = 1.99, the input x = 25 gives h(25) = \dfrac{1250}{628} \approx 1.9904 > 1.99. So 2 is the least upper bound, it is not a value, and h has no absolute maximum (Figure 3).
The minimum does exist. h(x) \ge 0 for every x and h(0) = 0, so the absolute minimum value is 0, at x = 0. The function is continuous everywhere; what fails is the bounded interval.
A jump on a closed interval
Let p(x) = x^2 for 0 \le x < 1 and p(x) = x - 2 for 1 \le x \le 2 (Figure 2, B). Find the absolute extreme values of p on [0, 2], if they exist.
SolutionThe interval is closed and bounded, but p is not continuous at 1: \lim_{x \to 1^-} p(x) = 1 while p(1) = -1. The theorem does not apply, so we look directly.
On [0, 1) the values x^2 fill [0, 1). On [1, 2] the values x - 2 fill [-1, 0]. Together the values of p are [-1, 1).
The least value -1 is reached at x = 1, so the absolute minimum value is -1. There is no absolute maximum: every value is below 1, and any value x^2 with x < 1 is beaten by the value at \displaystyle \frac{x + 1}{2}, which is closer to 1. The least upper bound 1 is the height the left piece approaches, and the jump carries the graph away before it gets there.
Ofelja’s positive function
Ofelja knows that f is continuous on [a, b] and that f(x) > 0 for every x in [a, b]. Show that there is a positive number m with f(x) \ge m for every x in [a, b]. Then show that the same claim is false for f(x) = x on (0, 1].
SolutionBy the extreme value theorem, f has an absolute minimum on [a, b], at some d. Put m = f(d). Then f(x) \ge m for every x in [a, b], and m > 0 because m is one of the values of f, all of which are positive.
The theorem gave m without saying what d is; the argument needs only that d exists.
On (0, 1], f(x) = x is positive and continuous. Suppose some m > 0 had x \ge m for every x in (0, 1]. Take x to be the smaller of \displaystyle \frac{m}{2} and 1. This x lies in (0, 1], yet \displaystyle x \le \frac{m}{2} < m. So no positive number is a lower bound, and the open end at 0 is exactly what breaks the claim: the values come as close to 0 as we like without reaching it.
A rational function on a closed interval
Find the absolute extreme values of g(x) = \dfrac{6x}{x^2 + 9} on [0, 5].
SolutionThe denominator is never zero, so g is continuous on the closed interval [0, 5], and the theorem promises both extremes. Here algebra can find them.
For the maximum, compare g(x) with 1. Since x^2 + 9 > 0,
which is always true, with equality only at x = 3. So g(x) \le 1 on [0, 5] and g(3) = \dfrac{18}{18} = 1: the absolute maximum value is 1, at x = 3, inside the interval.
For the minimum, x \ge 0 makes g(x) \ge 0, and g(0) = 0: the absolute minimum value is 0, at the end x = 0. At the other end g(5) = \dfrac{30}{34} = \dfrac{15}{17}, which is neither.
The general way to locate extremes, by derivatives, is Extrema on a closed interval.
Extremes without the hypotheses
Show that each function reaches an absolute maximum and an absolute minimum, although the extreme value theorem does not apply. (a) \lfloor x \rfloor on [0, 2]. (b) \sin\left(\dfrac{1}{x}\right) on (0, 1].
Solution(a) The floor function jumps at 1 and at 2, so it is not continuous on [0, 2]. Its values there are 0 on [0, 1), 1 on [1, 2) and 2 at x = 2. The absolute maximum value is 2, at x = 2, and the absolute minimum value is 0, reached at every x in [0, 1).
(b) The interval is not closed. Still, \displaystyle -1 \le \sin\left(\frac{1}{x}\right) \le 1, and both bounds are values: \displaystyle \frac{1}{x} = \frac{\pi}{2} gives \displaystyle x = \frac{2}{\pi} \approx 0.637 with value 1, and \displaystyle \frac{1}{x} = \frac{3\pi}{2} gives \displaystyle x = \frac{2}{3\pi} \approx 0.212 with value -1. Both inputs lie in (0, 1], so the maximum is 1 and the minimum is -1.
