Chapter 6 · Application of Derivatives

Example 19: Local minimum of $3+|x|$ at a corner

ExamplesFirst derivative test · Section 6.4

The question

Find local minimum value of the function $f$ given by $f(x) = 3 + |x|$ ⁠, $x \in \R$ ⁠.

Exam answer

For $x < 0$ ⁠, $f(x) = 3 - x$ ⁠, so $f'(x) = -1 < 0$ ⁠.

For $x > 0$ ⁠, $f(x) = 3 + x$ ⁠, so $f'(x) = 1 > 0$ ⁠.

$f$ is not differentiable at $x = 0$ ⁠, so $x = 0$ is a critical point, and $f'$ changes sign from negative to positive there.

So $x = 0$ is a point of local minima.

Hence the local minimum value is $f(0) = 3$ ⁠.

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

The one idea this problem is built on

The graph of $3 + |x|$ is a V with its corner at $(0, 3)$ ⁠. At a corner there is no derivative, so the Second Derivative Test, which needs $f'(0)$ and $f''(0)$ ⁠, cannot even start. The First Derivative Test still works, because it only needs $f'$ beside the point, not at it: the slope is $-1$ on the left and $+1$ on the right, so the graph falls into the corner and climbs out of it.

Given, to find, and the plan

Given: $f(x) = 3 + |x|$ for all real $x$ ⁠.
To find: The local minimum value of $f$ ⁠.
Plan: Remove the modulus by cases; show $f$ is continuous but not differentiable at 0, so 0 is a critical point; apply the First Derivative Test; confirm with Definition 4 and check there are no other local extrema.

Basics you need

1Absolute value by cases

$|x| = -x$ for $x < 0$ and $|x| = x$ for $x \ge 0$ ⁠. Geometrically, $|x|$ is the distance from $x$ to 0.

2Critical point (Note before Theorem 3)

A point $c$ where $f'(c) = 0$ or $f$ is not differentiable at $c$ ⁠. Local extrema can only occur at critical points (Theorem 2). Lesson: maxima and minima

3First and Second Derivative Tests (Theorems 3 and 4)

Theorem 3 needs $f$ continuous at the critical point $c$ and looks at the sign of $f'$ just left and right of $c$ ⁠: $-$ to $+$ means local minima. Theorem 4 needs $f$ twice differentiable at $c$ (Note after Theorem 4), with $f'(c) = 0$ ⁠. Lesson: first and second derivative tests

4Local minima (Definition 4(b))

$c$ is a point of local minima if there is an $h > 0$ with $f(c) \le f(x)$ for all $x$ in $(c-h, c+h)$ ⁠.

Solution

Write $f$ without the modulus

By Basic 1,
$f(x) = \begin{cases} 3 - x, & x < 0, \\ 3 + x, & x \ge 0. \end{cases}$
Both formulas give 3 at $x = 0$ ⁠: $\lim_{x\to0^-}(3 - x) = 3$ ⁠, $\lim_{x\to0^+}(3 + x) = 3$ and $f(0) = 3$ ⁠. So $f$ is continuous at 0.

The First Derivative Test (Theorem 3) requires $f$ to be continuous at the critical point, so this must be checked, not assumed. A jump at 0 would make the sign of $f'$ on the two sides meaningless for an extremum.
$f$ is not differentiable at 0, so 0 is a critical point

With $h > 0$ ⁠:
$\begin{aligned} \text{LHD} &= \lim_{h\to0}\frac{f(0-h) - f(0)}{-h} \\ &= \lim_{h\to0}\frac{(3+h) - 3}{-h} = -1, \\ \text{RHD} &= \lim_{h\to0}\frac{f(0+h) - f(0)}{h} \\ &= \lim_{h\to0}\frac{(3+h) - 3}{h} = 1. \end{aligned}$
(For the left side, $f(-h) = 3 - (-h) = 3 + h$ ⁠.) The two one-sided derivatives differ, so $f$ is not differentiable at $x = 0$ , and 0 is a critical point. Consequently the Second Derivative Test cannot be used: it needs $f''(0)$ ⁠, and even $f'(0)$ does not exist.

The book says "the second derivative test fails". More precisely, it does not apply: its hypothesis (twice differentiable at $c$ ⁠) is not met. The First Derivative Test has no such hypothesis at $c$ itself, only continuity.

