Chapter 6 · Application of Derivatives

Example 14: $x^2$ on $\R$ ⁠: minimum 0, no maximum

ExamplesMaxima and minima · Section 6.4

The question

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

f(x) = x^2, \quad x \in \R.

Exam answer

For all $x \in \mathbf{R}$ ⁠, $f(x) = x^2 \ge 0 = f(0)$ ⁠.

So the minimum value of $f$ is 0, at $x = 0$ ⁠.

For any $c \in \mathbf{R}$ ⁠,

\begin{aligned} f(|c| + 1) &= (|c| + 1)^2 \\ &> c^2 = f(c) \end{aligned}

So no point $c$ is a point of maximum value.

Hence the minimum value of $f$ is 0, at $x = 0$ ⁠, and $f$ has no maximum value.

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

The one idea this problem is built on

A maximum value must be an actual value $f(c)$ that no other value exceeds; a minimum value an actual value that no other value goes below. For $x^2$ ⁠, the fact "a square is never negative" hands us the minimum at once. For the maximum, we show that whatever point $c$ is proposed, some other point gives a bigger value, so no point can be the highest.

Given, to find, and the plan

Given: $f(x) = x^2$ for every real $x$ ⁠; the interval is $I = \R$ ⁠.
To find: The maximum value and the minimum value of $f$ on $\R$ ⁠, if they exist, and the points where they occur.
Plan: State Definition 3; prove $f(0) = 0$ is the minimum value; prove no maximum value exists by beating any proposed maximum; then redo the question on the smaller interval $[-2, 1]$ of the book's Note.

Basics you need

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

$f$ has a maximum value in I if there is a point $c$ in I with $f(c) \ge f(x)$ for all $x \in$ I; $f(c)$ is the maximum value and $c$ a point of maximum value. Minimum: $f(c) \le f(x)$ for all $x \in$ I. (This reprint prints the inequalities as $>$ and $<$ ⁠; they must be read as $\ge$ and $\le$ ⁠, since $x = c$ is itself in I and $f(c) > f(c)$ is impossible. The book's own working, " $f(x) \ge 0$ for all $x$ ", uses $\ge$ ⁠.) Lesson: maxima and minima

2Squares are never negative

For every real $x$ ⁠, $x^2 \ge 0$ ⁠, and $x^2 = 0$ only when $x = 0$ ⁠.

3Proving "there is no maximum"

To show no point $c$ is a point of maximum value, take an arbitrary $c$ and produce a point $x$ in the interval with $f(x) > f(c)$ ⁠. Then $c$ fails the definition, and since $c$ was arbitrary, every point fails.

4Absolute value

$|x|$ is the distance of $x$ from 0; $x^2 = |x|^2$ ⁠, and $|x| \le 2$ means $-2 \le x \le 2$ ⁠. Basics: interval notation

Solution

The minimum value is 0, at $x = 0$

For every real $x$ (Basic 2),
$f(x) = x^2 \ge 0 = f(0).$
So the point $c = 0$ satisfies $f(c) \le f(x)$ for all $x \in \R$ ⁠. By Definition 3(b), the minimum value of $f$ is 0, and the point of minimum value is $x = 0$ . It is the only such point, because $x^2 = 0$ only for $x = 0$ ⁠.

The book reads this from the graph (Fig 6.8). The inequality $x^2 \ge 0$ is the algebraic reason the graph never dips below the $x$ -axis and touches it only at the origin.

Definition 3(b): a minimum value is a value $f(c)$ that is $\le$ every value of $f$ on the interval. Lesson: maxima and minima
There is no maximum value

Suppose some point $c \in \R$ were a point of maximum value. Take the point $x = |c| + 1$ ⁠. Then
$\begin{aligned} f(|c|+1) &= (|c|+1)^2 \\ &= c^2 + 2|c| + 1 \\ &> c^2 = f(c), \end{aligned}$
because $2|c| + 1 \ge 1 > 0$ ⁠. So $f(c)$ is not $\ge$ every value of $f$ ⁠, and $c$ is not a point of maximum value. Since $c$ was arbitrary, $f$ has no maximum value, and no point of maximum value, in $\R$ .

"No maximum" is a statement about every point, so it needs an argument for an arbitrary $c$ (Basic 3). We used $|c|^2 = c^2$ ⁠. Put simply: the values $x^2$ grow without bound ( $f(10) = 100$ ⁠, $f(100) = 10000$ ⁠, …), so none of them is the largest.
The book's Note: restrict the domain to $[-2, 1]$

Now consider the same rule only for $-2 \le x \le 1$ ⁠.

