7.2 Monotonic Functions and Extreme Values

7.2.1 Monotonic Functions

Definition 2 Let \(f\) be a function on an interval \(I\).

1. \(f\) is said to be an increasing function if
   \[x_1,\;x_2 \in I\;\; \text{ and }\;\; x_1 < x_2\;\; \Longrightarrow \;\;f(x_1) < f(x_2).\]
2. \(f\) is said to be a decreasing function if
   \[x_1,\;x_2 \in I\;\; \text{ and }\;\; x_1 < x_2\;\; \Longrightarrow \;\;f(x_1) > f(x_2).\]

A function \(f\) that is either increasing or decreasing on an interval \(I\) is called a monotonic (monotone) function on \(I\).

Geometrically, the graph of an increasing (decreasing) function rises up (falls down) as \(x\) increases.

Theorem 8 (Increasing Decreasing Test) Let \(f\) be a continuous function on the closed interval \([a,b]\), and differentiable on the open interval \((a,b)\). Then

1. If \(f'(x) > 0\) for all \(x\in (a,b)\), then \(f\) is increasing on \([a,b]\).
2. If \(f'(x) < 0\) for all \(x\in (a,b)\), then \(f\) is decreasing on \([a,b]\).
3. If \(f'(x) = 0\) for all \(x\in (a,b)\), then \(f\) is constant on \([a,b]\).

Observation. Parts (a) and (b) were proven in the context of rates of change when solving related rate problems (see ??). Part (c) was proven in the context of antiderivatives (see ??). In fact, this statement is actually an “if and only if” statement. As it is the case that a constant function has zero derivative.

For the sake of clarity, we prove these results again.

Proof. (a) Let \(x_1,\;x_2 \in (a,b)\) and \(x_1 < x_2\). We must show that \(f(x_1) < f(x_2)\).

Note that \(f\) satisfies the assumptions of the MVT on the closed interval \([x_1,x_2]\). Thus, there exists \(c\in (x_1,x_2)\) such that

\[f(x_2)-f(x_1) = f'(c)(x_2 – x_1).\]

Since \(f'(c) > 0 \) and \(x_2 – x_1 > 0 \), then \(f(x_2)-f(x_1) > 0 \Longrightarrow f(x_2) > f(x_1)\).

(b) The proof is similar to (a) and is left as an exercise (see #12 in 7.2.2).

(c) Let \(x\) an arbitrary point in \((a,b]\). Then, by the MVT, there exists \(c\in (a,x)\) such that
\[\scriptsize{f(x)-f(a) = f'(c)(x – a)= 0\cdot(x-a)= 0. \;\;\Longrightarrow \;\; f(x) = f(a) \;\;\forall \; x \in [a,b].}\]

Thus, \(f\) is constant.

The following result about the persistence of the sign of \(f’\) follows immediately from DT (see Theorem Theorem 6 in 7.1.3.

Proposition 1 Let \(f\) be a differentiable function on an open interval \(I\), and assume that \(f'(x)\neq 0\) for all \(x\) in \(I\). Then, the sign of \(f’\) remains the same on \(I\).

Proof. Assume that there exist \(a,\, b\in I\), with \(a < b\) such that \(f'(a) < 0 < f'(b)\) (the case where the inequalities are reversed is similar. Since \(f\) is continuous on \([a,b]\), by the EVT, \(f\) attains an absolute maximum or an absolute minimum at some point \(c\in [a,b]\). By FT, \(c\) cannot be \(a\) nor \(b\). Thus, \(c\in (a,b)\). At this point, \(f'(c) = 0 \). This is a contradiction to the assumption that \(f'(x)\neq 0\) for all \(x\) in \(I\).

In view of Theorem Theorem 8 and the above proposition, it is convenient to introduce the following.

Definition 3 Let \(f\) be a function. A critical number (or critical point) of \(f\) is a number \(x\) in the interior of \(D_f\) such that \(f'(x) =0\) or \(f'(x)\) is undefined.

It now follows that in order to determine the intervals where a function \(f\) is increasing or decreasing, we need to determine the sign of \(f’\) on the intervals in \(D_f-\{\)critical numbers of \(f\}\).

Example 9
Given the function \(\displaystyle f(x) = \frac{x}{(x-2)^2}\), (a) find its critical numbers; (b) study the sign of its derivative to determine the intervals where \(f\) is increasing, and those where \(f\) is decreasing.

Solution. \(D_f= \left(-\infty, 2\right ) \cup \left (2, \infty\right )\). (a) To find the critical numbers of \(f\), let \(x\in D_f\), and compute

\[\small{f'(x) =\frac{(x-2)^2-x\cdot 2(x-2)}{(x-2)^4}=\frac{(x-2)(x-2-2x)}{(x-2)^4}= -\frac{x+2}{(x-2)^3}. }\]

It follows that \(x\) is a critical point of \(f\) if and only if

\[ \left \{
\renewcommand{\arraystretch}{1.5}
\begin{array}{l}
x\in \; \text{interior of }\; D_f\\
\left [\begin{array}{l}
f'(x)=0\\
f'(x) \; \; \text{undefined}
\end{array}\right.
\end{array}\right.
\Longleftrightarrow
\left \{
\renewcommand{\arraystretch}{1.5}
\begin{array}{l}
x (0, 2)\cup (2,\infty)\\
\left [\begin{array}{l}
x+2=0\\
x-2 =0
\end{array}\right.
\end{array}\right.
\Longleftrightarrow x=-2.
\]

(b) We now determine the signs of \(f’\) on the interior of \(D_f \) minus the critical points. That is, on \((-\infty, -2)\cup (-2, 2) \cup (2, \infty)\). Given that

\[f'(x) = -\frac{x+2}{(x-2)^3},\]

we see that if \(x > 2\), then \(x-2 > 0\) and \(x + 2 > 0 \). Thus, \(f'(x) < 0 \). Then, if \(-2 < x < 2\), then, only \(x- 2 \) changes sign, thus \(f'\) changes sign, and \(f'(x) > 0\). Finally, if \(x < -2\), then, only \(x + 2\) changes sign and \(f'(x) < 0 \). It follows that the signs of \(f'\) are as in Figure Figure 7.5(a). Thus, \(f\) is increasing on \((-2,2)\), and \(f\) is decreasing on \((-\infty, -2)\cup (2, \infty)\).