The theorem is a one-way guarantee: continuity on a closed interval is enough for extremes, not necessary.
Boundaries
- The interval must be closed and bounded, of the form [a, b]. On (a, b], [a, b), (a, b), [a, \infty) or (-\infty, \infty) the theorem says nothing.
- f must be continuous at every point of [a, b], one-sidedly at the ends. One break anywhere is enough to lose the guarantee, as Example 3 shows.
- The conclusion is existence only. It does not locate c or d, count them, or give the extreme values. An extreme may sit at an end or inside, and may be reached at many inputs.
- When the hypotheses fail, the theorem is silent, not negative. Extremes may still exist (Example 6) or may not (Example 3).
Where this goes next
The theorem promises a highest and lowest value; Extrema on a closed interval finds them, by showing that an extreme inside the interval can only sit at one of the Critical points and local versus absolute extrema, so the candidates are those points and the two ends. The same existence guarantee is the first step in the proof of The mean value theorem, and through it the mean value theorem. The wider picture of maxima and minima, absolute and local, is in Maxima, minima and critical points.
Exercises
Does the theorem apply?
For each function and interval, say whether the extreme value theorem guarantees an absolute maximum and an absolute minimum, and why. Where it does not apply, say whether each extreme exists.
- x^4 - 3x + 7 on [-2, 4]
- \dfrac{2}{x - 1} on [-3, 0]
- \dfrac{2}{x - 1} on [0, 2]
- \ln(x + 2) on (-2, 0]
- e^{-x} on [0, \infty)
- \tan x on \displaystyle \left[0, \frac{\pi}{4}\right]
Finding the extremes by algebra
Find the absolute maximum and minimum values and where they occur, without derivatives.
- (x + 2)^2 - 3 on [-3, 1]
- 3 - \lvert x - 1 \rvert on [-2, 3]
- \dfrac{1}{x^2 + 1} on [-1, 2]
- \dfrac{x^2 - 1}{x^2 + 1} on [-2, 1]
Bounded, or reaching the bound?
Find the least upper bound and the greatest lower bound of the values, and say which of them are reached.
- 2 - \dfrac{1}{x} on [1, \infty)
- x^2 on (-1, 2)
- \dfrac{x}{x + 1} on [0, \infty)
- \dfrac{3}{x^2 + 1} on (0, 2]
The converse is false
- Let k(x) = 1 - x^2 for -1 \le x < 0 and k(x) = x for 0 \le x \le 1. Show that k is not continuous on [-1, 1], and find its absolute extreme values there.
- Find the absolute extreme values of \cos x on the open interval (0, 4\pi) and every input where each is reached.
- Give a function that is continuous on the open interval (0, 1) and has an absolute maximum and an absolute minimum there, and one that is continuous on (0, 1) and has neither.
Reasoning
- Explain why a function continuous on [a, b] is bounded there, and give a function continuous on (0, 1) that is not bounded.
- Let f be continuous on [a, b] with absolute minimum value m and absolute maximum value M. Use the extreme value theorem and the intermediate value theorem to show that the values of f on [a, b] are exactly the numbers in [m, M].
- Let f be continuous on [-1, 1] with f(-1) = f(1). Show that f is constant or has an absolute maximum or an absolute minimum at some point of the open interval (-1, 1).
- Let f be continuous on [0, \infty) with f(1) = 2 and \lim_{x \to \infty} f(x) = 0. Show that f has an absolute maximum on [0, \infty), although the theorem does not apply directly.
Answers
- Guaranteed. A polynomial is continuous everywhere, and [-2, 4] is closed and bounded.
- Guaranteed. The only break is at x = 1, which is not in [-3, 0].
- Not guaranteed: 1 lies in [0, 2] and \displaystyle \frac{2}{x - 1} is undefined there. Neither extreme exists: \displaystyle \lim_{x \to 1^+} \frac{2}{x - 1} = \infty and \displaystyle \lim_{x \to 1^-} \frac{2}{x - 1} = -\infty.
