Chapter 6 · Application of Derivatives

Example 16: $f(x)=x$ on $(0,1)$ ⁠: no maximum, no minimum

ExamplesMaxima and minima · Section 6.4

The question

Find the maximum and the minimum values, if any, of the function given by

f(x) = x, \quad x \in (0, 1).

Exam answer

Let $x_0 \in (0, 1)$ ⁠. Then $\frac{x_0}{2} \in (0, 1)$ and

f\left(\frac{x_0}{2}\right) = \frac{x_0}{2} < x_0 = f(x_0)

So $x_0$ is not a point of minimum value.

Also $\frac{x_0 + 1}{2} \in (0, 1)$ and

f\left(\frac{x_0 + 1}{2}\right) = \frac{x_0 + 1}{2} > x_0

So $x_0$ is not a point of maximum value either.

Hence $f$ has neither a maximum value nor a minimum value in $(0, 1)$ ⁠.

Want the reason behind every step, the basics and the checks?

The one idea this problem is built on

The values of $f$ fill the open interval $(0,1)$ ⁠: they get as close as we like to 0 and to 1 but never equal them, because the end points 0 and 1 are not in the domain. A minimum would have to be a smallest value actually taken. But below any point we pick there is a smaller one (halfway to 0), and above any point a larger one (halfway to 1). So there is no smallest and no largest value.

Given, to find, and the plan

Given: $f(x) = x$ on the open interval $(0,1)$ ⁠, that is $0 < x < 1$ ⁠.
To find: The maximum and the minimum values of $f$ on $(0,1)$ ⁠, if any.
Plan: Note that $f$ is strictly increasing; then show, by beating an arbitrary candidate, that no point is a point of minimum value and no point is a point of maximum value; finally compare with the closed interval $[0,1]$ (the book's Remark).

Basics you need

1Maximum and minimum value (Definition 3, Section 6.4)

$c$ in I is a point of minimum value if $f(c) \le f(x)$ for all $x$ in I, and of maximum value if $f(c) \ge f(x)$ for all $x$ in I. The point $c$ must belong to I. (The reprint prints $<$ ⁠, $>$ ⁠; read $\le$ ⁠, $\ge$ ⁠, as in Example 14.) Lesson: maxima and minima

2Open and closed intervals

$(0,1)$ is the set of $x$ with $0 < x < 1$ ⁠: the end points are left out. $[0,1]$ includes them. Basics: interval notation

3Strictly increasing (Definition 1, Section 6.3)

$f$ is strictly increasing on I if $x_1 < x_2$ in I gives $f(x_1) < f(x_2)$ ⁠. Lesson: increasing and decreasing functions

4The midpoint lies strictly between

If $a < b$ ⁠, then $a < \frac{a+b}{2} < b$ ⁠: the average of two different numbers lies strictly between them.

Solution

$f$ is strictly increasing on $(0,1)$

If $0 < x_1 < x_2 < 1$ ⁠, then $f(x_1) = x_1 < x_2 = f(x_2)$ ⁠. So $f$ is strictly increasing (Definition 1). (Also $f'(x) = 1 > 0$ ⁠.)

For an increasing function, smaller inputs give smaller outputs, so a minimum would have to be at the smallest input and a maximum at the largest input. The whole question is therefore: does $(0,1)$ have a smallest or a largest number? It does not, and the next two steps prove it.
No minimum value

Suppose $x_0 \in (0,1)$ were a point of minimum value. Consider the point $\frac{x_0}{2}$ ⁠, halfway between 0 and $x_0$ (Basic 4):
$0 < \frac{x_0}{2} < x_0 < 1.$
So $\frac{x_0}{2}$ is in the domain, and $f\!\left(\frac{x_0}{2}\right) = \frac{x_0}{2}$ $< x_0 = f(x_0)$ ⁠. This contradicts $f(x_0) \le f(x)$ for all $x$ ⁠. Hence no point of $(0,1)$ is a point of minimum value: $f$ has no minimum value.

This is the book's argument ("we find $\frac{x_0}{2} < x_0$ for all $x_0 \in (0,1)$ "). Halving moves a point closer to 0 but never reaches 0, so it stays in the domain. For instance, if someone claims $0.001$ is the minimum, then $0.0005$ is smaller and still in $(0,1)$ ⁠.
No maximum value

Suppose $x_1 \in (0,1)$ were a point of maximum value. Consider the point $\frac{x_1+1}{2}$ ⁠, halfway between $x_1$ and 1 (Basic 4):
$x_1 < \frac{x_1+1}{2} < 1.$
It is also greater than 0, so it lies in $(0,1)$ ⁠, and $f\!\left(\frac{x_1+1}{2}\right) = \frac{x_1+1}{2}$ $> x_1 = f(x_1)$ ⁠. This contradicts $f(x_1) \ge f(x)$ for all $x$ ⁠. Hence $f$ has no maximum value in $(0,1)$ ⁠.