Figure 7.5: (a) Example 1

Figure 7.5: (b) Example 2

Example 10
Given the function \(\displaystyle f(x) = \frac{\sqrt{x} \, e^x}{3x-1}\), study the sign of its derivative to determine the intervals where \(f\) is increasing, and those where \(f\) is decreasing.

Solution. First note that the domain of \(f\) is given by
\(\displaystyle D_f= \left [0, \frac{1}{3}\right ) \cup \left (\frac{1}{3}, \infty\right )\).

We now find the critical numbers of \(f\). Let \(x\) be an interior point of \(D_f\). Then

\[\scriptsize{\begin{align*}
f'(x) &= \frac{(3x-1)\left (\displaystyle \frac{1}{2\,\sqrt{x} }\, e^x + \sqrt{x} \, e^x\right )- 3\sqrt{x} \, e^x}{(3x-1)^2}
= \frac{(3x-1)\left (e^x + 2x e^x\right )- 6x \, e^x}{2\,\sqrt{x}(3x-1)^2}=\\
&\underset{?}{=} \frac{(6 x^2-5 x-1) e^x}{2\,\sqrt{x}(3x-1)^2} = \frac{(6 x+1)(x-1) e^x}{2\,\sqrt{x}(3x-1)^2}.
\end{align*}}\]

It follows that \(x\) is a critical number of \(f\) if and only if

\[\small{\left \{
\renewcommand{\arraystretch}{1.5}
\begin{array}{l}
x\in \; \text{interior of }\; D_f\\
\left [\begin{array}{l}
f'(x)=0\\
f'(x) \; \; \text{undefined}
\end{array}\right.
\end{array}\right.
\Longleftrightarrow
\left \{
\renewcommand{\arraystretch}{1.5}
\begin{array}{l}
x\in (0, 1/3)\cup (1/3,\infty)\\
\left [\begin{array}{l}
(6 x+1)(x-1)=0\\
x=0
\end{array}\right.
\end{array}\right.
\Longleftrightarrow x=1.
}\]

Thus, we need to determine the signs of \(f’\) on \(\displaystyle \left (0,\frac{1}{3}\right ) \cup \left ( \frac{1}{3}, 1\right ) \cup \left (1, \infty\right )\). Since the factors \(6 x+1\), \(\sqrt{x}\), \(e^x\), \((3x-1)^2\) are all strictly positive on these intervals, it follows that the signs of \(f’\) are as in Figure Figure 7.5(b) (explain why). Hence, \(f\) is increasing on \((1,\infty)\), and \(f\) is decreasing on \(\displaystyle \left (0, \frac{1}{3}\right )\cup \left ( \frac{1}{3}, 1\right )\).

Example 11
Given the function \(\displaystyle f(x) = \frac{1}{2}\sin 2x + \sin x\),(a) find its critical numbers; (b) study the sign of its derivative to determine the intervals where \(f\) is increasing, and those where \(f\) is decreasing.

Solution.
\(f\) is periodic with period \(T=2\pi\), \(D_f= \mathbb{R}\), and its derivative is given by
\[\small{f'(x) = \cos 2x + \cos x = 2\cos^2x+\cos x -1 = (2\cos x -1) (\cos x + 1)}\]
(a) The critical points of \(f\) are the numbers in the interior of the domain of \(f\), where \(f’\) vanishes or is undefined. Since \(f\) is differentiable everywhere, the critical numbers are the numbers where \(f’\) vanishes.

\[f'(x) = 0 \Longleftrightarrow
\left [\renewcommand{\arraystretch}{1.5}
\begin{array}{l}
\displaystyle \cos x = \frac{1}{2}\\\cos x =-1
\end{array}\right .
\Longleftrightarrow
\left [\renewcommand{\arraystretch}{1.5}
\begin{array}{l}
\displaystyle x = \pm \frac{\pi}{3}+ 2\pi k, k\in \mathbb{Z}\\ x= \pi + 2\pi k, k \in \mathbb{Z}
\end{array}\right .
\]

(b) Since, \(f’\) is periodic with period \(T=2\pi\), we only need to determine the sign of \(f’\) on the interval \([0, 2\pi]\). The critical points on this interval are
\(\displaystyle \left \{ \frac{\pi}{3}, \pi, \frac{5\pi}{3}\right \}\). Thus, because of Proposition Proposition 1 in 7.2.1, we need to determine the sign of \(f’\) on the intervals

\[\left [0, \frac{\pi}{3}\right ),\;\; \left ( \frac{\pi}{3}, \pi\right ), \;\; \left (\pi, \frac{5\pi}{3}\right ) , \;\;\left ( \frac{5\pi}{3}, 2\pi\right ].\]

Given that

\[f'(x) = (2\cos x -1) (\cos x + 1),\]

and that \(\cos x + 1 > 0\) on these intervals (why), our task reduces to determine the sign of \(2\cos x -1\) on the intervals. Using the Unit Circle we can see that

\[\scriptsize{2\cos x -1 > 0 \Longleftrightarrow x \in \left [0, \frac{\pi}{3}\right ) \cup \left ( \frac{5\pi}{3}, 2\pi\right ] }\]

and

\[\scriptsize{2\cos x -1 < 0 \Longleftrightarrow x \in \left ( \frac{\pi}{3}, \pi\right )\cup \left (\pi, \frac{5\pi}{3}\right ).}\]

It follows that the signs of \(f’\) are as shown in Figure 7.6.

