Chapter 5 · Continuity and Differentiability

Exercise 5.2 · Q9: $|x-1|$ is not differentiable at $x=1$

Exercise 5.2Differentiability · Section 5.3

The question

Prove that the function $f$ given by

f(x) = |x - 1|, \quad x \in \R

is not differentiable at $x = 1$ ⁠.

Exam answer

Here $f(x) = |x - 1|$ ⁠, so $f(1) = 0$ and $f(1 + h) = |h|$ ⁠.

\begin{aligned} \text{LHD} &= \lim_{h\to0^-}\frac{f(1 + h) - f(1)}{h} \\ &= \lim_{h\to0^-}\frac{|h|}{h} = \lim_{h\to0^-}\frac{-h}{h} \\ &= -1 \\ \text{RHD} &= \lim_{h\to0^+}\frac{f(1 + h) - f(1)}{h} \\ &= \lim_{h\to0^+}\frac{|h|}{h} = \lim_{h\to0^+}\frac{h}{h} \\ &= 1 \end{aligned}

Since $\text{LHD} \ne \text{RHD}$ ⁠, $f$ is not differentiable at $x = 1$ ⁠. Hence proved.

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

The one idea this problem is built on

A function is differentiable at $c$ only if the slope of the chord from $c$ settles to one finite number whether we approach from the left or from the right. The graph of $|x - 1|$ is a V with its corner at $x = 1$ ⁠: every chord to the left has slope $-1$ ⁠, every chord to the right has slope $+1$ ⁠. Two different one-sided slopes mean no derivative.

Given, to find, and the plan

Given: $f(x) = |x - 1|$ for all real $x$ ⁠.
To show: $f$ is not differentiable at $x = 1$ ⁠.
Plan: Write $f$ piece by piece. Form the difference quotient $\frac{f(1+h) - f(1)}{h}$ ⁠. Find its limit as $h \to 0^-$ (left-hand derivative) and as $h \to 0^+$ (right-hand derivative) separately. Show they are different.

Basics you need

1Absolute value, piece by piece

$|a| = a$ if $a \ge 0$ and $|a| = -a$ if $a < 0$ ⁠. So $|x - 1| = x - 1$ for $x \ge 1$ and $1 - x$ for $x < 1$ ⁠. Basics: piecewise functions

2Differentiable at a point

$f$ is differentiable at $c$ if both one-sided limits of $\frac{f(c+h) - f(c)}{h}$ ⁠, as $h \to 0^-$ and $h \to 0^+$ ⁠, are finite and equal (Section 5.3). Lesson: differentiability

3One-sided limits

$h \to 0^-$ uses only negative $h$ ⁠; $h \to 0^+$ uses only positive $h$ ⁠. Basics: left-hand and right-hand limits

4Slope of a chord

$\frac{f(c+h) - f(c)}{h}$ is the slope of the line joining $(c, f(c))$ to $(c+h, f(c+h))$ ⁠; the derivative is the limit of these slopes. Basics: slope at a point

Solution

Write $f$ piece by piece and find $f(1)$
$f(x) = |x - 1| = \begin{cases} 1 - x, & x < 1 \\ x - 1, & x \ge 1 \end{cases}$
and $f(1) = |0| = 0$ ⁠.

The quantity inside the modulus, $x - 1$ ⁠, changes sign at $x = 1$ ⁠. That is where the formula changes, so that is the point where something can go wrong. Elsewhere $f$ is a single straight line.

Check one value: $x = -2$ gives $|-3| = 3$ ⁠, and $1 - (-2) = 3$ ⁠. Basics: piecewise functions
Note that $f$ is continuous at $1$
$\begin{aligned} \lim_{x\to1^-} f(x) &= \lim_{x\to1^-}(1 - x) = 0, \\ \lim_{x\to1^+} f(x) &= \lim_{x\to1^+}(x - 1) = 0, \\ f(1) &= 0. \end{aligned}$
The two one-sided limits and the value agree, so $f$ is continuous at $1$ ⁠.

Theorem 3 says a differentiable function is continuous. Here continuity holds, so we cannot use a break in the graph to rule out a derivative. We have to look at the slopes themselves. (This is the situation the book describes after Corollary 1: continuity does not imply differentiability.)

Continuity at a point, Definition 1, Section 5.2. Lesson: continuity at a point
Form the difference quotient at $1$

For $h \ne 0$ ⁠,
$\frac{f(1+h) - f(1)}{h} = \frac{|1 + h - 1| - 0}{h} = \frac{|h|}{h}.$
This is the slope of the chord from $(1, 0)$ to $(1+h, f(1+h))$ ⁠. The derivative at $1$ ⁠, if it exists, is the limit of this quotient as $h \to 0$ ⁠. Because $|h|$ behaves differently for negative and positive $h$ ⁠, we must take the two one-sided limits separately.