Maximum. Every $x$ in $[-2,1]$ satisfies $|x| \le 2$ ⁠, so
$f(x) = |x|^2 \le 2^2 = 4 = f(-2).$
Hence on $[-2,1]$ the maximum value is $4$ ⁠, at $x = -2$ . Equality $|x| = 2$ needs $x = \pm2$ ⁠, and only $-2$ is in the interval, so $-2$ is the only point of maximum value.

Minimum. Step 1 still applies, since $0 \in [-2,1]$ ⁠: the minimum value on $[-2,1]$ is $0$ ⁠, at $x = 0$ ⁠.

The parabola $y = x^2$ (thin blue) has a lowest point but rises without end, so it has no highest point. Over $[-2, 1]$ (thick gold) it has both: highest $(-2, 4)$ ⁠, lowest $(0, 0)$ ⁠. Equal scales on both axes.

Restricting the domain removes the far-away points that made the values unbounded, so a largest value can now exist. The point farthest from 0 in $[-2,1]$ is $-2$ (distance 2, while 1 is at distance 1), and $x^2$ is the square of that distance.

The Remark after Example 16 explains why this works in general: a continuous function on a closed interval always has a maximum and a minimum value. Example 16
How this fits with derivatives

The minimum on $\R$ is at $x = 0$ ⁠, an interior point where $f'(x) = 2x$ equals 0: the tangent there is horizontal. The maximum on $[-2,1]$ is at the end point $x = -2$ ⁠, where $f'(-2) = -4 \ne 0$ ⁠.

This previews two later facts: at an interior extreme point of a differentiable function the derivative is 0 (Theorem 2, Section 6.4), but at an end point of a closed interval it need not be, so end points must always be checked separately (Section 6.4.1).

Lesson: absolute extrema on a closed interval

Result

On $\R$ ⁠: the minimum value of $f$ is 0, at the point $x = 0$ ⁠; $f$ has no maximum value and no point of maximum value. On $[-2,1]$ (the book's Note): maximum value $(-2)^2 = 4$ at $x = -2$ ⁠, minimum value 0 at $x = 0$ ⁠. This agrees with the book's solution.

Check it with numbers

$x$	$-2$	$-1.5$	$-1$	0	0.5	1
$x^2$	4	2.25	1	0	0.25	1

On $[-2,1]$ the largest entry is 4 (at $-2$ ⁠) and the smallest 0 (at 0). Outside the interval, values keep growing: $f(3) = 9$ ⁠, $f(10) = 100$ ⁠.

Common mistakes

"The maximum is $\infty$ ⁠." $\infty$ is not a value of $f$ ⁠; a maximum value must be $f(c)$ for an actual point $c$ ⁠. The correct statement is "no maximum value".
Reading Definition 3 with a strict inequality. With $f(c) > f(x)$ for all $x$ ⁠, including $x = c$ ⁠, nothing could ever be a maximum. Use $\ge$ and $\le$ ⁠.
Taking the maximum on $[-2,1]$ at $x = 1$ because 1 is the right end. Compare $f(-2) = 4$ with $f(1) = 1$ ⁠: the left end wins.

What to take away

A minimum (maximum) value is proved by one inequality valid for every $x$ in the interval.
"No maximum" is proved by beating an arbitrary candidate point.
The answer depends on the interval: the same rule $x^2$ has no maximum on $\R$ but a maximum on $[-2,1]$ ⁠.

Every fact used, and where it comes from

Fact	Used in	Source
Maximum value, minimum value, point of maximum/minimum value	Steps 1–3	Definition 3, Section 6.4 (printed with $>$ ⁠, $<$ ⁠; read $\ge$ ⁠, $\le$ ⁠) (Lesson)
$x^2 \ge 0$ ⁠, with equality only at 0; $x^2 = \|x\|^2$	Steps 1–3	Algebra of real numbers
$(\|c\|+1)^2 = c^2 + 2\|c\| + 1$	Step 2	Expanding a square
Restriction to $[-2,1]$ ⁠: maximum 4 at $-2$	Step 3	Note after Example 14
Continuous function on a closed interval has a maximum and a minimum	Step 3	Remark after Example 16, Section 6.4
Interior extremum: $f' = 0$ or $f$ not differentiable	Step 4	Theorem 2, Section 6.4

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Example 14: x2x^2x2 on R\RR⁠: minimum 0, no maximum

Basics you need

Solution

The minimum value is 0, at x=0x = 0x=0

There is no maximum value

The book's Note: restrict the domain to [−2,1][-2, 1][−2,1]

How this fits with derivatives

Every fact used, and where it comes from

Example 14: $x^2$ on $\R$ ⁠: minimum 0, no maximum

The minimum value is 0, at $x = 0$

The book's Note: restrict the domain to $[-2, 1]$