Figure 7.6: Example 3

It follows that \(f\) is increasing on \(\displaystyle \left [0, \frac{\pi}{3}\right ) \cup \left ( \frac{5\pi}{3}, 2\pi\right ]\) and decreasing on \(\displaystyle \left ( \frac{\pi}{3}, \pi\right )\cup \left (\pi, \frac{5\pi}{3}\right )\).

7.2.2 Exercises

1. Each graph represents the graph of \(f’\) for some function \(f\). (i) Find the critical points of \(f\) (ii) determine the intervals where \(f\) is increasing and the intervals where decreasing.
   

   Figure 7.7: # 1 (a)

   

   Figure 7.7: # 1 (b)

In problems 2– 10, (a) find the critical points of the function; (b) study the sign of the derivative to determine the intervals where the function is monotonic increasing and those where it is monotonic decreasing. Note: these problems will continue in 7.2.4 with the same numbers.

1. \(f(x) = x^3 – 3 x\).
2. \(\displaystyle f(x) = \frac{x^2}{x-2}\).
3. \(f(x) = x \ln x^2\).
4. \(f(x) = \cos^2 x + \cos 2x\), \(x\in \left [0,2\pi\right ]\).
5. \(\displaystyle f(x) = \frac{e^{-2x}\sqrt{x}}{3-2x}\).
6. \(\scriptsize{f(x) = \arctan \left (x e^{-x^2}\right )}\).
7. \(\small{\displaystyle f(x) =\int_0^x t^2\left (\pi^2-9\arcsin^2 t\right )\,dt}\).
8. \(\displaystyle f(x) =\int_{2x^2}^0 t\ln \left (1- t\right ) \,dt\).
9. \(\displaystyle f(x) = \frac{1}{2}\cos 2x +\cos x\); \(x\in [-\pi,\pi]\).

11. Find the critical points of \(f”\) for each \(f\). (a) \(f(x) = \arctan (x^2-4)\); (b) \(\displaystyle f(x) = \int_1^{x^2} x^{x-1}\,dx\); (c) \(f(x) = 2^{-x^2}x\).
12. Find the values of \(b\) and \(c\) such that the function \(f(x)= 2x^2+bx + c\) has a critical point at \(c=2\) and \(f(2) = -1\).
13. Find a function \(f\) that satisfies the given conditions, and state its domain.
1. \(f”(x)=\csc^22x\), \(f\) has a critical point at \(\displaystyle c=\frac{\pi}{4}\) and \(f(c) = -3\);
2. \(f”(x)=\tan^2\pi x\) \(f\) has a critical point at \(c=1\) and \(f(c) = 4\).
14. Find the critical points of the function.
1. \(\displaystyle f(x) = \frac{1}{3}\sin 3x +\sin x\). Hint: show that \(f'(x) = 2\cos x \left (2\cos^2x-1\right )\).
2. \(\displaystyle f(x) = \frac{1}{3}\cos 3x +\cos x\). Hint: show that \(f'(x) = -4 \sin x \cos^2 x\).
15. Find a function \(f\) that satisfies the given conditions.
1. \(f”(x)=\sin^22x\), \(f\) has a critical point at \(\displaystyle c=\frac{\pi}{4}\) and \(f(c) = -3\);
2. \(f”(x)=\cos^2\pi x\), \(f\) has a critical point at \(\displaystyle c=\frac{1}{2}\) and \(f(c) = 4\).
16. Provide all necessary details to prove Theorem Theorem 8(b).

7.2.3 Relative Extrema

In 7.1.1, an absolute maximum (minimum) value of a function \(f\) on an interval \(I\) was defined as a value \(f(c)\) (\(f(d)\)) for some \(c\in I\) (\(d\in I\))such that \(f(c) \geq f(x)\) (\(f(d) \leq f(x)\)) for all \(x\in I\). Here we introduce a closely related concept.

Definition 4Let \(f\) be a real-valued function defined on the interval \(I\).

1. [(a)] \(f\) has an relative maximum (or local maximum) on \(I\) if there exists a point \(c\) in \(I\) and an open interval \((a,b)\) containing \(c\) such that
   \[f(c) \geq f(x)\;\;\; \forall \; x\in (a,b)\cap I.\]
   In this case, \(f(c)\) is called an relative maximum value (or local maximum value) of \(f\) on \(I\).
2. [(b)] (a) \(f\) has an relative minimum (or local minimum) on \(I\) if there exists a point \(c\) in \(I\) and an open interval \((a,b) \) containing \(c\) such that
   \[f(c) \leq f(x)\;\;\; \forall \; x\in (a,b)\cap I.\]
   In this case, \(f(c)\) is called an relative minimum value (or local minimum value) of \(f\) on \(I\).
3. [(c)] A number which is either a relative maximum or a relative minimum of \(f\) on \(I\) is called an extreme value or an extremum of \(f\).

Note that, in particular, any aboslute maximum or minimum of \(f\) on \(I\) is also a relative extrema of \(f\).

As an immediate consequence of Fermat’s Theorem we have the following.