The graph of $f(x) = x$ on $(0,1)$ ⁠: a segment whose end points (open circles) are missing. Gold dots are proposed extreme points; the green dot (halfway to 0) is lower than $x_0 = 0.4$ ⁠, the rose dot (halfway to 1) is higher than $x_1 = 0.7$ ⁠. Equal scales on both axes.

The difference is $\frac{x_1+1}{2} - x_1 = \frac{1 - x_1}{2}$ ⁠, which is positive because $x_1 < 1$ ⁠. Halving the gap to 1 always leaves a point closer to 1, but never 1 itself: if $0.999$ is claimed as the maximum, $0.9995$ beats it.
What changes on the closed interval $[0,1]$

The book's Remark: if the end points are added, so that $f(x) = x$ on $[0,1]$ ⁠, then $f(0) = 0 \le x = f(x)$ and $f(1) = 1 \ge x = f(x)$ for every $x$ in $[0,1]$ ⁠. So on $[0,1]$ the minimum value is 0 (at $x = 0$ ⁠) and the maximum value is 1 (at $x = 1$ ⁠).

On $(0,1)$ the numbers 0 and 1 are still bounds (every value lies between them), but they are not values of $f$ ⁠, because 0 and 1 are not in the domain. A maximum or minimum must be a value actually taken. Including the end points makes them values. The book records two general facts here: a monotonic function takes its maximum and minimum at the end points of its domain (when they belong to it), and every continuous function on a closed interval has a maximum and a minimum value.

This is why absolute extrema are always looked for on closed intervals $[a,b]$ in Section 6.4.1. Lesson: absolute extrema on a closed interval

Result

The function $f(x) = x$ ⁠, $x \in (0,1)$ ⁠, has neither a maximum value nor a minimum value in $(0,1)$ ⁠. (On the closed interval $[0,1]$ it would have minimum value 0 at $x = 0$ and maximum value 1 at $x = 1$ ⁠.) This agrees with the book's solution and Remark.

Check it with numbers

Candidate minimum $0.4$ ⁠: $\frac{0.4}{2} = 0.2$ is in $(0,1)$ and smaller. Candidate $0.001$ ⁠: $0.0005$ is smaller.
Candidate maximum $0.7$ ⁠: $\frac{0.7+1}{2} = 0.85$ is in $(0,1)$ and larger. Candidate $0.999$ ⁠: $0.9995$ is larger.

Every candidate is beaten, however close it is to 0 or 1.

Common mistakes

"The minimum is 0 and the maximum is 1." Those are values of $f$ only if 0 and 1 are in the domain; in $(0,1)$ they are not.
"The minimum is the smallest positive number." There is no smallest positive real number: half of any positive number is a smaller positive number.
Looking for $f'(x) = 0$ ⁠. $f'(x) = 1$ everywhere, so derivatives give no candidate at all here; the issue is the missing end points.

What to take away

A maximum or minimum value must be attained at a point of the domain; approaching a bound is not enough.
On open intervals, extreme values may fail to exist even for very simple functions.
On a closed interval, a continuous function always has both (Remark after this example; Section 6.4.1).

Every fact used, and where it comes from

Fact	Used in	Source
Strictly increasing	Step 1	Definition 1, Section 6.3 (Lesson)
Maximum / minimum value must be $f(c)$ for some $c$ in I	Steps 2–4	Definition 3, Section 6.4 (Lesson)
$a < \frac{a+b}{2} < b$ when $a < b$	Steps 2, 3	Arithmetic of inequalities
Monotonic functions take extreme values at the end points; continuous functions on closed intervals have both	Step 4	Remark after Example 16, Section 6.4

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Example 16: f(x)=xf(x)=xf(x)=x on (0,1)(0,1)(0,1)⁠: no maximum, no minimum

Basics you need

Solution

fff is strictly increasing on (0,1)(0,1)(0,1)

No minimum value

No maximum value

What changes on the closed interval [0,1][0,1][0,1]

Every fact used, and where it comes from

Example 16: $f(x)=x$ on $(0,1)$ ⁠: no maximum, no minimum

$f$ is strictly increasing on $(0,1)$

What changes on the closed interval $[0,1]$