- Not guaranteed: the interval is open at -2. The maximum is \ln 2 at x = 0, since \ln(x + 2) is increasing; there is no minimum, since \lim_{x \to -2^+} \ln(x + 2) = -\infty.
- Not guaranteed: the interval is unbounded. The maximum is 1 at x = 0, since e^{-x} is decreasing. There is no minimum: the values are positive and approach 0, and 0 is never reached.
- Guaranteed. \displaystyle \tan x = \frac{\sin x}{\cos x} is continuous where \cos x \ne 0, and \cos x > 0 on \displaystyle \left[0, \frac{\pi}{4}\right].
- Minimum -3 at x = -2, since (x + 2)^2 \ge 0. Maximum 6 at x = 1, where (x + 2)^2 is largest: \lvert x + 2 \rvert \le 3 on [-3, 1], with 3 only at x = 1 (at x = -3 the function is only -2).
- Maximum 3 at x = 1, since \lvert x - 1 \rvert \ge 0. Minimum 0 at x = -2, where \lvert x - 1 \rvert reaches its largest value 3 on [-2, 3] (at x = 3 it is only 2).
- Maximum 1 at x = 0, where x^2 + 1 is least. Minimum \displaystyle \frac{1}{5} at x = 2, where x^2 + 1 is largest (x^2 \le 4 on [-1, 2], equal to 4 only at x = 2).
- Write it as 1 - \dfrac{2}{x^2 + 1}. It is least when x^2 is least: minimum -1 at x = 0. It is largest when x^2 is largest, x^2 = 4 at x = -2: maximum \displaystyle 1 - \frac{2}{5} = \frac{3}{5}.
- Values [1, 2). Greatest lower bound 1, reached at x = 1 (absolute minimum). Least upper bound 2, not reached, since \displaystyle \frac{1}{x} > 0; no maximum.
- Values [0, 4). Greatest lower bound 0, reached at x = 0. Least upper bound 4, approached as x \to 2^- but not reached, since 2 is not in the interval; no maximum.
- Write it as 1 - \dfrac{1}{x + 1}. Values [0, 1). Greatest lower bound 0, reached at x = 0; least upper bound 1, not reached.
- Values \displaystyle \left[\frac{3}{5}, 3\right). Greatest lower bound \displaystyle \frac{3}{5}, reached at x = 2. Least upper bound 3, approached as x \to 0^+ but not reached; no maximum.
- \lim_{x \to 0^-} k(x) = 1 but k(0) = 0, so k is not continuous at 0. The left piece takes the values [0, 1) and the right piece [0, 1]. Maximum 1 at x = 1; minimum 0 at x = -1 and at x = 0.
- Maximum 1 at x = 2\pi only (the other places \cos x = 1, namely 0 and 4\pi, are left out). Minimum -1 at x = \pi and x = 3\pi.
- For example \sin(2\pi x), with maximum 1 at \displaystyle x = \frac{1}{4} and minimum -1 at \displaystyle x = \frac{3}{4} (a constant function also works). For neither: f(x) = x, whose values (0, 1) have no largest or smallest member.
- The theorem gives c and d with f(d) \le f(x) \le f(c) for every x in [a, b], so f(d) and f(c) are bounds. On (0, 1), \displaystyle \frac{1}{x} is continuous and unbounded.
- Every value lies between m and M by definition. Conversely, m = f(d) and M = f(c) for some c, d. For any N in [m, M], apply the intermediate value theorem to f on the closed interval between c and d, where f is continuous: some input gives N.
- The theorem gives an absolute maximum at some c and an absolute minimum at some d in [-1, 1]. If neither can be placed in (-1, 1), both values are reached only at the ends, so f(c) and f(d) both equal f(-1) = f(1). Then the maximum equals the minimum and f is constant.
- Since \lim_{x \to \infty} f(x) = 0, there is N > 1 with f(x) < 2 for all x > N. On [0, N], f is continuous, so it has an absolute maximum value M there, and M \ge f(1) = 2. Every x > N has f(x) < 2 \le M, so M is the absolute maximum on all of [0, \infty).