Corollary 1 If f is a continuous function over an interval \(I\), \(f(c)\) is a relative extremum of \(f\) on \(I\) for an interior point \(c\in I\), and \(f\) is differentiable at \(c\), then \(f'(c) = 0\).

Proof. Since \(c\) is assumed to be an interiror point of \(I\), there exists an open interval \((a,b)\) such that \(c\in (a,b)\subset I\). Apply FT to the function \(f\) on \((a,b)\).

Remark. (i) The point \(c\) is an interior point. To see that this condition is necessary, take \(f(x) =x\) on \(I=[-1,1]\). Then, \(f(-1)=-1\) and \(f(1)=1\) are extrema of \(f\), but \(f'(x)=1\) never vanishes.

(ii) It is important to note that the converse of the above result is not true. Namely, if \(f'(c)=0\) for some interior point \(c\in I\), this does not imply that \(f(c)\) is a relative extremum of \(f\) on \(I)\). A simple example of this fact is the function \(f(x)=x^3\). In this case, \(f'(0)=0\) and yet, \(f(0)=0\) is neither a relative extrema of \(f\) on \(\mathbb{R}\).

(iii) On the same token, the presence of a relative extremum does not guarantee that the derivative vanishes, as the derivative may not exist. In this case, a simple example is the function \(f(x) = |x|\). \(f(0)=0\) is an absolute, but \(f'(0)\) does not exist.

(iv) It follows that if a value \(f(c)\) is an extremum of \(f\), then, either \(c\) is an endpoint of \(D_f\) or \(c\) is a critical point of \(f\).

The following test clarifies when a critical point is an extremum of a function.

Proposition 2. (First Derivative Test.) Let \(f\) be a continuous function on an open interval \(I\), and let \(c\in I\) be a critical point.

1. [(1)] If the sign of \(f’\) changes from positive to negative at \(c\), then \(f(c)\) is a relative maximum of \(f\).
2. [(2)] If the sign of \(f’\) changes from negative to positive at \(c\), then \(f(c)\) is a relative minimum of \(f\).
3. [(3)] If the sign of \(f’\) does not change at \(c\), then \(f(c)\) is not an extreme value of \(f\).

Proof. (1) In this case, \(f\) is increasing on an interval of the form \((c-\delta, c)\), and decreasing on an interval of the form \((c, c+\delta)\), for some positive number \(\delta\). It follows that \(f(c) \geq f(x)\) for all \(x\in (c-\delta, c+\delta)\). That is, \(f(c)\) is a relative maximum value of \(f\) on \(I\).

The proofs of parts (2) and (3) are similar and are left as an exercise

Example 1. (Continuation of Example Example 9 in 7.2.1.) Given the function \(\displaystyle f(x) = \frac{x}{(x-2)^2}\), find its extreme values.

Solution. We have already established that
\(D_f= \left(-\infty, 2\right ) \cup \left (2, \infty\right )\), and that the only critical point of \(f\) is \( x=-2\). In addition, we have that \(f\) is increasing on \((-2,2)\), and \(f\) is decreasing on \((-\infty, -2)\cup (2, \infty)\) (see Figure 7.8(a)). Since
\(f'(x)\) is negative for \(x < -2\) and positive for \(-2 < x < 2\), it follows that \(\displaystyle f(-2)=-\frac{1}{8}\) is a relative minimum (see Figure 7.8(b)). Furthermore, since there are no other critical numbers, the only extreme value of \(f\) is the relative minimum \(\displaystyle f(-2)=-\frac{1}{8}\), and since \(f(x) > 0\) for \(x \in (2, \infty)\), \(f(-2)\) is an absolute minimum of \(f\) (see Figure 7.8(c)).

Figure 7.8: Example 1: (a)

Figure 7.8: Example 1: (b)

Figure 7.8: Example 1: (c)

Example 2
(Continuation of Example 9 in 7.2.1.)
Given the function \(\displaystyle f(x) = \frac{\sqrt{x} \, e^x}{3x-1}\), find its extreme values.

}

Solution. We already know that \(\displaystyle D_f= \left [0, \frac{1}{3}\right ) \cup \left (\frac{1}{3}, \infty\right )\), that \(f\) has one critical number \(x=1\), and that the signs of \(f’\) are as in Figure 7.9(a). Hence, \(f\) is increasing on \((1,\infty)\), and \(f\) is decreasing on \(\displaystyle \left (0, \frac{1}{3}\right )\cup \left ( \frac{1}{3}, 1\right )\). Thus, \(f\) has a relative minimum at \(x=1\). Furthermore, since \(f\) is decreasing on \(\displaystyle \left (0, \frac{1}{3}\right )\), it follows that \(f(0)\) is an relative maximum. However, \(x=0\) is not a critical point, as \(0\) is not an interior point of \(D_f\). Thus the extreme values of \(f\) are: the relative maximum \(f(0)=0\), and the relative minimum \(\displaystyle f(1) = \frac{e}{2}\) (see Figure 7.9(b)).

Figure 7.9: Example 2: (a)

Figure 7.9: Example 2: (b)

Example 3(Continuation of Example 11 in 7.2.1.) Given the function \(\displaystyle f(x) = \frac{1}{2}\sin 2x + \sin x\), find its extreme values.

