Calculus 1

Extrema on a closed interval

A function continuous on a closed interval is sure to have a largest and a smallest value there. With infinitely many inputs to search, how do we find them?

You need

Critical points and local versus absolute extremaCalculus 1 The extreme value theoremCalculus 1

Unlocks

OptimizationCalculus 1

Where you are 4 in Maxima, minima and critical points

Calculus8

Applications of derivatives11

Maxima, minima and critical points4

Critical points and local versus absolute extrema

Extrema on a closed interval

Increasing and decreasing from the derivative The first and second derivative tests

When you reach for it

The situation is always the same: a function f, a closed interval [a, b], and the question “what is the largest value of f on [a, b], the smallest, and where are they reached?” These are the absolute maximum and absolute minimum of f on [a, b], defined in Critical points and local versus absolute extrema.

If f is continuous on [a, b], the The extreme value theorem promises that both exist, but not where. Critical points are the interior inputs where f'(c) = 0 or f'(c) does not exist, and every local extremum sits at one. Together the two results turn an infinite search into a comparison of a few numbers. Unlike reading a graph (Turning points, maximums and minimums) or completing the square (Maximum, minimum and projectile problems), this works for any function we can differentiate, with proof that nothing was missed.

The procedure

The method rests on two facts, both already proved.

The extremes exist. Since f is continuous on [a, b], the extreme value theorem gives an input c where f takes its absolute maximum value M, and an input d where it takes its absolute minimum value m.

They can only be in a few places. Suppose the absolute maximum is at c and c is not an endpoint. Then c lies in the open interval (a, b), and f(c) \ge f(x) for every x in that open interval. That is exactly a local maximum at c, so c is a critical point: if f'(c) exists it equals 0 by Fermat’s theorem, and otherwise f'(c) does not exist. The same holds for the minimum. So the absolute extremes are at the endpoints or at critical points in (a, b), as Critical points and local versus absolute extrema showed.

Put together, every input where an absolute extreme occurs is on the candidate list: the two endpoints a and b, and the critical points of f that lie in (a, b). Usually the list is short. We evaluate f at every candidate, and

the largest of these values is the absolute maximum M, because M is one of them (its input is on the list) and M \ge f(x) for every x, in particular for every other candidate;
the smallest of these values is the absolute minimum m, for the same reason.

The answer is a value together with every input where it is reached: “the absolute maximum value is 32, at x = 0.”

The method never asks whether a critical point is a peak, a valley or neither, and it never needs the derivative at an endpoint. It does need continuity on the closed interval, checked first, and every critical point in (a, b), including those where f' does not exist. A critical point outside [a, b] is not a candidate.

Worked examples

Example 1

Rihards’s cubic

Rihards wants the absolute maximum and minimum values of f(x) = x^3 - 6x^2 + 32 on [-1, 5].

Solution

A polynomial is continuous everywhere, so both extremes exist. It is also differentiable everywhere, so its critical points are the solutions of f'(x) = 0:

\begin{aligned} f'(x) &= 3x^2 - 12x \\ &= 3x(x - 4). \end{aligned}

The critical points are 0 and 4, both inside (-1, 5). The candidates are these two and the endpoints.

x	-1	0	4	5
f(x)	25	32	0	7

For instance f(-1) = -1 - 6 + 32 = 25 and f(4) = 64 - 96 + 32 = 0. The absolute maximum value is 32, at x = 0, and the absolute minimum value is 0, at x = 4 (Figure 1). Both endpoints lose.

Figure 1The four candidates for f(x) = x^3 - 6x^2 + 32 on [-1, 5]. The highest of the four, (0, 32), is the absolute maximum, and the lowest, (4, 0), is the absolute minimum.

Example 2

Farai’s cusp

Farai looks for the absolute extremes of g(x) = x^{2/3}(5 - x) on [-1, 4].

Solution

g is continuous on [-1, 4], since \displaystyle x^{2/3} = \left(\sqrt[3]{x}\right)^2 is defined and continuous for every real x. Write g(x) = 5x^{2/3} - x^{5/3}. For x \ne 0 the power rule gives

g'(x) = \frac{10}{3}x^{-1/3} - \frac{5}{3}x^{2/3} = \frac{5(2 - x)}{3x^{1/3}},

where the second form puts both terms over 3x^{1/3}. The numerator is 0 at x = 2. At x = 0 the formula breaks, so we use the definition:

