Exercise 5.1 · Q34: Points of discontinuity of $|x|-|x+1|$

Basics you need

Solution

1. The domain is all of $\R$

   $|x|$ and $|x + 1|$ are defined for every real $x$⁠, so $f$ is defined everywhere. Every real number is a point where $f$ is either continuous or discontinuous, and we must decide which.

   A "point of discontinuity" must be a point of the domain (Section 5.2). Here there are no excluded points to set aside.

2. $|x|$ is continuous at every real number

   This is Example 7 of the textbook: away from 0, $|x|$ is $x$ or $-x$ near the point, a polynomial; at 0 both one-sided limits are 0, which equals $|0|$ (Example 3).

   We use this fact twice: directly for the first term, and as the outer layer of the second term.
   Modulus (Basic 1). Basics: piecewise functions

3. $|x + 1|$ is continuous at every real number

   Let $p(x) = x + 1$ (inner) and $g(u) = |u|$ (outer). Then $(g\circ p)(x) = |x + 1|$⁠. The polynomial $p$ is continuous at every $c$ (Example 14), and $g$ is continuous at every real number, in particular at $p(c) = c + 1$ (Step 2). By Theorem 2, $|x + 1|$ is continuous at every $c$⁠.

   $|x + 1|$ is "the modulus of a continuous function", the same pattern as Example 20 in the textbook. Its corner is at $x = -1$ (where $x + 1 = 0$⁠); that is not a break.
   Theorem 2 (Basic 3). Basics: composite functions

4. The difference is continuous everywhere: no discontinuities

   Take any real $c$⁠. Both $|x|$ and $|x + 1|$ are continuous at $c$ (Steps 2 and 3), so by Theorem 1, part (2), their difference is continuous at $c$⁠:

   $$\lim_{x\to c}\big(|x| - |x + 1|\big) = |c| - |c + 1| = f(c).$$

   Since $c$ was arbitrary, $f$ is continuous at every real number. So $f$ has no point of discontinuity.

   This answers the question completely. Theorem 1 says differences of continuous functions are continuous, and both parts are continuous everywhere, so there is nowhere left for a discontinuity to hide.
   Theorem 1(2): $f - g$ is continuous where $f$ and $g$ are (Basic 2). Lesson: algebra of continuous functions

5. Cross-check: write $f$ without modulus signs

   The modulus signs change formula where $x = 0$ and where $x + 1 = 0$⁠, i.e. at $x = -1$⁠. These two points split the line into three parts:

   $$\begin{aligned} x < -1&:\ \ (-x) - (-(x + 1)) = 1,\\ -1 \le x < 0&:\ \ (-x) - (x + 1) = -2x - 1,\\ x \ge 0&:\ \ x - (x + 1) = -1. \end{aligned}$$

   (For $x < -1$ both $x$ and $x + 1$ are negative; for $-1 \le x < 0$⁠, $x < 0$ but $x + 1 \ge 0$⁠; for $x \ge 0$ both are $\ge 0$⁠.) Inside each part $f$ is a polynomial, hence continuous. Now test the two joints.

   At $x = -1$⁠: $f(-1) = |-1| - |0| = 1$⁠.

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

   At $x = 0$⁠: $f(0) = |0| - |1| = -1$⁠.

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

   At each joint the left-hand limit, right-hand limit and value agree, so $f$ is continuous at $-1$ and at $0$ too, exactly as Step 4 said.

   The theorem-based argument (Steps 2 to 4) is quicker, but the explicit form shows why it works and gives a picture. Two corners appear, at $-1$ and $0$⁠, and corners are not discontinuities.
   Definition 1 at each joint (Basic 4). Lesson: continuity at a point

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