Solution. We have already seen that \(f\) is periodic with period \(T=2\pi\), \(D_f= \mathbb{R}\), and its derivative is given by

\[\small{f'(x) = \cos 2x + \cos x = 2\cos^2x+\cos x -1 = (2\cos x -1) (\cos x + 1).}\]

The critical points of \(f\) are given by

\[f'(x) = 0 \;\; \Longleftrightarrow\;\; x \in \left \{ \pm \frac{\pi}{3}+ 2\pi k \;\; \text{ or } \;\; \pi + 2\pi k \,:\, k \in \mathbb{Z}
\right \}.\]

And, since, \(f’\) is periodic with period \(T=2\pi\), we only need to determine the sign of \(f’\) on the interval \([0, 2\pi]\). The signs of \(f’\) are as shown in Figure 7.10(a). Thus,

\(f\) is increasing on \(\displaystyle \left [0, \frac{\pi}{3}\right ) \cup \left ( \frac{5\pi}{3}, 2\pi\right ]\) and \(f\) is decreasing on \(\displaystyle \left ( \frac{\pi}{3}, \pi\right )\cup \left (\pi, \frac{5\pi}{3}\right )\).

If follows that on \((0, 2\pi)\)

\(\displaystyle f\left (\frac{\pi}{3}\right ) =\frac{3\sqrt{3}}{4}\) is the only local maximum value of \(f\), and

\(\displaystyle f\left (\frac{5\pi}{3}\right ) =-\frac{3\sqrt{3}}{4}\) is the only local minimum value of \(f\).

Furthermore, note that even though \(x=\pi\) is a critical number (\(f'(\pi)=0\)), \(f(\pi)\) is neither a local maximum value nor a local minimum value of \(f\).

By periodicity,

\(\displaystyle f\left (\frac{\pi}{3} + 2\pi k\right ) =\frac{3\sqrt{3}}{4}\), \(k \in \mathbb{Z}\), are all the local maximum values of \(f\), and

\(\displaystyle f\left (\frac{5\pi}{3} + 2\pi k\right ) =-\frac{3\sqrt{3}}{4}\), \(k \in \mathbb{Z}\), are all the local minimum values of \(f\).

(see Figure 7.10(b)).

Figure 7.10: Example 3: (a)

Figure 7.10: Example 3: (b)

7.2.4 Exercises

1. Each graph in Figure 7.7 represents the graph of \(f’\) for some function \(f\). Determine the relative maxima and minima of \(f\).
   

   Figure 7.11: # 1 (a)

   

   Figure 7.11: # 1 (b)

In problems 2– 10, find the extrema of the function. Note: these problems correspond to the problems in 7.2.2 with the same number. Use the information from those problems as needed.

2. \(f(x) = x^3 – 3 x\).
3. \(\displaystyle f(x) = \frac{x^2}{x-2}\).
4. \(f(x) = x \ln x^2\).
5. \(f(x) = \cos^2 x + \cos 2x\), \(x\in \left [0,2\pi\right ]\).
6. \(\displaystyle f(x) = \frac{e^{-2x}\sqrt{x}}{3-2x}\).
7. \(f(x) = \arctan \left (x e^{-x^2}\right )\).
8. \(\small{\displaystyle f(x) =\int_0^x t^2\left (\pi^2-9\arcsin^2 t\right )\,dt}\).
9. \(\displaystyle f(x) =\int_{2x^2}^0 t\ln \left (1- t\right ) \,dt\).
10. \(\displaystyle f(x) = \frac{1}{2}\cos 2x +\cos x\); \(x\in [-\pi,\pi]\).

In problems 11– 12, (a) find the critical numbers of the function; (b) find the intervals where \(f\) is increasing and where \(f\) is decreasing; (c) find the extrema of the function (if any).

11. \(\displaystyle f(x) =\frac{e^{2x}}{1-e^{2x}} \).
12. \(\displaystyle f(x) =x+ \cos 2x \), \(-\pi \leq x \leq \pi\).
13. \(\displaystyle f(x) = 2^{-4x^2} x \).
14. \(\displaystyle f(x) = \arcsin \frac{x}{4x^2+1}\).
15. \(\displaystyle f(x) =\frac{1}{1+\cos^2x} \).
16. \(\displaystyle f(x) =\ln \left (4x^2+1\right ) \).

17. Carry out the proofs of parts (2) and (3) in Proposition 2.
18. Let \(I= (a,b)\) be an open interval, and \(c \in I\).
1. Show that if \(f(c) = 0\) and \(f'(x) \geq 0\) for \(x \in I\), then \(f(x) \geq 0\) for \(x \ in (a,c)\), and \(f(x) \leq 0\) for \(x \ in (c,b)\).
2. Show that if \(f(c) = g(c)\) and \(f'(x) \geq g'(x)\) for \(x\) in an open interval \(I\), then \(f(x) \geq g(x)\) for \(x \ in I\).
19. Let \(I= (a,b)\) be an open interval, and \(c \in I\).
    1. Show that if \(f(c) = 0\), \(f'(c) = 0\), and \(f”(x) > \geq \) for \(x \in I\), then \(f(x) \geq 0\) for \(x \ in I\).
    2. Show that if \(f(c) = k\), \(f'(c) = 0\), and \(f”(x) \geq 0 \) for \(x \in I\), then \(f(x) \geq k\) for \(x \ in I\).
    3. Show that if \(f(c) = g(c)\), \(f'(c) = g'(c)\), and \(f”(x) \geq g”(x)\) for \(x \in I\), then \(f(x) \geq g(x)\) for \(x \ in I\).

