Extreme Values of Functions on Closed Intervals

What are the absolute maximum and minimum of a function?

A function f with domain D has an absolute maximum value at c if f(x)\le f(c) for all x in D, and an absolute minimum value at c if f(x)\ge f(c) for all x in D. Maximum and minimum values are called extreme values (or extrema) of the function, and absolute maxima and minima are also called global maxima and minima. The value is a number, f(c), and the place where it is taken is a point, c: the maximum of y=x^2 on [0,2] is 4, taken at x=2.

The extremes depend on the domain and not only on the formula. On the closed interval \left[-\fracπ2,\fracπ2\right] the function \cos x takes the absolute maximum 1 (once, at 0) and the absolute minimum 0 (twice, at both ends), while \sin x takes the maximum 1 and the minimum -1. A value can be taken at more than one place.

Example 1. All four functions below have the same rule y=x^2, but the domains differ.

So a function might have no maximum or minimum if the domain is not bounded or does not contain an endpoint. On (0,2] the values 0.01, 0.0001, 0.000001, … keep getting smaller, and there is no smallest one: 0 would be the smallest, but 0 is not in the domain. The next section says exactly when this cannot happen.

When is a maximum or a minimum guaranteed?

The Extreme Value Theorem says that if f is continuous on a closed interval [a,b], then f attains both an absolute maximum value M and an absolute minimum value m in [a,b]. That is, there are numbers x_1 and x_2 in [a,b] with f(x_1)=m, f(x_2)=M, and m\le f(x)\le M for all x in [a,b]. The proof needs a careful knowledge of the real numbers (their completeness) and is not given here. It also does not say where the extremes are: they may be at interior points, at the endpoints, or one of each.

The three requirements, closed, bounded and continuous, are all essential. Without one of them the conclusion can fail: - not closed: y=x^2 on (0,2] has no minimum, and on (0,2) neither; - not bounded: y=x^2 on (-∞,∞) has no maximum, and y=e^x on (-∞,∞) has neither; - not continuous: the function y=x for 0\le x<1 and y=0 at x=1 is continuous at every point of [0,1] except x=1, yet it has no highest point: the values come as near to 1 as we like and never reach it.

The theorem gives a guarantee and not an equivalence. A function that fails a hypothesis may still have extremes (the function above has the minimum 0), but only when the three hold are we sure. Look at the six cases below, and at which condition each one breaks.

What are local extreme values?

A function f has a local maximum at c if f(x)\le f(c) for all x of the domain in some open interval around c, and a local minimum if f(x)\ge f(c) there; local extrema are also called relative extrema. The word local means that only the neighbours of c are compared, and not the whole domain. At an endpoint x=a of [a,b] the neighbours are the x in a half-open interval [a,a+\delta), and at x=b those in (b-\delta,b].

An absolute maximum is also a local maximum, because it is the largest value overall and so also the largest in its neighbourhood; the list of local maxima therefore includes the absolute maximum if there is one, and the same holds for minima. The converse is false. For f(x)=x^3-3x on [-2,3] the local maximum f(-1)=2 is far below the absolute maximum f(3)=18, and the local minimum f(1)=-2 equals the endpoint value f(-2)=-2, so the absolute minimum is taken twice.

A function can have many local extrema, even on a finite interval: f(x)=\sin\frac1x on (0,1] has infinitely many.

What is a critical point, and why do extremes sit on them?

First Derivative Theorem for local extreme values: if f has a local maximum or minimum at an interior point c of its domain, and f' is defined at c, then f'(c)=0. The tangent line there has to be horizontal, because a slope that is not zero would let f climb on one side of c.

Proof. Suppose f has a local maximum at c, so f(x)-f(c)\le0 for x near c. Because c is an interior point, f'(c)=\lim_x→ c(f(x)-f(c))/(x-c) is a two-sided limit: the right-hand and left-hand limits both exist and equal f'(c). From the right, x-c>0 and f(x)-f(c)\le0, so the quotient is \le0: f'(c)=\lim_x→ c^+(f(x)-f(c))/(x-c)\le0. From the left, x-c<0 and f(x)-f(c)\le0, so the quotient is \ge0: f'(c)=\lim_x→ c^-(f(x)-f(c))/(x-c)\ge0. Together f'(c)=0. For a local minimum reverse the inequalities. ∎

Where extremes can be. Since all the domains here are intervals, a function can have an extreme value (local or absolute) only at 1. interior points where f'=0, 2. interior points where f' does not exist, 3. endpoints of the interval.

Definition. An interior point of the domain of f where f' is zero or does not exist is a critical point of f. So the only places where a continuous function on a closed, bounded interval can take an extreme value are its critical points and its endpoints. The second kind of critical point is a real possibility: the derivative does not exist at a corner, a cusp or a vertical tangent line, and the derivative function may be undefined elsewhere too.

