Often, we drop the word “absolute” and speak simply of the min or max on an interval \(I\). When no interval is mentioned, it is understood that we refer to the extreme values on the entire domain of the function.

DEFINITION Extreme Values on an Interval

Let \(f(x)\) be a function on an interval \(I\) and let \(d \in I\). We say that \(f(d)\) is the

• Absolute minimum of \(f(x)\) on I if \(f(d) \leq f(x)\) for all \(x \in I\).
• Absolute maximum of \(f(x)\) on \(I\) if \(f(d) \geq f(x)\) for all \(x \in I\).

REMINDER A closed, bounded interval is an interval \(I = [a, b]\) (endpoints included), where \(a\) and \(b\) are both finite. Often, we drop the word “bounded” and refer to \(I\) more simply as a closed interval. An open interval \((a, b)\) (endpoints not included) may have one or two infinite endpoints.

THEOREM 1 Existence of Extrema on a Closed Interval

A continuous function \(f(x)\) on a closed (bounded) interval \(I = [a, b]\) takes on both a minimum and a maximum value on \(I\).

CONCEPTUAL INSIGHT

Why does Theorem 1 require a closed interval? Think of the graph of a continuous function as a string. If the interval is closed, the string is pinned down at the two endpoints and cannot fly off to infinity (or approach a min/max without reaching it) as in Figure 4.9(B). Intuitively, therefore, it must have a highest and lowest point. However, a rigorous proof of Theorem 1 relies on the completeness property of the real numbers (see Appendix D).

Adapted from “Stories About Maxima and Minima,” V. M. Tikhomirov, AMS (1990)

DEFINITION Local Extrema

We say that \(f(x)\) has a

• Local minimum at \(x = c\) if \(f(c)\) is the minimum value of \(f\) on some open interval (in the domain of \(f\)) containing \(c\).
• Local maximum at \(x = c\) if \(f(c)\) is the maximum value of \(f(x)\) on some open interval (in the domain of \(f\)) containing \(c\).

DEFINITION Interior Point

A point \(c\) in the domain of a function \(f\) is an interior point if there is an open interval containing \(c\) which is also in the domain of \(f\).

DEFINITION Critical Point

An interior point \(c\) in the domain of a function \(f\) is called a critical point if either \(f′(c) = 0\) or \(f′(c)\) does not exist.

EXAMPLE 1

Find the critical points of \(f(x) = x^{3} - 9x^{2} + 24x - 10\).

Solution The function \(f(x)\) is differentiable everywhere (Figure 4.12), so the critical points are the solutions of \(f'(x) = 0\):

Figure 4.12: Graph of \(f(x) = x^{3} - 9x^{2} + 24x - 10\).

The critical points are the roots \(c = 2\) and \(c = 4\).

EXAMPLE 2 Nondifferentiable Function

Find the critical points of \(f(x) = |x|\).

Solution

Figure 4.13: Graph of \(f(x) = |x|\).

THEOREM 2 Fermat’s Theorem on Local Extrema

If \(f(c)\) is a local min or max, then \(c\) is a critical point of \(f(x)\).

Proof

Suppose that \(f(c)\) is a local minimum (the case of a local maximum is similar). If \(f'(c)\) does not exist, then \(c\) is a critical point and there is nothing more to prove. So assume that \(f'(c)\) exists. We must then prove that \(f'(c) = 0\).

Because \(f(c)\) is a local minimum, we have \(f(c + h) \geq f(c)\) for all sufficiently small \(h \neq 0\). Equivalently, \(f(c + h) - f(c) \geq 0\). Now divide this inequality by \(h\):

Figure 4.14 shows the graphical interpretation of these inequalities. Taking the one-sided limits of both sides of (1) and (2), we obtain

Figure 4.14: ---

Thus \(f'(c) \geq 0\) and \(f'(c) \leq 0\). The only possibility is \(f'(c) = 0\) as claimed.

CONCEPTUAL INSIGHT

Fermat’s Theorem does not claim that all critical points yield local extrema. “False positives” may exist—that is, we might have \(f'(c) = 0\) without \(f(c)\) being a local min or max. For example, \(f(x) = x^{3}\) has derivative \(f'(x) = 3x^{2}\) and \(f'(0) = 0\), but \(f(0)\) is neither a local min nor max (Figure 4.15). The origin is a point of inflection (studied in Section 4.4), where the tangent line crosses the graph.

Figure 4.15: The tangent line at \((0, 0)\) is horizontal, but \(f(0)\) is not a local min or max.

