• In many practical problems, we need to find the greatest (maximum) value or the least (minimum) value—can be more than one of each—of a function.
• The maximum and minimum values of a function are called extreme values or extrema of the function.
• Extremum is the singular form of extrema. The plural form of maximum and minimum are maxima and minima.
• Differentiation can help us locate the extreme values of a function.

In calculus, there are two types of “maximum” and “minimum”, which are distinguished by the two prefixes: absolute and relative.

Absolute maxima and minima

The concepts of absolute maximum and minimum were introduced in Chapter 4. Let’s review the definitions.

Let the function $$f$$ be defined on a set $$E$$. We say $$f$$ has an absolute maximum on $$E$$ at a point $$p$$ if $f(x)\leq f(p)\quad\text{for all }x\text{ in }E,$ and an absolute minimum value on $$E$$ at $$q$$ if $f(x)\geq f(q)\quad\text{for all }x\text{ in }E.$

• Absolute maxima and absolute minima (plural forms of maximum and minimum) are also referred to as global maxima and global minima.

We learned that:

The Extreme Value Theorem: If $$f$$ is continuous on a closed interval $$[a,b]$$, then $$f$$ attains both its absolute maximum $$M$$ and absolute minimum $$m$$ in $$[a,b]$$. That is, there are numbers $$p$$ and $$q$$ in $$[a,b]$$ such that $$f(p)=M$$ and $$f(q)=m$$.

We should emphasize that:

1. The continuity of the function on an open interval (instead of a closed interval) is not sufficient to guarantee the existence of the absolute maximum and minimum of the function.
2. If the function fails to be continuous even at one point in the interval $$[a,b]$$, the extreme value theorem may fail to be true (Although a discontinuous function may have max and min).

Local (or relative) maxima and minima

Figure 1 shows the graph of function $$f$$ with absolute maximum at $$x=b$$ and absolute minimum at $$x=p$$. The point $$(c,f(c))$$ is higher than all nearby points on the curve, although it is not as high as $$(b,f(b))$$. That is, if we consider only values of $$x$$ sufficiently close to $$c$$, then $$f(c)$$ is the largest value of those values of $$f$$ (Figure 2). In this case, we say $$f$$ has a local (or relative) maximum at $$x=c$$.

Similarly, we say $$f$$ has a local (or relative) minimum at $$x=d$$ because $$f(x)\geq f(d)$$ for all $$x$$ very close to $$d$$ (Figure 1). Other local maxima and minima are denoted on Figure 3.

• Geometrically speaking local maxima and local minima are respectively the “peaks” and “valleys” of the curve.

The precise definitions are as follows.

A function $$f$$ is said to have a local (or relative) maximum at a point $$c$$ within its domain $$D$$ if there is some open interval $$I$$ containing $$c$$ such that $f(x)\leq f(c)\quad\text{for all }x\in\ensuremath{I}.$ The concept of local (or relative minimum) is similarly defined by reversing the inequality.
• Every absolute maximum or minimum that is not an endpoint of an interval is a local maximum and local minimum, respectively. An endpoint is precluded from being a local extremum, because we cannot find an open interval around an endpoint that is contained in the domain of the function.
Example 1

The graph of a function $$f$$ is illustrated in Figure 4. Find its absolute and local extrema.

Solution 1

The lowest point of the graph is $$(x_{3},f(x_{3}))$$, and therefore, the function has an absolute minimum at $$x_{1}$$. Because $$x_{3}$$ is an interior point of the interval $$[a,b]$$, $$f(x_{3})$$ is also a local minimum. The absolute maximum occurs at $$b$$, but because $$b$$ is an endpoint and the function is not defined on its right side, $$f(b)$$ is not a local maximum. The other local minima occur at $$x_{1}$$ and $$x_{5}$$.

Because $$x_{2}\not\in Dom(f)$$ , the function cannot have a maximum there. Moreover, it is evident that $$f(x_{4})$$ is a local maximum. We claim that $$f(x_{6})$$ is a local maximum because if we zoom in, we realize that for all $$x$$ close enough to $$x_{6}$$, $f(x)\leq f(x_{6}).$

All the absolute and local extrema are shown in Figure 6.

Example 2

The graph of a function $$f$$ is illustrated in the following figure. Specify where its extrema occur.