A critical point is only a candidate. y=x^3 has y'=3x^2=0 at x=0, and y=x^1/3 has y'=\frac13x^-2/3 undefined at x=0, yet neither has a local extreme value at 0: each is increasing through the origin, and each has a point of inflection there. Conversely, an extreme value can sit where there is no derivative at all: f(x)=|x| has an absolute minimum at x=0 although it is not differentiable there, which is consistent with the theorem, because the theorem only speaks about points where f' is defined.

How do you find the absolute extrema on a closed interval?

To find the absolute extrema of a continuous function f on a finite closed interval, find all the critical points of f in the interval, evaluate f at all critical points and at the endpoints, and take the largest and the smallest of these values. The Extreme Value Theorem makes sure that the extremes exist, and the First Derivative Theorem makes sure that no other place has to be examined.

Example 2. Find the absolute maximum and minimum of f(x)=x^2 on [-2,1]. The function is differentiable everywhere, so the only critical point is where f'(x)=2x=0, namely x=0. The values are f(0)=0 f(-2)=4 f(1)=1, so the absolute maximum is 4 at x=-2 and the absolute minimum is 0 at x=0.

Example 3. Find the absolute maximum and minimum of f(x)=10x(2-\ln x) on \left[1,e^2\right]. The derivative is f'(x)=10(2-\ln x)+10x\left(-\frac1x\right) f'(x)=10(1-\ln x), which is 0 only where \ln x=1, at x=e, inside the interval. The values are f(e)=10e f(1)=20 f\left(e^2\right)=10e^2(2-2)=0. Since 10e≈27.2, the absolute maximum is 10e at the critical point x=e, and the absolute minimum is 0 at the right endpoint x=e^2.

Example 4. Find the absolute maximum and minimum of f(x)=x^2/3 on [-2,3]. The derivative f'(x)=\frac23x^-1/3 has no zero, but it is undefined at the interior point x=0, which is therefore the only critical point. The values are f(0)=0 f(-2)=4^(1/3)≈1.59 f(3)=9^(1/3)≈2.08. The absolute maximum is 9^(1/3)≈2.08 at the right endpoint x=3 and the absolute minimum is 0 at x=0, where the graph has a cusp.

Try the method yourself below: type a function and a closed interval, or choose one of the examples, and read the table of candidates.

What is the biggest box, the peak current and the highest point?

The closed interval method answers questions that ask for the best value of something: cut squares of side x from the corners of a 10 by 16 sheet and fold up the sides, and the volume is V(x)=x(10-2x)(16-2x) for 0\le x\le5, whose largest value is 144 at x=2. The derivative is V'(x)=12x^2-104x+160 V'(x)=4(3x-20)(x-2), so the critical points are x=2 and x=(20)/(3); the second is outside [0,5] and does not count. The values at the candidates are V(0)=0, V(2)=144 and V(5)=0. The absolute maximum is 144 cubic units, from squares of side 2 and a box 6 by 12 by 2, and the extreme values at the endpoints are the smallest: no box at all. Change the sheet below; for a sheet w by l the critical point inside is x=((w+l)-√(w^2-wl+l^2))/(6).

The peak of an alternating current. If i=2\cos t+2\sin t amperes at time t seconds, then i'=-2\sin t+2\cos t=0 when \tan t=1, that is t=\fracπ4 and t=(5π)/(4) in [0,2π]. The values are i\left(\fracπ4\right)=2√2, i\left((5π)/(4)\right)=-2√2 and i(0)=i(2π)=2, so the peak current, the largest magnitude, is 2√2≈2.83 A.

The highest point of a body. A body moving vertically has the height s=-\frac12gt^2+v_0t+s_0, with s in meters and t in seconds. Then s'=-gt+v_0=0 at t=(v_0)/(g), the moment the velocity is zero, and the maximum height is s_0+(v_0^2)/(2g). With v_0=20 m/s, s_0=5 m and g=9.8 m/s², that is 5+(400)/(19.6)≈25.4 m, reached after about 2.04 s.

Beyond closed intervals. If the interval is not closed or not bounded, an extreme value may not exist at all, as the four domains of y=x^2 show. Such problems need more than the three steps above: the sign of f' on both sides of each critical point, or the limits at the open ends.

How do you find extreme values in code?