This is the same corner as $|x|$ at 0 in Example 15; adding the constant 3 lifts the graph without changing any slope.
First Derivative Test at 0

To the left of 0, $f(x) = 3 - x$ ⁠, so $f'(x) = -1 < 0$ ⁠. To the right of 0, $f(x) = 3 + x$ ⁠, so $f'(x) = 1 > 0$ ⁠.

So $f'$ changes sign from negative to positive as $x$ increases through 0, and $f$ is continuous at 0 (Step 1). By Theorem 3(ii), $x = 0$ is a point of local minima, and the local minimum value is
$f(0) = 3 + |0| = 3.$

The graph of $f(x) = 3 + |x|$ ⁠: the line $y = 3 - x$ (slope $-1$ ⁠) for $x < 0$ and $y = 3 + x$ (slope 1) for $x \ge 0$ ⁠, meeting at the corner $(0, 3)$ ⁠. Equal scales on both axes.

Falling (slope $-1$ ⁠) until 0 and rising (slope 1) after: the corner is the bottom of a valley, just a sharp one rather than a rounded one.

Theorem 3(ii), Section 6.4. Lesson: first and second derivative tests
Confirm with the definition, and check there is nothing else

Definition 4(b). For every real $x$ ⁠, $f(x) = 3 + |x| \ge 3 = f(0)$ ⁠. So the inequality holds on every interval $(-h, h)$ ⁠, and 0 is a point of local minima. (It is even the minimum on all of $\R$ ⁠.)

No other local extrema. At any $x \ne 0$ ⁠, $f$ is differentiable with $f'(x) = -1$ or $1$ ⁠, never 0. So no $x \ne 0$ is a critical point, and by Theorem 2 none is a local extremum. In particular $f$ has no local maximum.

The question asks for "the local minimum value", which presumes there is exactly one; this step justifies that.

Result

$f$ is not differentiable at $x = 0$ (so the second derivative test cannot be used), but $f'(x) = -1$ for $x < 0$ and $f'(x) = 1$ for $x > 0$ ⁠. By the first derivative test, $x = 0$ is a point of local minima, and the local minimum value is $f(0) = 3$ . This agrees with the book's solution.

Check it with numbers

$f(-0.5) = 3.5$ ⁠, $f(-0.1) = 3.1$ ⁠, $f(0) = 3$ ⁠, $f(0.1) = 3.1$ ⁠, $f(0.5) = 3.5$ ⁠. The value at 0 is below all its neighbours, and the drop on each side is exactly $|x|$ (slope 1 in size).

Common mistakes

Differentiating $|x|$ at 0 as if it were 0 or 1. $|x|$ has no derivative at 0; the one-sided derivatives are $-1$ and $1$ ⁠.
Trying $f'(x) = 0$ ⁠. It has no solution; the critical point here is of the "not differentiable" kind.
Using the second derivative test. It requires $f'(0)$ and $f''(0)$ ⁠, which do not exist.
Skipping continuity. The first derivative test is stated for a critical point where $f$ is continuous; here it is, but it must be said.

What to take away

Critical points include points where $f$ is not differentiable; corners of modulus functions are the standard example.
At such points use the first derivative test (with continuity checked), not the second.
Adding a constant shifts the extreme value but not the extreme point.

Every fact used, and where it comes from

Fact	Used in	Source
$\|x\| = -x$ for $x < 0$ ⁠, $x$ for $x \ge 0$	Step 1	Definition of absolute value
Continuity at a point by one-sided limits	Step 1	Section 5.2 (Chapter 5)
Not differentiable when the one-sided derivatives differ	Step 2	Section 5.3 (Chapter 5); Example 15
Critical point; Theorem 2	Steps 2, 4	Note before Theorem 3; Theorem 2, Section 6.4 (Lesson)
Second Derivative Test needs $f$ twice differentiable at $c$	Step 2	Theorem 4 and the Note after it, Section 6.4
First Derivative Test	Step 3	Theorem 3(ii), Section 6.4 (Lesson)
Local minima	Step 4	Definition 4(b), Section 6.4

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Example 19: Local minimum of 3+∣x∣3+|x|3+∣x∣ at a corner

Basics you need

Solution

Write fff without the modulus

fff is not differentiable at 0, so 0 is a critical point

First Derivative Test at 0

Confirm with the definition, and check there is nothing else

Every fact used, and where it comes from

Example 19: Local minimum of $3+|x|$ at a corner

Write $f$ without the modulus

$f$ is not differentiable at 0, so 0 is a critical point