Example 15: $|x|$ on $\R$⁠: minimum 0 at a corner, no maximum

Basics you need

Solution

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

   For every real $x$⁠, $f(x) = |x| \ge 0 = f(0)$ (Basic 2). So $c = 0$ satisfies $f(c) \le f(x)$ for all $x \in \R$⁠, and by Definition 3(b) the minimum value of $f$ is 0, at the point $x = 0$. No other point gives 0, since $|x| = 0$ only for $x = 0$⁠.

   The book reads this off the graph (Fig 6.9): the V touches the $x$-axis only at its corner, the origin. The inequality $|x| \ge 0$ is the reason.

2. There is no maximum value

   Take any $c \in \R$ and the point $x = |c| + 1$⁠. Then

   $$f(|c|+1) = |c| + 1 > |c| = f(c).$$

   So $c$ is not a point of maximum value. As $c$ was arbitrary, $f$ has no maximum value in $\R$ and no point of maximum value.

   Whatever point is proposed as "highest", moving one unit further from the origin gives a higher point on the graph. A largest value would have to beat all of these, and none does.

3. Note (i): restrict the domain to $[-2, 1]$

   For $-2 \le x \le 1$ we have $|x| \le 2$⁠, so $f(x) \le 2 = |-2| = f(-2)$⁠. Hence on $[-2,1]$⁠, the maximum value is 2, at $x = -2$ (the only point there with $|x| = 2$⁠). The minimum value is still 0, at $x = 0$⁠.

   On $[-2,1]$ the farthest point from 0 is $-2$⁠, at distance 2, and $|x|$ is exactly that distance.

4. Note (ii): $f$ is not differentiable at the minimum point

   At $c = 0$⁠, for $h \ne 0$ the difference quotient is $\frac{|0+h| - |0|}{h} = \frac{|h|}{h}$⁠.

   From the right ($h > 0$⁠): $\frac{|h|}{h} = \frac{h}{h} = 1$⁠, so the right-hand limit is $1$⁠.

   From the left ($h < 0$⁠): $\frac{|h|}{h} = \frac{-h}{h} = -1$⁠, so the left-hand limit is $-1$⁠.

   The one-sided limits differ, so $f$ is not differentiable at $x = 0$, the very point where the minimum occurs.

   The book's Note (ii) says "the function $f$ in Example 27 is not differentiable at $x = 0$". That reference is a leftover from an older numbering: Example 27 in this reprint is $2x^3 - 15x^2 + 36x + 1$ on $[1, 5]$⁠, a polynomial that is differentiable everywhere. The Note is about the present function $|x|$ (it follows Example 15 and speaks of non-differentiability at 0), and the book proves this fact in Section 5.3 (right after Corollary 1).
   Why it matters: looking only for points where $f'(x) = 0$ would miss this minimum completely, since $f'(x) = \pm1$ wherever it exists. This is why a critical point (Note before Theorem 3) is a point where $f'(c) = 0$ or $f$ is not differentiable. Lesson: first and second derivative tests

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