In code the closed interval method is three loops: scan for the critical points, evaluate at every candidate, and keep the largest and the smallest. SymPy does it exactly: sp.solveset(df, x, interval) returns every solution of f'=0 in the interval (the older sp.solve can return only one of the periodic solutions of \cos x-\sin x=0), sp.singularities(df, x) finds where f' does not exist, and maximum and minimum from sympy.calculus.util give the answer directly. Without a CAS, scan for sign changes of f' and refine them by bisection, which is what the C++ program at the end does.

import sympy as sp
from sympy.calculus.util import maximum, minimum

x = sp.symbols('x', real=True)


def closed_interval(f, a, b):
    """The closed interval method: the critical points inside (a, b), then compare the values."""
    df = sp.diff(f, x)
    crit = set(sp.solveset(df, x, sp.Interval.open(a, b)))              # f'(c) = 0, every solution in the interval
    crit |= {c for c in sp.singularities(df, x) if a < c < b}          # f' does not exist
    values = {p: sp.simplify(f.subs(x, p)) for p in sorted({a, b, *crit}, key=float)}
    return values, max(values.values(), key=float), min(values.values(), key=float)


examples = (
    (x**2, -2, 1),                                       # Example 2
    (10 * x * (2 - sp.log(x)), 1, sp.E**2),              # Example 3
    (x * (10 - 2 * x) * (16 - 2 * x), 0, 5),             # the box of Exercise 67
    (2 * sp.cos(x) + 2 * sp.sin(x), 0, 2 * sp.pi),       # Exercise 70, the peak current
    (x**3, -1, 1),                                       # a critical point that is not an extreme value
)
for f, a, b in examples:
    print(f, '->', closed_interval(f, a, b))
# x**2 -> ({-2: 4, 0: 0, 1: 1}, 4, 0)
# 10*x*(2 - log(x)) -> ({1: 20, E: 10*E, exp(2): 0}, 10*E, 0)
# x*(10 - 2*x)*(16 - 2*x) -> ({0: 0, 2: 144, 5: 0}, 144, 0)
# 2*sin(x) + 2*cos(x) -> ({0: 2, pi/4: 2*sqrt(2), 5*pi/4: -2*sqrt(2), 2*pi: 2}, 2*sqrt(2), -2*sqrt(2))
# x**3 -> ({-1: -1, 0: 0, 1: 1}, 1, -1)

print(maximum(x**2, x, sp.Interval(-2, 1)), minimum(x**2, x, sp.Interval(-2, 1)))      # 4 0   (SymPy's own answer)
print(sp.singularities(sp.diff(x ** sp.Rational(2, 3), x), x))                         # {0}   (Example 4: f' does not exist at 0)

const best = (f, a, b, critical) => {
  const pts = [a, ...critical, b].map((x) => [x, f(x)])
  return [pts.reduce((m, p) => (p[1] > m[1] ? p : m)), pts.reduce((m, p) => (p[1] < m[1] ? p : m))]   // [max, min]
}
best((x) => x * x, -2, 1, [0])                                       // [[-2, 4], [0, 0]]
best((x) => 10 * x * (2 - Math.log(x)), 1, Math.E ** 2, [Math.E])    // [[2.718281828459045, 27.18281828459045], [7.3890560989306495, 0]]
best((x) => Math.cbrt(x) ** 2, -2, 3, [0])                           // [[3, 2.080083823051904], [0, 0]]: 0 is where f' does not exist
best((x) => x ** 3, -1, 1, [0])                                      // [[1, 1], [-1, -1]]: 0 is critical but not an extreme

#include <iostream>
#include <cstdio>
#include <cmath>
using namespace std;

double f(double x) { return 10 * x * (2 - log(x)); }     // Example 3, on [1, e^2]
double df(double x) { return 10 * (1 - log(x)); }

double clean(double v) { return fabs(v) < 1e-9 ? 0 : v; }  // round noise like 1e-15 to 0

double bisect(double lo, double hi) {                       // a zero of f' between lo and hi
  for (int i = 0; i < 100; i++) {
    double mid = (lo + hi) / 2;
    if (df(lo) * df(mid) <= 0) hi = mid; else lo = mid;
  }
  return (lo + hi) / 2;
}

int main() {
  double a = 1, b = exp(2.0), cand[12];
  int n = 0;
  cand[n++] = a;                                            // the endpoints are candidates
  double step = (b - a) / 1000;
  for (int i = 0; i < 1000; i++) {                          // step 1: the critical points, where f' changes sign
    double x0 = a + i * step, x1 = x0 + step;
    if (df(x0) * df(x1) < 0) cand[n++] = bisect(x0, x1);
  }
  cand[n++] = b;
  double hiX = cand[0], loX = cand[0];
  for (int i = 0; i < n; i++) {                             // steps 2 and 3: evaluate f and compare
    printf("x = %.4f   value %.4f\n", cand[i], clean(f(cand[i])));
    if (f(cand[i]) > f(hiX)) hiX = cand[i];
    if (f(cand[i]) < f(loX)) loX = cand[i];
  }
  printf("maximum %.4f at x = %.4f; minimum %.4f at x = %.4f\n", f(hiX), hiX, clean(f(loX)), loX);