Solution 2

Because there is no greatest and no least value, the function does not have an absolute maximum or absolute minimum. For every $$x_{0}\in(a,b)$$, there is some neighborhood $$I$$ of $$x_{0}$$ that is completely contained in $$(a,b)$$ (Figure 8(a)) and for each $$x\in I$$, we have $$f(x)\leq f(x_{0})$$ and $$f(x)\geq f(x_{0})$$ (because $$f(x)=f(x_{0})$$).1 Therefore, the function has at every $$x$$ in $$(a,b)$$ a local maximum and a local minimum. The function has a local maximum at $$x=a$$ and local minimum at $$x=b$$ (Figure 8(b)).

It is evident from Figure 9 that at a local extremum, the tangent line is either parallel to the $$x$$-axis (slope = 0) or has no tangent line. The following theorem helps us locate all the possible values of $$c$$ for which there is a local extremum.

Theorem. (Fermat’s Theorem): Suppose $$f$$ is a function that is defined on an open interval containing the point $$c$$. If $$f(c)$$ is a local maximum or minimum, then either $$f$$ is not differentiable at $$c$$ (meaning $$f'(c)$$ does not exist) or $$f'(c)=0$$.

• Notice that differentiability, or even continuity, of $$f$$ at other points is not required.
• The geometrical interpretation of the above theorem is: At a local max or min, $$f$$ either has no tangent, or f has a horizontal tangent.

Show the proof …

We shall give the proof for the case of a local minimum at $$x=c$$. According to the definition, we have $f(c)\leq f(c+h)$ or $0\leq f(c+h)-f(c)$ for all $$h$$ sufficiently close to zero (that is, when $$c+h$$ is near $$c$$). If $$f'(c)$$ does not exist, there is nothing else to prove. So suppose $f'(c)=\lim_{h\to0}\frac{f(c+h)-f(c)}{h}$ exists as a definite number. We need to show $$f'(c)=0$$. When $$h$$ is small, we have $\frac{f(c+h)-f(c)}{h}\geq0\quad\text{if }h>0$ and $\frac{f(c+h)-f(c)}{h}\leq0\quad\text{if }h<0$ because the numerator in both cases is either positive or zero ($$f(c+h)-f(c)\geq0$$). If we let $$h\to0^{+}$$, from the first case, we have $f'(c)\geq0,$ and if we let $$h\to0^{-}$$, from the second case, we have $f'(c)\leq0.$ Because we have assumed that $$f'(c)$$ exists, we must have the same limit in both cases, so $0\leq f'(c)\leq0.$ This can happen only when $$f'(c)=0$$. The proof for the case of a local maximum is similar.

The above theorem states a necessary condition for a local extremum. That the condition is not sufficient is evident from a glance at the point $$(r,f(r))$$ in Figure 7. The graph of $$f$$ has a horizontal tangent at this point, but $$f$$ does not have an extreme value at $$x=r$$. As another example, consider $$f(x)=x^{3}$$ $f(x)=x^{3}\Rightarrow f'(x)=3x^{2}$ $f'(0)=0$ but $$x=0$$ does not give either a local maximum or a local minimum of $$f$$, as is obvious from the graph of $$y=x^{3}$$ (Figure 10(a)). If $$g(x)=\sqrt[3]{x}$$, then $g(x)=x^{1/3}\Rightarrow g'(x)=\frac{1}{3}x^{1/3-1}=\frac{1}{3}x^{-2/3}=\frac{1}{3\sqrt[3]{x^{2}}}$ and $$g'(0)$$ is not defined (we may say $$g'(0)=+\infty$$), but $$g(0)=0$$ is not a local extremum (Figure 10(b)).

Critical points

A number in the domain of the function at which the derivative is zero or the derivative does not exist has a special name. It is called a critical number.

Definition. Critical point: A point $$c$$ in the domain of a function $$f$$ is called a critical point (or critical number) of $$f$$ if $f'(c)=0\quad\text{or}\quad f'(c)\text{ does not exist.}$
The number $$f(c)$$ is called a critical value of $$f$$.

• Recall that if $$f'(c)=+\infty$$ or $$f'(c)=-\infty$$, we say $$f'(c)$$ does not exist because $$+\infty$$ and $$-\infty$$ are not numbers.

By the above definition, we can reword Fermat’s theorem as:

Fermat’s Theorem: If $$f(c)$$ is a local maximum or a local minimum, then $$x=c$$ is a critical number of $$f$$.

• According to the above theorem every single local extreme value is a critical value, but not every critical value is necessarily a local extreme value.

• We mentioned that every absolute extreme value, with the exception of an absolute extreme value that occurs at an endpoint, is also a local extreme value. Hence:An absolute maximum or minimum of a function occurs either at a critical point or at an endpoint of its domain.

This provides us a method to find the absolute maximum and the absolute minimum of a differentiable function on a finite closed interval $$[a,b]$$.

