Miscellaneous · Q20: Yes: $f(x)=|x|+|x-1|$ is continuous everywhere, not differentiable only at 0 and 1

Basics you need

Solution

1. Choose the function and write it piecewise

   Let $f(x) = |x| + |x - 1|$ for all real $x$⁠. The insides $x$ and $x - 1$ change sign at 0 and 1, so:

   $$f(x) = \begin{cases} -x + (1 - x) = 1 - 2x, & x < 0, \\ x + (1 - x) = 1, & 0 \le x \le 1, \\ x + (x - 1) = 2x - 1, & x > 1. \end{cases}$$
   On each piece both absolute values have a fixed sign, so they can be replaced by ordinary expressions. At the junctions both neighbouring formulas give the same value ($f(0) = 1$ and $f(1) = 1$⁠), which already hints at continuity.
   $|u| = u$ for $u \ge 0$ and $-u$ for $u < 0$ (Basic 4). Basics: piecewise functions

2. $f$ is continuous at every real number

   By the theorems. $|x|$ is continuous (Example 7). $|x - 1|$ is the composite of the continuous functions $x \mapsto x - 1$ and $u \mapsto |u|$⁠, so it is continuous (Theorem 2). Their sum $f$ is continuous (Theorem 1).

   Directly at the junctions. At $x = 0$⁠:

   $$\begin{aligned} \lim_{x\to0^-} f(x) &= \lim_{x\to0^-}(1 - 2x) = 1, \\ \lim_{x\to0^+} f(x) &= \lim_{x\to0^+} 1 = 1, \\ f(0) &= |0| + |-1| = 1. \end{aligned}$$

   At $x = 1$⁠:

   $$\begin{aligned} \lim_{x\to1^-} f(x) &= \lim_{x\to1^-} 1 = 1, \\ \lim_{x\to1^+} f(x) &= \lim_{x\to1^+}(2x - 1) = 1, \\ f(1) &= |1| + |0| = 1. \end{aligned}$$

   All three numbers agree at each junction, and inside each piece $f$ is a polynomial, which is continuous (Example 14). So $f$ is continuous everywhere.

   The theorem argument is already a complete proof; the direct check shows the same thing through the definition, at the only two places where something could go wrong.
   Definition 1, Section 5.2; Theorems 1 and 2, Section 5.2.1; Examples 7 and 14 (Basics 1, 2). Lesson: continuity at a point

3. $f$ is differentiable at every $x$ other than 0 and 1

   Take $c < 0$⁠. For $|h| < |c|$⁠, both $c$ and $c + h$ are negative, so

   $$\frac{f(c+h) - f(c)}{h} = \frac{(1 - 2c - 2h) - (1 - 2c)}{h} = -2.$$

   Hence $f'(c) = -2$⁠.

   In the same way $f'(c) = 0$ for $0 < c < 1$ (where $f$ is the constant 1) and $f'(c) = 2$ for $c > 1$ (where $f(x) = 2x - 1$⁠).

   Around any point that is not a junction there is a small open interval on which $f$ is one straight line, so the difference quotient is that line's slope for all small $h$⁠. Only at 0 and 1 do the two sides of a point use different formulas.
   Derivative as the limit of the difference quotient (Basic 3). Basics: slope at a point

4. Not differentiable at $x = 0$

   Here $f(0) = 1$⁠. For small $h < 0$⁠, $f(h) = 1 - 2h$⁠; for small $h > 0$ ($h < 1$⁠), $f(h) = 1$⁠.

   $$\begin{aligned} \text{LHD} &= \lim_{h\to0^-}\frac{(1 - 2h) - 1}{h} = -2, \\ \text{RHD} &= \lim_{h\to0^+}\frac{1 - 1}{h} = 0. \end{aligned}$$

   $-2 \ne 0$⁠, so $f$ is not differentiable at 0.

   The graph arrives at $(0, 1)$ with slope $-2$ and leaves with slope 0: a corner. Note that the term $|x - 1|$ is the straight line $1 - x$ near 0, so it adds the same slope $-1$ on both sides and cannot cancel the corner of $|x|$⁠.
   Left-hand and right-hand derivatives, Section 5.3 (Basic 3). Lesson: differentiability

5. Not differentiable at $x = 1$

   Here $f(1) = 1$⁠. For small $h < 0$ ($h > -1$⁠), $f(1 + h) = 1$⁠; for small $h > 0$⁠, $f(1 + h) = 2(1 + h) - 1 = 1 + 2h$⁠.

   $$\begin{aligned} \text{LHD} &= \lim_{h\to0^-}\frac{1 - 1}{h} = 0, \\ \text{RHD} &= \lim_{h\to0^+}\frac{(1 + 2h) - 1}{h} = 2. \end{aligned}$$

   $0 \ne 2$⁠, so $f$ is not differentiable at 1.

   Same picture as at 0: slope 0 on the left, slope 2 on the right. Near 1 the term $|x|$ is just $x$⁠, a line, so the corner of $|x - 1|$ survives.
   Compare Exercise 5.2, Q9: $|x - 1|$ is not differentiable at 1. Exercise 5.2 Q9

6. Conclude

   $f(x) = |x| + |x - 1|$ is continuous at every real number (Step 2), differentiable at every real number except 0 and 1 (Step 3), and not differentiable at 0 or at 1 (Steps 4, 5). So it is continuous everywhere and fails to be differentiable at exactly two points. The answer to the question is yes.

   "Exactly two" needs both halves: at least these two points fail (Steps 4, 5) and no other point fails (Step 3). Many other examples work the same way, for instance $|x - p| + |x - q|$ for any two different numbers $p$⁠, $q$⁠.
   Theorem 3 and its remark: continuity does not imply differentiability. Lesson: differentiability

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