7.2.5 Absolute Mini-Max Values

Absolute Mini-Max Values of Continuous Functions on Closed Intervals}

Let \(f\) be a continuous function on the closed interval \([a,b]\). By the EVT, there exist \(c, \, d\in [a.b]\) such that

\[f(x) \leq f(c)\;\; \text{ and } \;\; f(d) \leq f(x) \;\; \forall \; x\in [a,b].\]

There are two options for \(c\) (or for \(d\)), either \(c\) is one of the endpoints \(a\) or \(b\), or \(c\) is an interior point \(c\in (a,b)\). In the latter case, by FT, \(c\) is a critical point of \(f)\). Thus, in order to find \(c\) and \(d\), it is only necessary to follow these steps:

1. Find the values \(f(a)\) and \(f(b)\).
2. Find the critical point of \(f\), and evaluate \(f\) at these points.
3. The largest value among the values in 1 and 2 is the absolute maximum value of \(f\) on \([a,b]\), and the smallest value among the values in 1 and 2 is the absolute minimum value of \(f\) on \([a,b]\).

Example 1
Find the absolute maximum value and absolute minimum value of \(\displaystyle f(x) = x^5- \frac{5}{3}x^3\), \(x\in [-1,2]\).

Solution. Given tha t\(f\) is a polynomials function, \(f\) is continuous everywhere.

1. Values at the end points: \(\displaystyle f(-1) = \frac{2}{3}\) and \(\displaystyle f(2) =\frac{56}{3}\).
2. Values at the critical numbers.
   \[\scriptsize{f'(x) = 5 x^2 (x^2-1)= 0\;\; \Longleftrightarrow \;\; x\in \left \{-1, 0,1\right \} \;\; \Longrightarrow \;\; f(0) = 0\; \text{ and } \; f(1) = -\frac{2}{3}.}\]
   (Note that \(-1\) is and endpoint and not an interior point of the interval \([-1,2]\), thus, it is not a critical number of \(f\) on \([-1,2]\).)
3. It follows that the absolute maximum value of \(f\) is \(\displaystyle f(2) =\frac{56}{3}\), and the absolute minimum value of \(f\) is \(\displaystyle f(1) =-\frac{2}{3}\).

Example 2 Find the absolute maximum value and absolute minimum value of \(\displaystyle f(x) = \frac{\ln^2\sqrt{x}}{x}\), \(\displaystyle x\in \left [\frac{1}{2},4\right ]\).

Solution. \(f\) is continuous on its domain: \(D_f= (0, \infty)\), and it can be rewritten as

\[f(x) = \frac{\ln^2\sqrt{x}}{x} f(x) = \frac{\left (\ln x^{1/2} \right )^2 }{x}= \frac{\ln^2 x}{4x}.\]

1. Values at the end points: \(\displaystyle f\left ( \frac{1}{2}\right )= \frac{\ln^22}{2}\) and \(\displaystyle f\left (4\right )= \frac{\ln^22}{4}\).
2. Values at the critical numbers.
   \[\small{f'(x) = \frac{(2-\ln x )\ln x}{4x^2}= 0\;\; \Longleftrightarrow \;\; x\in \left \{1, e^2\right \} \;\; \Longrightarrow \;\; f(1) = 0.}\]
   Note that \(\displaystyle e^2 \notin \left [\frac{1}{2},4\right ]\).
3. Since \(\displaystyle \frac{\ln^22}{2} > \frac{\ln^22}{4} > 0\), the absolute maximum value of \(f\) is \(\displaystyle f\left ( \frac{1}{2}\right )= \frac{\ln^22}{2}\), and the absolute minimum value of \(f\) is \(\displaystyle f\left (1\right ) = 0\).

Absolute Mini-Max Values of Continuous Functions on Open Intervals

There are contexts (see Example 3) in 7.3.1) where it is necessary to find the absolute maximum/minimum value of a function on an open interval. Naturally, there is no guarantee that such a value exists. However, a function may have a relative maximum/minimum value. In this case, it is convenient to have a test to determine is such a value is actually an absolute maximum/minimum value.

Proposition 3 Let \(f\) be a differentiable function defined on an open interval \((a,b)\), and let \(c\in (a,b)\) be a critical point of \(f\). Then

1. [(a)] \(f(c)\) is an absolute minimum value of \(f\) if \(f'(x) < 0\) for \(a < x < c \) and \(f'(x) > 0 \) for \(c < x < b\);
2. [(b)] \(f(c)\) is an absolute maximum value of \(f\) if \(f'(x) > 0\) for \(a < x < c \) and \(f'(x) < 0 \) for \(c < x < b\).

Proof(a) \(f\) is a decreasing function for \(a < x < c \) and an increasing function for \(c < x < b\). Thus, the result follows immediately. The proof of (b) is similar.

Example 3 Find the absolute maximum value and absolute minimum value of the function

\[\small{\displaystyle f(r) = \pi r^2+ \frac{2000 \pi}{r}, \;\; r > 0.}\]

Solution. \(f\) is continuous on its domain: \(D_f= (0, \infty)\). And

\[\small{f'(r) = 2\pi r – \frac{2000 \pi }{r^2} = \frac{2\pi \left (r^3 – 1000\right ) }{r^2} = \frac{2\pi (r-10)\left (r^2+10r + 100\right ) }{r^2}.}\]

Thus

\[\scriptsize{f'(r)=0 \;\;\Longleftrightarrow \;\; r^3 – 1000 = (r-10)\left (r^2+10r + 100\right ) =0 \;\;\Longleftrightarrow \;\; r=10.}\]

This is the only critical number of \(f\). Since the factors \(2\pi \left (r^2+10r + 100\right )\) and \(r^2\) are strictly positive on \((0, \infty)\), it is immediate that
\[f'(r) < 0 \;\; \forall \; x\in (0,10)\;\;\; \text{ and } \;\;\; f'(r) > 0 \;\; \forall \; x\in (10, \infty).
\]

Thus, by Proposition 3(a),

\[f(10)= \pi (10)^2+ \frac{2000\,\pi}{10}= 300\,\pi \]

is an absolute minimum value of \(\displaystyle f(r) = \pi r^2+ \frac{2000 \pi}{r}\) on \((0, \infty)\).