\frac{g(h) - g(0)}{h} = \frac{h^{2/3}(5 - h)}{h} = \frac{5 - h}{h^{1/3}},

which grows without bound as h \to 0^+ and decreases without bound as h \to 0^-. So g'(0) does not exist, and since 0 is interior it is a critical point (the graph has a cusp there). The candidates are -1, 0, 2 and 4:

\begin{aligned} g(-1) &= 1 \cdot 6 = 6, \\ g(0) &= 0, \\ g(2) &= 3\sqrt[3]{4} \approx 4.76, \\ g(4) &= \sqrt[3]{16} \approx 2.52. \end{aligned}

The exact comparison is quick: \displaystyle 3\sqrt[3]{4} < 6 because \displaystyle \sqrt[3]{4} < 2, which holds since 4 < 8. The absolute maximum value is 6, at the endpoint x = -1. The absolute minimum value is 0, at the cusp x = 0; indeed x^{2/3} \ge 0 and 5 - x > 0 on [-1, 4]. Leaving out the input where g' does not exist would have lost it.

Example 3

A trigonometric function over one turn

Find the absolute extremes of f(x) = 2\sin x - x on [0, 2\pi].

Solution

f is continuous and differentiable everywhere, with f'(x) = 2\cos x - 1. Setting f'(x) = 0 gives \displaystyle \cos x = \frac{1}{2}, whose solutions in (0, 2\pi) are \displaystyle x = \frac{\pi}{3} and \displaystyle x = \frac{5\pi}{3}. Using \displaystyle \sin \frac{\pi}{3} = \frac{\sqrt{3}}{2} and \displaystyle \sin \frac{5\pi}{3} = -\frac{\sqrt{3}}{2}, the candidates give

\begin{aligned} f(0) &= 0, \\ f\left(\tfrac{\pi}{3}\right) &= \sqrt{3} - \tfrac{\pi}{3} \approx 0.685, \\ f\left(\tfrac{5\pi}{3}\right) &= -\sqrt{3} - \tfrac{5\pi}{3} \approx -6.968, \\ f(2\pi) &= -2\pi \approx -6.283. \end{aligned}

The absolute maximum value is \displaystyle \sqrt{3} - \frac{\pi}{3}, at \displaystyle x = \frac{\pi}{3}, and the absolute minimum value is \displaystyle -\sqrt{3} - \frac{5\pi}{3}, at \displaystyle x = \frac{5\pi}{3}. The minimum beats the endpoint value -2\pi by less than 1: exact values or careful decimals are needed when candidates are close.

Example 4

An exponential product

Find the absolute extremes of f(x) = (x^2 - 8)e^x on [-5, 3].

Solution

f is a product of functions continuous and differentiable everywhere. By the product rule,

\begin{aligned} f'(x) &= 2x\,e^x + (x^2 - 8)e^x \\ &= e^x(x + 4)(x - 2). \end{aligned}

Since e^x > 0 for every x, f'(x) = 0 exactly when x = -4 or x = 2, both in (-5, 3). The values are

\begin{aligned} f(-5) &= 17e^{-5} \approx 0.115, \\ f(-4) &= 8e^{-4} \approx 0.147, \\ f(2) &= -4e^2 \approx -29.56, \\ f(3) &= e^3 \approx 20.09. \end{aligned}

The only negative value is -4e^2, so the absolute minimum value is -4e^2, at x = 2. Among the positive values, e^3 beats 8e^{-4} because e^7 > 2^7 > 8, and it beats 17e^{-5} because e^8 > 2^8 > 17. The absolute maximum value is e^3, at x = 3. The critical point -4 is a candidate that wins nothing.

Example 5

When the interval decides

Let k(x) = x^2 + \dfrac{16}{x}. Find its absolute extremes on (a) [-4, -1] and (b) [1, 4]. (c) Explain why the method cannot be used on [-1, 1].

Solution

k is defined and continuous for every x \ne 0, and

k'(x) = 2x - \frac{16}{x^2} = \frac{2(x^3 - 8)}{x^2}.

The numerator is 0 only at x = 2, the one real solution of x^3 = 8. The derivative also fails to exist at 0, but 0 is not in the domain, so it is not a critical point. The only critical point of k is 2.

(a) 0 is not in [-4, -1], so k is continuous there. The critical point 2 lies outside the interval, so the only candidates are the endpoints: k(-4) = 16 - 4 = 12 and k(-1) = 1 - 16 = -15. The absolute maximum value is 12, at x = -4, and the absolute minimum value is -15, at x = -1.

(b) Now 2 is inside, and the candidates give k(1) = 17, k(2) = 4 + 8 = 12 and k(4) = 16 + 4 = 20. The absolute maximum value is 20, at x = 4, and the absolute minimum value is 12, at x = 2.