In this section, we restrict our attention to closed intervals because in this case extreme values are guaranteed to exist (Theorem 1). Optimization on open intervals is discussed in Section 4.7.

THEOREM 3 Extreme Values on a Closed Interval

Assume that \(f(x)\) is continuous on \([a, b]\) and let \(f(c)\) be the minimum or maximum value on \([a, b]\). Then \(c\) is either a critical point or one of the endpoints \(a\) or \(b\).

Proof

If \(c\) is one of the endpoints \(a\) or \(b\), there is nothing to prove. If not, then \(c\) belongs to the open interval \((a, b)\). In this case, \(f(c)\) is also a local min or max because it is the min or max on \((a, b)\). By Fermat’s Theorem, \(c\) is a critical point.

EXAMPLE 3

Find the extrema of \(f(x) = 2x^{3} - 15x^{2} + 24x + 7\) on \([0, 6]\).

Solution The extreme values occur at critical points or endpoints by Theorem 3, so we can break up the problem neatly into two steps.

Step 1. Find the critical points.

The function \(f(x)\) is differentiable, so we find the critical points by solving

The critical points are \(c = 1\) and \(4\).

Step 2. Compare values at the critical points and endpoints.

The maximum of \(f(x)\) on \([0, 6]\) is the largest of the values in this table, namely \(f(6) = 43\). Similarly, the minimum is \(f(4) = -9\). See Figure 4.16.

Figure 4.16: Extreme values of \(f(x) = 2x^{3} - 15x^{2} + 24x + 7\) on \([0, 6]\).

EXAMPLE 4 Function with a Cusp

Find the max of \(f(x) = 1 - (x - 1)^{\frac{2}{3}}\) on \([-1, 2]\).

Solution First, find the critical points:

The equation \(f'(x) = 0\) has no solutions because \(f'(x)\) is never zero. However, \(f'(x)\) does not exist at \(x = 1\), so \(c = 1\) is a critical point (Figure 4.17).

Figure 4.17: Extreme values of \(f(x) = 1 - (x - 1)^{\frac{2}{3}}\) on \([-1, 2]\).

\(x\)-value	Value of \(f\)
\(1\) (critical point)	\(f(1) = 1\) max
\(-1\) (endpoint)	\(f(-1)\approx-0.59\) min
\(2\) (endpoint)	\(f(2)=0\)

Next, compare values at the critical points and endpoints.

EXAMPLE 5 Logarithmic Example

Find the extreme values of the function \(f(x) = x^{2} - 8 \ln x\) on \([1, 4]\).

Solution First, solve for the critical points:

The only critical point in the interval \([1, 4]\) is \(c = 2\). Next, compare the values of \(f(x)\) at the critical points and endpoints (Figure 4.18):

Figure 4.18: Extreme values of \(f(x) = x^{2} - 8 \ln x\) on \([1, 4]\).

\(x\)-value	Value of \(f\)
\(2\) (critical point)	\(f(2)\approx -1.55\) min
\(1\) (endpoint)	\(f(1)= 1\)
\(2\) (endpoint)	\(f(4)\approx4.9\) max

We see that the min on \([1, 4]\) is \(f(2) \approx -1.55\) and the max is \(f(4) \approx 4.9\).

EXAMPLE 6 Trigonometric Function

Find the min and max of the function \(f(x) = \sin x + \cos^{2}x\) on \([0, 2\pi ]\) (Figure 4.19).

Solution First, solve for the critical points:

Figure 4.19: \(f(x)\) attains a max at \(\frac{\pi}{6}\) and \(\frac{5\pi}{6}\) and a min at \(\frac{3\pi}{2}\).

Then compare the values of \(f(x)\) at the critical points and endpoints: \[\begin{array}{llc} \hline x\text{-value}&\text{Value of }f&\\ \hline \tfrac{\pi}{2}\text{ (critical point)}&f\left(\tfrac{\pi}{2}\right) = 1+0^2 = 1&\\ \tfrac{3\pi}{2}\text{ (critical point)}&f\left(\tfrac{3\pi}{2}\right) = -1+0^2 = -1&\text{min}\\ \tfrac{\pi}{6}\text{ (critical point)}&f\left(\tfrac{\pi}{6}\right) = \tfrac{1}{2}+\left(\tfrac{\sqrt{3}}{2}\right)^2 = \tfrac{5}{4}&\text{max}\\ \tfrac{5\pi}{6}\text{ (critical point)}&f\left(\tfrac{5\pi}{6}\right) = \tfrac{1}{2}+\left(-\tfrac{\sqrt{3}}{2}\right)^2 = \tfrac{5}{4}&\text{max}\\ 0\text{ and }2\pi\text{ (endpoints)}&f(0)=f(2\pi)=1&\\ \hline \end{array}\]

THEOREM 4 Rolle’s Theorem

Assume that \(f(x)\) is continuous on \([a, b]\) and differentiable on \((a, b)\). If \(f(a) = f(b)\), then there exists a number \(c\) between \(a\) and \(b\) such that \(f'(c) = 0\).