Strategy for finding absolute extrema of a continuous function $$f$$ on a finite closed interval $$[a,b]$$:

• Step 1: Find $$f'(x)$$
• Step 2: Find all critical values: Set $$f'(x)=0$$ and solve it for $$x$$. Also find every value of $$x$$ for which $$f'(x)$$ does not exist. Evaluate $$f$$ at each of these numbers that lie between $$a$$ and $$b$$.
• Step 3: Evaluate $$f(a)$$ and $$f(b)$$.
• Step 4: The largest value of $$f$$ from Steps 2 and 3 is the absolute maximum of $$f$$ and the least value of $$f$$ from these steps is the absolute minimum of $$f$$ on $$[a,b]$$.
Example 3

Find the absolute maximum and minimum value of the function $f(x)=\frac{1}{3}x^{3}-4x$ on the interval $$[-3,4]$$.

Solution 3

Step 1: Finding the derivative of $$f$$ $f(x)=\frac{1}{3}x^{3}-4x\Rightarrow f'(x)=x^{2}-4$ Step 2: Finding the critical values of $$f$$. The function is differentiable everywhere, so all the critical points are obtained by setting $$f'(x)=0$$ $f'(x)=x^{2}-4=0$ $x^{2}=4$ $x=\pm2$ Because both $$x=2$$ and $$x=-2$$ lie between $$-3$$ and $$4$$, we evaluate $$f$$ at both of these numbers $f(2)=\frac{1}{3}(2^{3})-4(2)=-\frac{16}{3}\approx-5.333$ $f(-2)=\frac{1}{3}(-2)^{3}-4(-2)=\frac{16}{3}\approx5.333$ Step 3: Evaluating $$f$$ at the endpoints $f(-3)=\frac{1}{3}(-3)^{3}-4(-3)=3$ $f(4)=\frac{1}{3}(4^{3})-4(4)=\frac{16}{3}\approx5.33$ Step 4: Comparing the critical values and the endpoint values.

$$x$$ $$-3$$ $$-2$$ $$2$$ $$4$$
$$f(x)$$ $$3$$ $$\frac{16}{3}$$ $$-\frac{16}{3}$$ $$\frac{16}{3}$$
max min max

The absolute maximum of $$f$$ on $$[-3,4]$$ is $$16/3$$ that occurs at $$x=-2$$ and $$x=4$$, and its absolute minimum on this interval is $$-16/3$$ that occurs at $$x=-2$$. The graph of $$f$$ is shown in Figure 11.

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Example 4

Find the extrema of $$f(x)=x-3(x-1)^{2/3}$$ on $$[-1,2]$$.

Solution 4

Step 1: Finding the derivative of $$f$$ $f(x)=x-3(x-1)^{2/3}\Rightarrow f'(x)=1-3\cdot\frac{2}{3}(x-1)^{-1/3}$ so $f'(x)=1-\frac{2}{\sqrt[3]{x-1}}.$ Step 2: Finding the critical values of $$f$$ $f'(x)=0$ $1-\frac{2}{\sqrt[3]{x-1}}=0$ $\frac{1}{\sqrt[3]{x-1}}=\frac{1}{2}$ $\sqrt[3]{x-1}=2$ $x-1=2^{3}=8$ $x=9$ But $$x=9$$ does not lie between $$-1$$ and $$2$$.
We notice that $$f'(x)$$ does not exists when $$x=1$$. Therefore, $$x=1$$ is another critical point of $$f$$, and $f(1)=1-3(1-1)^{2/3}=1.$ Step 3: Evaluating $$f$$ at the endpoints of the given interval \begin{aligned} f(-1) & =-1-3(-2)^{2/3}\\ & =-1-3\sqrt[3]{4}\approx-5.76\end{aligned} \begin{aligned} f(2) & =2-3(2-1)^{2/3}\\ & =2-3\\ & =-1\end{aligned} Step 4: Comparing the critical values and the endpoint values.

$$x$$ $$-1$$ $$1$$ $$2$$
$$f(x)$$ $$-1-3\sqrt[3]{4}$$ $$1$$ $$-1$$
min max

Thus the absolute max of $$f$$ on $$[-1,2]$$ is $$1$$ that occurs at $$x=1$$ and the absolute min of $$f$$ on that interval is $$1-3\sqrt[3]{4}$$ that occurs at $$x=-1$$. The graph of $$f$$ is shown in Figure 12.

