Chapter 5 · Continuity and Differentiability

Example 12: $x+2$ for $x<0$ ⁠, $-x+2$ for $x>0$ ⁠: continuous on its domain

ExamplesContinuity of piecewise functions · Section 5.2

The question

Discuss the continuity of the function defined by

f(x) = \begin{cases} x + 2, & \text{if } x < 0 \\ -x + 2, & \text{if } x > 0 \end{cases}

Exam answer

$f$ is not defined at $x = 0$ ⁠, so its domain is all real $x \ne 0$ ⁠. (So $x = 0$ is not a point of discontinuity.)

For $x < 0$ ⁠, $f(x) = x + 2$ ⁠, a polynomial. So $f$ is continuous at every $c < 0$ ⁠.

For $x > 0$ ⁠, $f(x) = -x + 2$ ⁠, a polynomial. So $f$ is continuous at every $c > 0$ ⁠.

$\therefore\ f$ is continuous at every point of its domain, so $f$ is a continuous function.

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

The one idea this problem is built on

Continuity is only tested at points of the domain. Neither line of the definition covers $x = 0$ ⁠, so 0 is not in the domain of $f$ ⁠. Every point that is in the domain lies strictly inside one of the two pieces, where $f$ is a straight line and hence continuous. So $f$ is continuous at every point of its domain, which by Definition 2 makes it a continuous function, even though its graph has a gap at $x = 0$ ⁠.

Given, to find, and the plan

Given: $f(x) = x + 2$ for $x < 0$ and $f(x) = -x + 2$ for $x > 0$ ⁠; no rule for $x = 0$ ⁠.
To find: At which points $f$ is continuous, and whether $f$ is a continuous function.
Plan: Find the domain $D_1 \cup D_2$ ⁠. Show continuity at every point of $D_1$ and of $D_2$ ⁠. Then explain why $x = 0$ is not examined, and apply Definition 2.

Basics you need

1The domain of a piecewise function

A piecewise function is defined exactly at the inputs covered by one of its conditions. A strict inequality such as $x < 0$ does not include 0. Basics: piecewise functions

2Limit of a straight line

$\lim_{x\to c}(mx + a) = mc + a$ for constants $m$ ⁠, $a$ (algebra of limits, Class XI). Basics: what a limit is

3Definition 1 and Definition 2

Definition 1: for $c$ in the domain, $f$ is continuous at $c$ if $\lim_{x\to c} f(x) = f(c)$ ⁠. Definition 2: $f$ is a continuous function if it is continuous at every point of its domain. Lesson: continuity at a point

4One-sided limits

Used in Step 4 only, to see what happens near the excluded point 0: the left-hand limit uses $x < 0$ ⁠, the right-hand limit uses $x > 0$ ⁠. Basics: left-hand and right-hand limits

Solution

Find the domain

The first line covers every $x < 0$ ⁠, the second every $x > 0$ ⁠. The number 0 satisfies neither condition, and no line gives a value for it. So the domain is
$D_1 \cup D_2, \quad D_1 = \{x \in \R : x < 0\},$ $D_2 = \{x \in \R : x > 0\}.$
In other words, $f$ is defined at all real numbers except 0.

Before asking "is $f$ continuous at $c$ ⁠?" we must know which $c$ are allowed. Definition 1 only speaks about points $c$ in the domain, so finding the domain first tells us exactly which points to test.

$x < 0$ and $x > 0$ are strict inequalities: neither includes 0. Compare Example 13, where " $x \ge 0$ " does include 0. Basics: piecewise functions
Points of $D_1$ ( $c < 0$ ⁠)

Fix $c < 0$ ⁠. Then $f(c) = c + 2$ ⁠. Every $x$ with $|x - c| < |c|$ satisfies $x < c + |c| = 0$ ⁠, so near $c$ we have $f(x) = x + 2$ and
$\lim_{x\to c} f(x) = \lim_{x\to c}(x + 2) = c + 2 = f(c).$
So $f$ is continuous at every point of $D_1$ .

The distance from $c$ to 0 is $|c| = -c$ ⁠. Inside that distance we never reach 0 or the positive numbers, so near $c$ the function is just the line $x + 2$ ⁠, and the limit of a line is found by substituting.

Basics: what a limit is
Points of $D_2$ ( $c > 0$ ⁠)

Fix $c > 0$ ⁠. Then $f(c) = -c + 2$ ⁠. Every $x$ with $|x - c| < c$ satisfies $x > c - c = 0$ ⁠, so near $c$ we have $f(x) = -x + 2$ and
$\lim_{x\to c} f(x) = \lim_{x\to c}(-x + 2) = -c + 2 = f(c).$
So $f$ is continuous at every point of $D_2$ .

Same reasoning on the positive side: within distance $c$ of $c$ all inputs are positive, so only the rule $-x + 2$ is used.
What about $x = 0$ ⁠?