There is a slight variation of Proposition 3 in terms of the second derivative

Proposition 4 Let \(f\) be a function defined on an open interval \((a,b)\), such that \(f”(x)\) exists for all \(x\in (a,b)\), and let \(f'(c) = 0\) for some \(c\in (a,b)\). Then

1. [(a)] \(f(c)\) is an absolute minimum value of \(f\) if \(f”(x) > 0\) for all \(x\in (a, b) \);
2. [(b)] \(f(c)\) is an absolute maximum value of \(f\) if \(f”(x) < 0\) for for all \(x\in (a, b) \).

Proof(a) Since \(f”(x) > 0\), then \(f'(x)\) is an increasing function on \((a,b) \), thus, \(f'(x) < 0\) for \(a < x < c\) and \(f'(x) > 0\) for \(c < x < b\). The result follows immediately from Proposition 3(a). The proof of (b) is similar.

Example 4(Continuation.) Find the absolute maximum value and absolute minimum value of the function

\[\displaystyle f(r) = \pi r^2+ \frac{2000 \pi}{r}, \;\; r > 0.\]

Solution. We have already seen in Example 3 that
\(\displaystyle f'(r) = 2\pi r – \frac{2000 \pi }{r^2} \)
, and that \(r=10\) is the only critical number of \(f\). Since

\[f”(r) = 2\pi + \frac{4000 \pi }{r^3} > 0 \;\; \forall \; r \in (0, \infty),\]

by Proposition 4(a),

\[f(10)= \pi (10)^2+ \frac{2000\,\pi}{10}= 300\,\pi \]

is an absolute minimum value of \(\displaystyle f(r) = \pi r^2+ \frac{2000 \pi}{r}\) on \((0, \infty)\).

Basic Applications}

Example 5 The sum of two numbers is 10, find the absolute minimum value of the sum of their cubes.

Solution. Let \(x\) and \(y\) be two arbitrary numbers. Then, the quantity to be minimized is the sum of their cubes:

\[\hspace{1in}\hphantom{(*)} S=x^3+y^3,\hspace{1in}(*)\]

and the constraint is

\[\hspace{1in}\hphantom{(**)} x+ y = 10.\hspace{1in}(**)\]

Solving for \(y\) in \((**)\) and substituting in \((*)\), we have

\[S= S(x) = x^3+ (10-x)^3, \;\; x\in \mathbb{R}.\]

\(S\) is continuous everywhere, and

\[S'(x) = 60 (x-5)\]

(verify this). Thus, \(x=5\) is the only critical number \(S\). Furthermore

\[S'(x) > 0, \;\text{ if } \; x < 5, \;\; \text{ and }\;\; S'(x) < 0, \;\text{ if } \; x > 5.\]

Thus, by Proposition 3(a),

\[S(5) = 5^2 +(10-5)^2 = 50\]

is an absolute minimum value of \(\displaystyle S(x) = x^3+ (10-x)^3, \;\; x\in \mathbb{R}.\).

Example 6
The area of a rectangle is 12 m\(^2\). Find the dimensions of the rectangle with such area that has the smallest possible perimeter.

Solution. Let \(l\) and \(w\) denote the length and width of the rectangle. Then, we have to minimize

\[\hspace{1in}\hphantom{(*)} P=2l+ 2w,\hspace{1in}(*)\]

under the constraints

\[\hspace{.85in}\hphantom{(**)} lw = 12, l, \, w > 0 .\hspace{.85in}(**)\]

Solving for \(l\) in \((**)\) and substituting in \((*)\), we have

\[P = P(w) = 2\cdot\frac{12}{w}+ 2w, \;\; w > 0.\]

\(P\) is continuous, and

\[P'(w) = -\frac{24}{w^2} + 2 = \frac{2\left (w^2-12\right )}{w^2}.\]

Thus, \(x=2\sqrt{3}\) is the only critical number \(P\) on \((0, \infty)\). Furthermore

\[P'(w) < 0, \;\text{ if } \; w < 2\sqrt{3}, \;\; \text{ and }\;\; P'(w) < 0, \;\text{ if } \; w > 2\sqrt{3}.\]

It follows, by Proposition 3(a), that

\[P\left (2\sqrt{3}\right ) = \frac{24}{2\sqrt{3}}+ 2\cdot 2\sqrt{3} = 8\sqrt{3} \; (\text{m})\]

is an absolute minimum value of \(\displaystyle P(w) = 2\cdot\frac{12}{w}+ 2w, \;\; w > 0.\). Hence, this is the minimum perimeter of a rectangle with area 12 m\(^2\). And its length is give by \(\displaystyle l = \frac{12}{2\sqrt{3}} = 2\sqrt{3}\) m, and its width by \(w= 2\sqrt{3}\) m. That is, the rectangle is a square.

Example 7 Find the point on the line \(\mathcal{L}\) with equation \(4 x – 3 y = 2\) that is closest to the point \(A(2, -3)\).