(c) k is not defined at 0, which lies in [-1, 1], so k is not continuous there and the extreme value theorem does not apply. There are no extremes at all: k grows without bound as x \to 0^+ and decreases without bound as x \to 0^-. Comparing k(-1) and k(1) would give two wrong answers.

Example 6

Farai’s tie

Farai finds the absolute extremes of f(x) = x^3 - 6x^2 + 9x - 1 on [0, 4].

Solution

f'(x) = 3x^2 - 12x + 9 = 3(x - 1)(x - 3), so the critical points are 1 and 3, both in (0, 4). The values are

\begin{gathered} f(0) = -1, \quad f(1) = 3, \\ f(3) = -1, \quad f(4) = 3. \end{gathered}

The largest value, 3, appears twice, and so does the smallest, -1. The absolute maximum value is 3, reached at x = 1 and at x = 4; the absolute minimum value is -1, reached at x = 0 and at x = 3 (Figure 2). An absolute extreme value is one number, but it may be reached at several inputs, and the answer names all of them. Expanding confirms 3 - f(x) = -(x - 1)^2(x - 4) and f(x) + 1 = x(x - 3)^2, both \ge 0 on [0, 4].

Figure 2For x^3 - 6x^2 + 9x - 1 on [0, 4] the maximum value 3 is reached twice, at an interior peak and at an end, and so is the minimum value -1.

Example 7

Moussa’s water depth

Moussa models the water depth at a marker post, in centimeters, by

D(t) = 0.05t^3 - 1.05t^2 + 6t + 50,

for 0 \le t \le 12, where t is the number of months after the start of January. Find the greatest and least depths over the year, and when they occur.

Solution

D is a polynomial on the closed interval [0, 12], so both extremes exist. Its derivative is

\begin{aligned} D'(t) &= 0.15t^2 - 2.1t + 6 \\ &= 0.15(t - 4)(t - 10), \end{aligned}

so the critical points are t = 4 and t = 10. The candidates give

\begin{gathered} D(0) = 50, \quad D(4) = 60.4, \\ D(10) = 55, \quad D(12) = 57.2. \end{gathered}

For example D(4) = 3.2 - 16.8 + 24 + 50 = 60.4. The greatest depth is 60.4 centimeters, at t = 4 (the start of May). The least depth is 50 centimeters, at t = 0 (the start of January). The dip to 55 centimeters at t = 10 is not the least depth: the water was lower at the start of the year.

Boundaries

f must be continuous on the whole closed interval. One break and the extremes may not exist: \displaystyle k(x) = x^2 + \frac{16}{x} has none on [-1, 1] (Example 5). The candidate list still produces numbers, which is why continuity is checked first.
The interval must be closed and bounded. On (a, b), [a, b) or [a, \infty) the The extreme value theorem promises nothing, and the largest candidate value can be wrong. For x^3 on (-1, 2) the only candidate is 0, yet there is no extreme: the values approach -1 and 8 at the open ends without reaching them.
Every critical point in the interval counts, and only those. An input where f' does not exist is a candidate (Example 2). A critical point outside [a, b] is not (Example 5).
It finds absolute extremes only. Whether a critical point that loses the comparison is a local maximum, a local minimum or neither is a separate question, settled by The first and second derivative tests.

Where this goes next

In Optimization the function must first be built from a situation, and the closed interval comes from its physical limits; after that, the comparison is the one done here. Increasing and decreasing from the derivative and The first and second derivative tests describe what happens between the candidates. The argument in Exercise 23, that a function equal at both ends must have an interior extreme, is the argument behind The mean value theorem, read here from the candidate list. The wider family of maxima and minima is in Maxima, minima and critical points.

Exercises

Polynomials and powers

For Exercises 1 to 14, find the absolute maximum and minimum values of the function on the interval, and every input where each occurs.

f(x) = x^3 - 3x^2 - 9x + 1 on [-2, 4]
f(x) = x^4 - 2x^2 + 5 on [-2, 3]
f(x) = 12x - x^3 on [0, 3]
f(x) = x^3 - 48x + 5 on [5, 6]
f(x) = x^{2/3} on [-8, 1]
f(x) = x^{4/3} - 16x^{1/3} on [-1, 8]
\displaystyle f(x) = \sqrt{x}\,(6 - x) on [0, 9]
f(x) = (x^2 - 1)^{2/3} on [-1, 3]