Derivative of $f$ at $c$ ⁠: $\lim_{h\to0}\frac{f(c+h) - f(c)}{h}$ ⁠, Section 5.3. Basics: slope at a point
Left-hand derivative ( $h < 0$ ⁠)

For $h < 0$ ⁠, $|h| = -h$ ⁠, so
$\begin{aligned} \lim_{h\to0^-}\frac{|h|}{h} &= \lim_{h\to0^-}\frac{-h}{h} \\ &= \lim_{h\to0^-}(-1) = -1. \end{aligned}$
For every negative $h$ the quotient is exactly $-1$ (for example $h = -0.01$ ⁠: $\frac{0.01}{-0.01} = -1$ ⁠). A quantity that is constantly $-1$ has limit $-1$ ⁠. Geometrically, every chord to the left lies along the arm $y = 1 - x$ ⁠, which has slope $-1$ ⁠.

Left-hand limit: only $h < 0$ is used. Basics: one-sided limits
Right-hand derivative ( $h > 0$ ⁠)

For $h > 0$ ⁠, $|h| = h$ ⁠, so
$\begin{aligned} \lim_{h\to0^+}\frac{|h|}{h} &= \lim_{h\to0^+}\frac{h}{h} \\ &= \lim_{h\to0^+}1 = 1. \end{aligned}$
Every chord to the right lies along the arm $y = x - 1$ ⁠, of slope $+1$ ⁠, so the quotient is exactly $1$ for every positive $h$ ⁠.
Compare and conclude

Left-hand derivative $= -1$ ⁠, right-hand derivative $= 1$ ⁠. They are finite but not equal, so $\lim_{h\to0}\frac{f(1+h)-f(1)}{h}$ does not exist. Hence $f$ is not differentiable at $x = 1$ .

The definition asks for both one-sided limits to be finite and equal; one failure is enough. The graph has a corner at $(1, 0)$ ⁠: no single line can be "the tangent" there, because the left arm and the right arm leave the corner in different directions.

Same argument as the book's for $|x|$ at $0$ (Section 5.3, after Corollary 1): the graph of $|x - 1|$ is the graph of $|x|$ shifted 1 unit right. Lesson: differentiability

The graph of

y = |x - 1|

(same scale on both axes). Left of

1

it is the line

y = 1 - x

(slope

-1

⁠, gold); right of

1

it is

y = x - 1

(slope

+1

⁠, blue). The two arms meet at the corner

(1, 0)

with different slopes, so there is no tangent there.

Result

At $x = 1$ ⁠: left-hand derivative $-1$ ⁠, right-hand derivative $1$ ⁠. Since $-1 \ne 1$ ⁠, $f(x) = |x - 1|$ is not differentiable at $x = 1$ , even though it is continuous there. (The answer key gives no answer for this proof question; the conclusion is the one asked for.)

Check it with numbers

$h$	$f(1+h)$	$\frac{f(1+h) - f(1)}{h}$
$-0.1$	$0.1$	$-1$
$-0.001$	$0.001$	$-1$
$0.001$	$0.001$	$1$
$0.1$	$0.1$	$1$

However small $h$ is, the quotient is $-1$ on the left and $1$ on the right; it never settles to one value.

Common mistakes

Differentiating each piece and "plugging in" $x = 1$ ⁠. The pieces' derivatives $-1$ and $1$ suggest the answer, but the proof must use the limits of the difference quotient at $1$ itself.
Concluding from continuity. $f$ is continuous at $1$ ⁠; that does not make it differentiable. Continuity is necessary, not sufficient.
Writing $\frac{|h|}{h} = 1$ ⁠. That is true only for $h > 0$ ⁠; for $h < 0$ the quotient is $-1$ ⁠.

What to take away

To prove non-differentiability at a corner, compute the left-hand and right-hand derivatives from the difference quotient and show they differ.
$|x - c|$ is continuous at $c$ but not differentiable there; everywhere else its derivative is $-1$ (left of $c$ ⁠) or $1$ (right of $c$ ⁠).

Every fact used, and where it comes from

Fact	Used in	Source
$\|a\| = a$ for $a \ge 0$ ⁠, $\|a\| = -a$ for $a < 0$	Steps 1, 4, 5	Definition of absolute value (Basics: piecewise functions)
Continuity at a point	Step 2	Definition 1, Section 5.2
Differentiable implies continuous (not conversely)	Step 2	Theorem 3 and Corollary 1, Section 5.3
Differentiable at $c$ iff both one-sided limits of the difference quotient are finite and equal	Steps 3 to 6	Section 5.3 (Lesson: differentiability)
$\|x\|$ is not differentiable at $0$	Step 6	Section 5.3, after Corollary 1

Found an error, or want more?

Tell us if something on this page is wrong or unclear, or if you would like a fuller explanation, another method or a new feature. Send feedback on this problem →

Exercise 5.2 · Q9: ∣x−1∣|x-1|∣x−1∣ is not differentiable at x=1x=1x=1

Basics you need

Solution

Write fff piece by piece and find f(1)f(1)f(1)

Note that fff is continuous at 111

Form the difference quotient at 111

Left-hand derivative (h<0h < 0h<0⁠)

Right-hand derivative (h>0h > 0h>0⁠)

Compare and conclude

Every fact used, and where it comes from

Exercise 5.2 · Q9: $|x-1|$ is not differentiable at $x=1$

Write $f$ piece by piece and find $f(1)$

Note that $f$ is continuous at $1$

Form the difference quotient at $1$

Left-hand derivative ( $h < 0$ ⁠)

Right-hand derivative ( $h > 0$ ⁠)