0 is not in the domain, so $f(0)$ does not exist and Definition 1 does not apply at 0. In this book's sense, 0 is neither a point of continuity nor a point of discontinuity: it is simply not a point where the question is asked.

It is still instructive to look at the one-sided limits there, because a limit does not need $f(0)$ ⁠:
$\lim_{x\to 0^-} f(x) = \lim_{x\to 0^-}(x + 2) = 2,$ $\lim_{x\to 0^+} f(x) = \lim_{x\to 0^+}(-x + 2) = 2.$
So $\lim_{x\to 0} f(x) = 2$ exists, but there is no value $f(0)$ to compare it with.

This is the heart of the example. The graph has a gap at $(0, 2)$ ⁠, so to draw it we must lift the pen. The book's point is that lifting the pen only at a point where $f$ is not defined does not spoil continuity of the function: continuity is a statement about points of the domain only, exactly as for $\frac{1}{x}$ in Example 9.

If we added the value $f(0) = 2$ ⁠, the new function would be $2 - |x|$ ⁠, continuous at 0 as well. But that is a different function from the one in the question. Basics: one-sided limits
Conclusion

By Steps 2 and 3, $f$ is continuous at every point of its domain $D_1 \cup D_2$ ⁠. By Definition 2, $f$ is a continuous function.

The graph of $f$ ⁠, same scale on both axes (compare Fig 5.6). The blue line $y = x + 2$ for $x < 0$ and the violet line $y = -x + 2$ for $x > 0$ both head for $(0, 2)$ ⁠, drawn as an open dot because $f$ has no value at 0.

Definition 2 asks for continuity at every point of the domain, and Steps 2 and 3 cover all of it. The excluded point 0 is not part of the requirement.

Lesson: continuity at a point

Result

$f$ is defined at every real number except 0 and is continuous at every point of its domain $\{x : x < 0\} \cup \{x : x > 0\}$ ⁠. Hence $f$ is a continuous function; the point 0, not being in the domain, is not a point of discontinuity. This agrees with the textbook's solution.

Check it with numbers

At $c = -1$ ⁠: $f(-1) = 1$ ⁠, $f(-1.001) = 0.999$ ⁠, $f(-0.999) = 1.001$ ⁠, all close to 1. At $c = 3$ ⁠: $f(3) = -1$ ⁠, $f(2.999) = -0.999$ ⁠, $f(3.001) = -1.001$ ⁠. Near the gap: $f(-0.001) = 1.999$ and $f(0.001) = 1.999$ ⁠, both close to 2, while $f(0)$ is undefined.

Common mistakes

Calling 0 a point of discontinuity. A point of discontinuity must be in the domain. 0 is not, so this function has no points of discontinuity at all.
Reading $x < 0$ as $x \le 0$ ⁠. Then $f(0)$ would be 2. The book's definition uses strict inequalities on both lines.
Equating "continuous" with "drawn without lifting the pen". That picture works on an interval. Here the domain itself has a hole, and the pen is lifted only there.

What to take away

Always find the domain first: continuity questions are asked only at points of the domain.
A gap in the graph at a point outside the domain does not make the function discontinuous (Definition 2).
Same lesson as Example 9 ( $\frac{1}{x}$ ⁠) and later Example 18 ( $\tan x$ ⁠): continuous functions can have graphs in several pieces.

Every fact used, and where it comes from

Fact	Used in	Source
A piecewise function is defined only where a condition holds	Step 1	Definition of a piecewise function (Basics: piecewise functions)
$\lim_{x\to c}(\pm x + 2) = \pm c + 2$	Steps 2, 3, 4	Algebra of limits, Class XI (Basics: limits)
Continuity is defined at points of the domain	Steps 2–4	Definition 1, Section 5.2 (Lesson: continuity at a point)
Continuous function = continuous at every point of the domain	Step 5	Definition 2, Section 5.2
One-sided limits	Step 4	Section 5.2 (Basics: one-sided limits)
A function undefined at 0 can still be continuous	Step 4	Example 9, Section 5.2
Graph of $f$	Step 5	Fig 5.6, Section 5.2

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Example 12: x+2x+2x+2 for x<0x<0x<0⁠, −x+2-x+2−x+2 for x>0x>0x>0⁠: continuous on its domain

Basics you need

Solution

Find the domain

Points of D1D_1D1​ (c<0c < 0c<0⁠)

Points of D2D_2D2​ (c>0c > 0c>0⁠)

What about x=0x = 0x=0⁠?

Conclusion

Every fact used, and where it comes from

Example 12: $x+2$ for $x<0$ ⁠, $-x+2$ for $x>0$ ⁠: continuous on its domain

Points of $D_1$ ( $c < 0$ ⁠)

Points of $D_2$ ( $c > 0$ ⁠)

What about $x = 0$ ⁠?