  // Example 4: x^(2/3) on [-2, 3]; the only critical point is 0, where f' does not exist
  printf("%.4f %.4f %.4f\n", pow(fabs(-2.0), 2.0 / 3), 0.0, pow(3.0, 2.0 / 3));
  // the box of Exercise 67: V(x) = x (10 - 2x)(16 - 2x) at 0, at the critical point 2 and at 5
  printf("%.0f %.0f %.0f\n", 0.0, 2 * (10 - 4.0) * (16 - 4.0), 5 * (10 - 10.0) * (16 - 10.0));
  return 0;
}

/* output:
x = 1.0000   value 20.0000
x = 2.7183   value 27.1828
x = 7.3891   value 0.0000
maximum 27.1828 at x = 2.7183; minimum 0.0000 at x = 7.3891
1.5874 0.0000 2.0801
0 144 0
*/

Where do maxima and minima come from?

The search for the biggest and the smallest is older than the derivative, and the derivative began as a tool for it.

• Kepler, 1615. Johannes Kepler, looking for the best shape of a wine barrel, noticed that near the largest volume the volume changes very little when the shape changes a little: the flat top of a curve, the horizontal tangent of the theorem above.
• Fermat, about 1629. Pierre de Fermat found maxima and minima by comparing f(x) with f(x+e), dividing the (nearly) zero difference by e and then putting e=0. That is the condition f'(x)=0 before there was a derivative, and the First Derivative Theorem is still called Fermat's theorem on stationary points.
• Newton and Leibniz. With the derivative as a limit of difference quotients, the argument of the proof above became a few lines. The distinction between a critical point and an extreme, between x^3 and x^2, was made clear by the second-derivative and sign tests that followed.
• Bolzano and Weierstrass. The Extreme Value Theorem looks obvious from a graph but cannot be proved without the completeness of the real numbers. Bernard Bolzano gave the first arguments in manuscripts of the 1830s, and Karl Weierstrass proved it in his lectures in the 1860s; this is why the proof is not part of an introductory course.

What mistakes do people make with extreme values?

The most common mistakes with extreme values are the nine below, each with the correct statement.

Practice: test your understanding

These questions mix everything above. A wrong answer costs nothing here: read the hint and try again.

Summary: extreme values at a glance

• Absolute extremes: f(c)\ge f(x) (maximum) or \le (minimum) for all x in the domain; they depend on the domain.
• Extreme Value Theorem: f continuous on a closed, bounded [a,b] has both an absolute maximum and an absolute minimum there.
• First Derivative Theorem: at an interior local extreme where f' exists, f'=0.
• Critical point: an interior point where f'=0 or f' does not exist; a candidate only (x^3 at 0).
• Closed interval method: critical points in (a,b), values at them and at a and b, then the largest and the smallest.
• Watch for: endpoints, corners and cusps (|x|, x^2/3), critical points outside the interval, breaks inside it.
• Code: sp.solveset(df, x, interval), sp.singularities(df, x), or a scan for sign changes of f' with bisection.

Glossary: key terms

Absolute maximum and minimum

The largest and the smallest value that a function takes on its whole domain; they are also called global extremes and depend on the domain.

Extreme Value Theorem

The theorem that a function continuous on a closed, bounded interval attains both an absolute maximum and an absolute minimum there.

Local extreme value

A value that is the largest or the smallest only among the values at points near it; also called a relative extreme.

Critical point

An interior point of the domain of a function where the derivative is zero or does not exist; the only places, besides the endpoints, where an extreme value can occur.

Closed interval method

The procedure that finds the critical points of a continuous function in a closed interval, evaluates it there and at the endpoints, and takes the largest and the smallest value.

Cusp

A point of a graph where the curve has a vertical tangent and the two branches meet at a point, such as x to the two thirds at zero; the derivative does not exist there.

Frequently asked questions

References

1. Joel Hass, Christopher Heil, Maurice D. Weir et al. (2023). Thomas' Calculus: Early Transcendentals in SI Units (15th ed.), Section 4.1: Extreme Values of Functions on Closed Intervals. Pearson.
2. James Stewart (2015). Calculus (8th ed.), section 4.1: Maximum and Minimum Values. Cengage Learning.
3. Pierre de Fermat (1679). Methodus ad disquirendam maximam et minimam. Toulouse (written about 1629).
4. Johannes Kepler (1615). Nova stereometria doliorum vinariorum. Linz.

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