Chapter 5 · Continuity and Differentiability

Exercise 5.1 · Q16: Continuity of $-2$ ( $x \le -1$ ⁠), $2x$ ( $-1 < x \le 1$ ⁠), $2$ ( $x > 1$ ⁠)

Exercise 5.1Continuity of piecewise functions · Section 5.2

The question

Discuss the continuity of the function $f$ ⁠, where $f$ is defined by

f(x) = \begin{cases} -2, & \text{if } x \le -1 \\ 2x, & \text{if } -1 < x \le 1 \\ 2, & \text{if } x > 1 \end{cases}

Exam answer

For $x < -1$ ⁠, $f(x) = -2$ ⁠; for $-1 < x < 1$ ⁠, $f(x) = 2x$ ⁠; for $x > 1$ ⁠, $f(x) = 2$ ⁠. These are polynomials, so $f$ is continuous at every $x$ other than $-1$ and $1$ ⁠.

At $x = -1$ ⁠:

\begin{aligned} \text{LHL} &= \lim_{h\to0} f(-1 - h) \\ &= \lim_{h\to0}(-2) = -2 \\ \text{RHL} &= \lim_{h\to0} f(-1 + h) \\ &= \lim_{h\to0} 2(-1 + h) = -2 \end{aligned}

and $f(-1) = -2$ ⁠. So $\text{LHL} = \text{RHL} = f(-1)$ ⁠: $f$ is continuous at $x = -1$ ⁠.

At $x = 1$ ⁠:

\begin{aligned} \text{LHL} &= \lim_{h\to0} f(1 - h) \\ &= \lim_{h\to0} 2(1 - h) = 2 \\ \text{RHL} &= \lim_{h\to0} f(1 + h) \\ &= \lim_{h\to0} 2 = 2 \end{aligned}

and $f(1) = 2(1) = 2$ ⁠. So $\text{LHL} = \text{RHL} = f(1)$ ⁠: $f$ is continuous at $x = 1$ ⁠.

Hence $f$ is continuous at every real number.

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

The one idea this problem is built on

The middle piece $2x$ climbs from $2(-1) = -2$ to $2(1) = 2$ ⁠, and these are exactly the heights of the flat outer pieces. So the line ramps up from the lower floor to the upper floor without a gap. Test the two junctions $x = -1$ and $x = 1$ with one-sided limits and values: everything agrees, and $f$ is continuous at every real number.

Given, to find, and the plan

Given: $f(x) = -2$ for $x \le -1$ ⁠, $f(x) = 2x$ for $-1 < x \le 1$ ⁠, $f(x) = 2$ for $x > 1$ ⁠.
To find: At which points $f$ is continuous and at which it is not.
Plan: (1) Domain and junctions. (2) Points strictly inside each piece. (3) The junction $x = -1$ ⁠. (4) The junction $x = 1$ ⁠. (5) Conclude.

Basics you need

1Reading a piecewise function

$x = -1$ satisfies $x \le -1$ (not $-1 < x$ ⁠), so $f(-1) = -2$ ⁠. $x = 1$ satisfies $-1 < x \le 1$ ⁠, so $f(1) = 2 \cdot 1 = 2$ ⁠. Basics: piecewise functions

2A limit only looks near the point

If $f$ equals a polynomial (or a constant) on an interval around $c$ ⁠, its limit at $c$ is that polynomial's value. Basics: what a limit is

3Left-hand and right-hand limits

At a junction, compute each side with its own formula. Basics: one-sided limits

4Continuity at a point (Definition 1)

Left-hand limit $=$ right-hand limit $= f(c)$ ⁠.

Solution

Domain and junctions

Every real $x$ satisfies exactly one of $x \le -1$ ⁠, $-1 < x \le 1$ ⁠, $x > 1$ ⁠, and each formula is defined everywhere. So the domain is $\R$ ⁠, and the junctions are $x = -1$ and $x = 1$ ⁠.

The two junctions are the only places where a jump could come from switching formulas; everywhere else one formula governs a whole neighbourhood.

Regions: $(-\infty, -1]$ ⁠, $(-1, 1]$ ⁠, $(1, \infty)$ ⁠. Basics: piecewise functions
Points strictly inside a piece

If $c < -1$ ⁠: near $c$ (within distance $-1 - c$ ⁠), $f(x) = -2$ ⁠, so $\lim_{x\to c} f(x) = -2 = f(c)$ ⁠.