Proof

Since \(f(x)\) is continuous and \([a, b]\) is closed, \(f(x)\) has a min and a max in \([a, b]\). Where do they occur? If either the min or the max occurs at a point \(c\) in the open interval \((a, b)\), then \(f(c)\) is a local extreme value and \(f'(c) = 0\) by Fermat’s Theorem (Theorem 2). Otherwise, both the min and the max occur at the endpoints. However, \(f(a) = f(b)\), so in this case, the min and max coincide and \(f(x)\) is a constant function with zero derivative. Therefore, \(f'(c) = 0\) for all \(c \in (a, b)\).

EXAMPLE 7 Illustrating Rolle’s Theorem

Verify Rolle’s Theorem for

Solution The hypotheses of Rolle’s Theorem are satisfied because \(f(x)\) is differentiable (and therefore continuous) everywhere, and \(f(2) = f(-2)\):

We must verify that \(f'(c) = 0\) has a solution in \((-2, 2)\), so we solve \(f'(x) = 4x^{3} - 2x = 2x(2x^{2} - 1) = 0\). The solutions are \(c = 0\) and \(c=\pm\frac{1}{\sqrt{2}}\approx\pm0.707\). They all lie in \((-2, 2)\), so Rolle’s Theorem is satisfied with three values of \(c\).

EXAMPLE 8 Using Rolle’s Theorem

Show that \(f(x) = x^{3} + 9x - 4\) has precisely one real root.

Solution First, we note that \(f(0) = -4\) is negative and \(f(1) = 6\) is positive. By the Intermediate Value Theorem (Section 2.8), \(f(x)\) has at least one root \(a\) in \([0, 1]\). If \(f(x)\) had a second root \(b\), then \(f(a) = f(b) = 0\) and Rolle’s Theorem would imply that \(f'(c) = 0\) for some \(c \in (a, b)\). This is not possible because \(f'(x) = 3x^{2} + 9 \geq 9\), so \(f'(c) = 0\) has no solutions. We conclude that \(a\) is the only real root of \(f(x)\) (Figure 4.21).

Figure 4.21: Graph of \(f(x) = x^{3} + 9x - 4\). This function has one real root.

We can hardly expect a more general method… This method never fails and could be extended to a number of beautiful problems; with its aid we have found the centers of gravity of figures bounded by straight lines or curves, as well as those of solids, and a number of other results which we may treat elsewhere if we have the time to do so.

HISTORICAL PERSPECTIVE

Sometime in the 1630’s, in the decade before Isaac Newton was born, the French mathematician Pierre de Fermat invented a general method for finding extreme values. Fermat said, in essence, that if you want to find extrema, you must set the derivative equal to zero and solve for the critical points, just as we have done in this section. He also described a general method for finding tangent lines that is not essentially different from our method of derivatives. For this reason, Fermat is often regarded as an inventor of calculus, together with Newton and Leibniz.

At around the same time, René Descartes (1596-1650) developed a different but less effective approach to finding tangent lines. Descartes, after whom Cartesian coordinates are named, was a profound thinker—the leading philosopher and scientist of his time in Europe. He is regarded today as the father of modern philosophy and the founder (along with Fermat) of analytic geometry. A dispute developed when Descartes learned through an intermediary that Fermat had criticized his work on optics. Sensitive and stubborn, Descartes retaliated by attacking Fermat’s method of finding tangents and only after some third-party refereeing did he admit that Fermat was correct. He wrote:

…Seeing the last method that you use for finding tangents to curved lines, I can reply to it in no other way than to say that it is very good and that, if you had explained it in this manner at the outset, I would have not contradicted it at all.

However, in subsequent private correspondence, Descartes was less generous, referring at one point to some of Fermat’s work as “le galimatias le plus ridicule”—the most ridiculous gibberish. Today Fermat is recognized as one of the greatest mathematicians of his age who made far-reaching contributions in several areas of mathematics.