Trigonometric, exponential and rational functions

f(x) = 2\cos x + \sin 2x on [0, \pi]
f(x) = \cos 2x + 2\sin x on [0, \pi]
f(x) = (x - 2)e^x on [0, 3]
f(x) = e^x - 3x on [0, 2]
f(x) = \dfrac{x}{x^2 - x + 1} on [0, 3]
f(x) = x + \dfrac{4}{x} on \displaystyle \left[\frac{1}{2}, 4\right]

In context

Farai’s remote-control car moves along a straight track with speed v(t) = t^3 - 9t^2 + 24t feet per second, t seconds after it starts, for 0 \le t \le 5. Find its greatest and least speeds and when they occur.
Rihards reads that the concentration of a medicine in the blood t hours after a dose is modeled by C(t) = \dfrac{5t}{t^2 + 4} milligrams per liter, for 0 \le t \le 6. Find the highest and lowest concentrations in that time and when they occur.

Open intervals and broken hypotheses

Find the absolute extremes of f(x) = x^2 - 4x on the open interval (0, 3), or explain why one does not exist.
Moussa evaluates f(x) = \dfrac{1}{x - 2} at the ends of [0, 3] and reports a minimum of \displaystyle -\frac{1}{2} and a maximum of 1. Explain why the method does not apply, and decide whether f has absolute extremes on [0, 3].
Find the absolute extremes of f(x) = x^3 - 3x on the open interval (-2, 2), and prove your answer by factoring.

Reasoning

Explain why the method never needs to decide whether a critical point is a peak, a valley or neither.
Give a function continuous on [0, 1] with no critical points, find its absolute extremes, and explain what this shows about the endpoints.
Let f be continuous on [a, b] and differentiable on (a, b) with f'(x) \ne 0 for every x in (a, b). Show that the absolute maximum and minimum of f on [a, b] are at the endpoints, one at each.
Let f be continuous on [a, b] with f(a) = f(b), and not constant. Show that f has a critical point in (a, b).