Solution.
Let \(P(x,y)\) be an arbitrary point on \(\mathcal{L}\). We need to minimize the distance from \(P\) to \(A)\). Or, equivalently, the square of the distance from \(P\) to \(A)\), that is, if \(s\) denotes the square of the distance \(|AP|\), then, we need to minimize
\[\small{\hspace{1in}\hphantom{(*)} s = |AP|^2 = (x-2)^2+(y+3)^2,\hspace{1in}(*)}\]
under the constraint that \(P\) belongs to the line
\[\small{\hspace{1.in}\hphantom{(**)} 4x – 3 y = 2 \;\; \Longleftrightarrow \;\;y = \frac{4x-2}{3}.\hspace{1.in}(**)}\]

Substituting \(y\) as in \((**)\) in \((*)\), we have
\[\scriptsize{s= s(x) = (x-2)^2+\left (\frac{4x-2}{3}+3\right )^2= (x-2)^2+\frac{1}{9}\left ( 4x+7\right )^2, \;\; x \in \mathbb{R}.}\]
\(s\) is continuous on \(\mathbb{R}\), and
\[\scriptsize{s'(x) = 2(x-2) + \frac{8}{9}(4x+7)=\frac{2}{9}(9x-18+16x+28) = \frac{10}{9}(5x+2).}\]
Thus, \(\displaystyle x=-\frac{2}{5}\) is the only critical number \(s\). Furthermore
\[s'(x) < 0, \;\text{ if } \; x < - \frac{2}{5}, \;\; \text{ and }\;\; s'(x) > 0, \;\text{ if } \; x > -\frac{2}{5}.\]
It follows, by Proposition 3(a), that
\[s\left (-\frac{2}{5}\right ) \underset{?}{=} 9\]
is an absolute minimum value of \(\displaystyle s(x) = (x-2)^2+\frac{1}{9}\left ( 4x+7\right )^2, \;\; x\in \mathbb{R}\). Therefore, the \(x\)-coordinate of the point on \(\mathcal{L}\) closest to \(A)\) is \(\displaystyle – \frac{2}{5}\). Hence, the \(y\)-coordinate is given by (using \((**)\))
\[ y =\frac{\displaystyle 4 \left ( – \frac{2}{5}\right )-2}{3} =-\frac{18}{15}= -\frac{6}{5},\]
and the point closest to \(A\) is the point \(\displaystyle Q\left (-\frac{2}{5}, -\frac{6}{5}\right )\).

Observation. Example 6 can be solved by elementary methods using equations of lines. In fact, \(Q\) corresponds to the point of intersection of \(\mathcal{L}\) and the line perpendicular to \(\mathcal{L}\) passing through \(A\).

7.2.6 Exercises

In problems 1– 6, find the absolute maximum value and absolute minimum value of the function (if any) on the given interval.

1. \(\displaystyle f(x) = \frac{1}{1+\sin^2x}\), \(x\in [0,\pi]\).
2. \(\displaystyle f(x) = x^2\ln^2\left (x^3\right )\), \(\displaystyle x\in \left [\frac{1}{4},2\right ]\).
3. \(\displaystyle f(x) =27 x^2 + \frac{8}{x} \), \(x \in \left (0, \infty \right )\).
4. \(\displaystyle f(x) =\arctan \frac{2x}{x^2+1} \), \(x\in (-\infty, \infty)\).
5. \(\displaystyle f(x) = x\sqrt[3]{25-x^2}\), \(x\in \left [-5 , 5\right ]\).
6. \(\displaystyle f(x) =\frac{x^2}{x^2-1} \), \(x\in \left [-2 ,2 \right ]\).
7. \(\displaystyle f(x) =\arcsin \frac{x^2}{x^2+1} \), \(x\in \left [-1 ,1 \right ]\).
8. \(\displaystyle f(x) =\frac{(x+1) \sqrt{x^2+64}}{x}\), \(x\in (0, \infty)\).
9. \(\displaystyle f(x) = e^{-x}\cos x\), \(x\in [0,2\pi]\).
10. \(\displaystyle f(\theta) = (\cot \theta + 1) \sec \theta\), \(\theta \in [0,2\pi]\).

11. The sum of two positive numbers is 4, find the absolute minimum value of (a) the sum of their reciprocals; (b) the product of their reciprocals.
12. The perimeter of a rectangle is 20 cm. Find the dimensions and the area of such a rectangle with maximum area.
13. Find the point on the line \(\mathcal{L}\) with the equation \(2 x – 4 y= 1\) that is closest to the point \(\displaystyle A\left ( \frac{1}{2},-1\right )\).
    Solve this problem in two different ways: (a) using equations of lines (see observation at the end of Example 6); (b) via calculus.
14. A piece of wire of length 6 m is to be cut into two pieces. One piece is bent into a square, and the other into an equilateral triangle. Where should the wire be cut (or not cut) to enclose (a) the maximum; (b) the minimum possible area by the shape(s). Determine the exact area in each case.
15. The sum of two non-negative numbers is 8. Find the numbers such that the sum of the square of one and the cube of the other is (a) an absolute maximum; (b) an absolute minimum (if any).
16. The difference of two numbers is 4. Find the numbers such that the product of one with the square of the reciprocal of the other is (a) an absolute maximum; (b) an absolute minimum (if any).
17. Find the point on the graph of \(y = \sqrt{x}+2\) which is closest to the point \(A(1, 2)\).
18. Figure 7.12 shows a rectangle inscribed in the graph of a function \(f\) with base along the \(x\)-axis. Find the dimensions and the exact area of the rectangle with maximum area if (a) \(f(x) = x(4-x)\), \(x\in [0,4]\); (b) \(f(x) = e^{-x^2}\), \(x\in [-2,2]\).


Figure 7.12: # 18: (a)



Figure 7.12: # 18: (b)

1. Show that among all rectangles with fixed perimeter, a square has the largest area.
2. Show that among all rectangles with fixed area, a square has the smallest perimeter.
20. Briefly explain what is that problems 11– 19 have in common.
21. (a) Prove Proposition 3(b); (b) prove Proposition 4(b).