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Example 5

Find the absolute maximum and the absolute minimum of $$f(x)=\sin2x+2\cos x$$

Solution 5

The domain of $$f$$ is $$(-\infty,\infty)$$, but because $$f$$ is periodic, we can examine it for one period. The fundamental period of $$\sin2x$$ is $$\frac{2\pi}{2}=\pi$$ and the fundamental period of $$\cos x$$ is $$2\pi$$ (Figure 13). Hence the fundamental period of $$f$$ is $$2\pi$$ and we can find the absolute max and min on $$[0,2\pi]$$ (or $$[-\pi,\pi]$$).

Step 1: $f(x)=\sin2x+2\cos x\Rightarrow f'(x)=2\cos2x-2\sin x$ Step 2: $f'(x)=2\cos2x-2\sin x=0$ We can express $$\cos2x$$ in terms of $$\sin x$$: $$\cos2x=1-2\sin^{2}x$$ (See Section 3.3.3). Thus $f'(x)=2(1-\sin^{2}x)-2\sin x=0$ $-4\sin^{2}x-2\sin x+2=0$ This is a quadratic equation in terms of $$\sin x$$. Let $$u=\sin x$$. Thus $-4u^{2}-2u+2=0$ $\Rightarrow u=\frac{2\pm\sqrt{2^{2}-4(-4)2}}{2(-4)}=\frac{2\pm\sqrt{36}}{-8}$ $u=-\frac{1}{2},\quad u=-1$ We have to solve $\sin x=-\frac{1}{2},\quad\sin x=-1$ when $$0\leq x\leq2\pi$$. $\sin x=-\frac{1}{2}\Rightarrow x=\frac{\pi}{6},\quad x=\pi-\frac{\pi}{6}=\frac{5\pi}{6}$ and $\sin x=-1\Rightarrow x=\frac{3\pi}{2}.$ (See Figure 14)

Critical points (or critical numbers) are $x=\frac{\pi}{6},\quad x=\frac{5\pi}{6},\quad x=\frac{3\pi}{2}.$ Let’s evaluate $$f$$ at these points. \begin{aligned} f\left(\frac{\pi}{6}\right) & =\sin\frac{\pi}{3}+2\cos\frac{\pi}{6}\\ & =\frac{\sqrt{3}}{2}+2\left(\frac{\sqrt{3}}{2}\right)\\ & =\frac{3\sqrt{3}}{2}.\end{aligned} \begin{aligned} f\left(\frac{5\pi}{6}\right) & =\sin\frac{5\pi}{3}+2\cos\frac{5\pi}{6}\\ & =\sin\left(2\pi-\frac{\pi}{3}\right)+2\cos\left(\pi-\frac{\pi}{6}\right)\\ & =\sin\left(-\frac{\pi}{3}\right)-2\cos\frac{\pi}{6}\\ & =-\sin\frac{\pi}{3}-2\cos\frac{\pi}{6}\\ & =-\frac{\sqrt{3}}{3}-2\frac{\sqrt{3}}{2}\\ & =-\frac{3\sqrt{3}}{2}.\end{aligned} Here we have used the identities $$\sin(2\pi+\theta)=\sin\theta$$ and $$\cos(\pi-\theta)=-\cos\theta$$ (See Section 3.3.3). \begin{aligned} f\left(\frac{3\pi}{2}\right) & =\sin3\pi+2\cos\frac{3\pi}{2}\\ & =0+0\\ & =0.\end{aligned} Step 3: Evaluation of $$f$$ at the endpoints $f(0)=\sin(2\cdot0)+2\cos0=0+2(1)=2$

$f(2\pi)=\sin(4\pi)+2\cos(2\pi)=0+2(1)=2.$ Step 4:

$$x$$ $$0$$ $$\frac{\pi}{6}$$ $$\frac{5\pi}{6}$$ $$\frac{3\pi}{2}$$ $$2\pi$$
$$f(x)$$ $$2$$ $$\frac{3\sqrt{3}}{2}\approx2.598$$ $$-\frac{3\sqrt{3}}{2}\approx-2.598$$ $$0$$ $$2$$
max min

The above table shows that the maximum value of $$f$$ is $$3\sqrt{3}/2$$ and its absolute minimum is $$-3\sqrt{3}/2$$ that occur at $$x=\pi/6$$ and $$x=5\pi/6$$, respectively.

1. $$x\leq y$$ means $$x<y$$ or $$x=y$$. So we can write $$2\leq2$$. Here $$f(x)=f(x_{0})$$ for all $$x$$ in $$I$$, and therefore we can write $$f(x)\leq f(x_{0}$$) or $$f(x)\geq f(x_{0}).$$↩︎