Answers

f'(x) = 3x^2 - 6x - 9 = 3(x - 3)(x + 1); candidates give f(-2) = -1, f(-1) = 6, f(3) = -26, f(4) = -19. Maximum 6 at x = -1; minimum -26 at x = 3.
f'(x) = 4x(x - 1)(x + 1); f(-2) = 13, f(-1) = 4, f(0) = 5, f(1) = 4, f(3) = 68. Maximum 68 at x = 3; minimum 4 at x = -1 and x = 1.
f'(x) = 3(2 - x)(2 + x); -2 is outside. f(0) = 0, f(2) = 16, f(3) = 9. Maximum 16 at x = 2; minimum 0 at x = 0.
f'(x) = 3(x^2 - 16) has zeros \pm 4, both outside [5, 6]. f(5) = -110, f(6) = -67. Maximum -67 at x = 6; minimum -110 at x = 5.
f'(x) = \dfrac{2}{3x^{1/3}} is never 0 and does not exist at 0, a critical point. f(-8) = 4, f(0) = 0, f(1) = 1. Maximum 4 at x = -8; minimum 0 at x = 0.
For x \ne 0, f'(x) = \dfrac{4(x - 4)}{3x^{2/3}}, zero at 4; f'(0) does not exist. f(-1) = 17, f(0) = 0, \displaystyle f(4) = -12\sqrt[3]{4} \approx -19.05, f(8) = -16. Maximum 17 at x = -1; minimum \displaystyle -12\sqrt[3]{4} at x = 4.
For x > 0, \displaystyle f'(x) = \dfrac{3(2 - x)}{2\sqrt{x}}, zero at 2; 0 is an endpoint, a candidate as such. f(0) = 0, \displaystyle f(2) = 4\sqrt{2}, f(9) = -9. Maximum \displaystyle 4\sqrt{2} at x = 2; minimum -9 at x = 9.
f'(x) = \dfrac{4x}{3(x^2 - 1)^{1/3}}: zero at 0, does not exist at 1 (and at the endpoint -1). f(-1) = 0, f(0) = 1, f(1) = 0, f(3) = 4. Maximum 4 at x = 3; minimum 0 at x = -1 and x = 1.
f'(x) = -2\sin x + 2\cos 2x, or -2(2\sin x - 1)(\sin x + 1), zero in (0, \pi) at \displaystyle \frac{\pi}{6} and \displaystyle \frac{5\pi}{6}. f(0) = 2, \displaystyle f\left(\frac{\pi}{6}\right) = \frac{3\sqrt{3}}{2}, \displaystyle f\left(\frac{5\pi}{6}\right) = -\frac{3\sqrt{3}}{2}, f(\pi) = -2. Maximum \displaystyle \frac{3\sqrt{3}}{2} at \displaystyle x = \frac{\pi}{6}; minimum \displaystyle -\frac{3\sqrt{3}}{2} at \displaystyle x = \frac{5\pi}{6}.
f'(x) = 2\cos x(1 - 2\sin x), zero at \displaystyle \frac{\pi}{6}, \displaystyle \frac{\pi}{2}, \displaystyle \frac{5\pi}{6}. Values 1, \displaystyle \frac{3}{2}, 1, \displaystyle \frac{3}{2}, 1 at 0, \displaystyle \frac{\pi}{6}, \displaystyle \frac{\pi}{2}, \displaystyle \frac{5\pi}{6}, \pi. Maximum \displaystyle \frac{3}{2} at \displaystyle x = \frac{\pi}{6} and \displaystyle x = \frac{5\pi}{6}; minimum 1 at x = 0, \displaystyle x = \frac{\pi}{2} and x = \pi.
f'(x) = (x - 1)e^x, zero at 1. f(0) = -2, f(1) = -e, f(3) = e^3; -e < -2 since e > 2. Maximum e^3 at x = 3; minimum -e at x = 1.
f'(x) = e^x - 3, zero at \ln 3 \approx 1.10. f(0) = 1, f(\ln 3) = 3 - 3\ln 3 \approx -0.296, f(2) = e^2 - 6 \approx 1.389. Maximum e^2 - 6 at x = 2; minimum 3 - 3\ln 3 at x = \ln 3.
x^2 - x + 1 > 0 always, and f'(x) = \dfrac{1 - x^2}{(x^2 - x + 1)^2}, zero in the interval at 1. f(0) = 0, f(1) = 1, \displaystyle f(3) = \frac{3}{7}. Maximum 1 at x = 1; minimum 0 at x = 0.
f'(x) = 1 - \dfrac{4}{x^2}, zero at 2 (-2 is outside). \displaystyle f\left(\frac{1}{2}\right) = \frac{17}{2}, f(2) = 4, f(4) = 5. Maximum \displaystyle \frac{17}{2} at \displaystyle x = \frac{1}{2}; minimum 4 at x = 2.
v'(t) = 3(t - 2)(t - 4); v(0) = 0, v(2) = 20, v(4) = 16, v(5) = 20. Greatest speed 20 feet per second, at t = 2 and t = 5; least speed 0 at t = 0.
C'(t) = \dfrac{5(4 - t^2)}{(t^2 + 4)^2}, zero at t = 2. C(0) = 0, \displaystyle C(2) = \frac{5}{4}, \displaystyle C(6) = \frac{3}{4}. Highest \displaystyle \frac{5}{4} milligrams per liter at t = 2; lowest 0 at t = 0.
f'(x) = 2x - 4, zero at 2, f(2) = -4, and f(x) + 4 = (x - 2)^2 \ge 0: minimum -4 at x = 2. No maximum: f(x) = x(x - 4) < 0 on (0, 3), but f(x) \to 0 as x \to 0^+, so no value is largest.
f is undefined at 2, which is in [0, 3], so f is not continuous there and the theorem does not apply. No extremes: \displaystyle \lim_{x \to 2^+} \frac{1}{x - 2} = \infty and \displaystyle \lim_{x \to 2^-} \frac{1}{x - 2} = -\infty.
Critical points \pm 1, with f(-1) = 2 and f(1) = -2. Since 2 - f(x) = -(x + 1)^2(x - 2) \ge 0 for x < 2 and f(x) + 2 = (x - 1)^2(x + 2) \ge 0 for x > -2: maximum 2 at x = -1; minimum -2 at x = 1.
The largest candidate value is the absolute maximum whatever kind of point gives it, since the maximum exists and its input is on the list. An interior input carrying the largest value is automatically a local maximum, so a point that is neither can never win.
f(x) = 3 - 2x: f'(x) = -2, never 0, so there are no critical points. Maximum 3 at x = 0, minimum 1 at x = 1. With no critical points, only the endpoints remain, and they must be on the list.
There are no critical points in (a, b): f' exists and is not 0. The extremes exist by the extreme value theorem and are at critical points or endpoints, so both are at a or b. If f(a) = f(b), then the maximum and minimum are equal, f is constant and f' = 0, a contradiction; so one end holds the maximum and the other the minimum.
Some x_0 has f(x_0) \ne f(a). If f(x_0) > f(a), the absolute maximum (which exists) exceeds f(a) = f(b), so it is at some c in (a, b); that is a local maximum, so c is critical. If f(x_0) < f(a), use the minimum.