If $-1 < c < 1$ ⁠: near $c$ (within distance $\min(c + 1, 1 - c)$ ⁠), $f(x) = 2x$ ⁠, so $\lim_{x\to c} f(x) = 2c = f(c)$ ⁠.

If $c > 1$ ⁠: near $c$ (within distance $c - 1$ ⁠), $f(x) = 2$ ⁠, so $\lim_{x\to c} f(x) = 2 = f(c)$ ⁠.

So $f$ is continuous at every $c \ne -1, 1$ ⁠.

A point strictly inside one region has a small interval around it inside the same region, on which $f$ is a constant or the polynomial $2x$ ⁠. The limit uses only that interval, so it equals the value.

Constants and polynomials are continuous (Examples 5 and 14). Lesson: continuity at a point
The junction $x = -1$

Value. $-1 \le -1$ ⁠, so $f(-1) = -2$ ⁠.

Left-hand limit. For $x < -1$ ⁠, $f(x) = -2$ ⁠:
$\lim_{x\to -1^-} f(x) = \lim_{x\to -1^-} (-2) = -2.$
Right-hand limit. For $-1 < x < 1$ ⁠, $f(x) = 2x$ ⁠:
$\lim_{x\to -1^+} f(x) = \lim_{x\to -1^+} 2x = 2(-1) = -2.$
Compare. Left-hand limit $= -2$ ⁠, right-hand limit $= -2$ ⁠, $f(-1) = -2$ ⁠. All equal, so $f$ is continuous at $x = -1$ .

Approaching $-1$ from the right means taking $x$ slightly bigger than $-1$ ⁠, such as $-0.999$ ⁠, where the middle formula $2x$ applies. Its values approach $2(-1) = -2$ ⁠, the same as the flat piece on the left.

Signs: $x \to -1^+$ means $x > -1$ ⁠, for example $x = -0.999$ ⁠. Basics: one-sided limits
The junction $x = 1$

Value. $1$ satisfies $-1 < x \le 1$ ⁠, so $f(1) = 2 \cdot 1 = 2$ ⁠.

Left-hand limit. For $-1 < x < 1$ ⁠, $f(x) = 2x$ ⁠:
$\lim_{x\to 1^-} f(x) = \lim_{x\to 1^-} 2x = 2 \cdot 1 = 2.$
Right-hand limit. For $x > 1$ ⁠, $f(x) = 2$ ⁠:
$\lim_{x\to 1^+} f(x) = \lim_{x\to 1^+} 2 = 2.$
Compare. Left-hand limit $= 2$ ⁠, right-hand limit $= 2$ ⁠, $f(1) = 2$ ⁠. All equal, so $f$ is continuous at $x = 1$ .

At this junction the value comes from the middle piece. The line $2x$ reaches height $2$ exactly where the upper floor at height $2$ begins.

Definition 1: all three numbers must agree. Lesson: continuity at a point
Conclude

Steps 2, 3 and 4 cover every real number, and at each one $f$ is continuous. So $f$ is continuous at every real number: it is a continuous function with no point of discontinuity.

The graph (below) is one unbroken path: a floor, a ramp, a floor. It has corners at $(-1, -2)$ and $(1, 2)$ ⁠, but corners are about differentiability (Section 5.3), not continuity.

Blue:

y = -2

for

x \le -1

⁠. Green:

y = 2x

for

-1 < x \le 1

⁠, from

(-1, -2)

(1, 2)

⁠. Gold:

y = 2

for

x > 1

⁠. The filled dots are

f(-1) = -2

and

f(1) = 2

⁠; the graph is unbroken. One square is one unit on both axes.

Result

$f$ is continuous at every $x \ne \pm 1$ (constant or linear pieces). At $x = -1$ ⁠: left-hand limit $-2$ ⁠, right-hand limit $-2$ ⁠, $f(-1) = -2$ ⁠. At $x = 1$ ⁠: left-hand limit $2$ ⁠, right-hand limit $2$ ⁠, $f(1) = 2$ ⁠. $f$ is continuous at every real number. (Textbook answer: "Continuous".)

Check it with numbers

Around $-1$ ⁠: $f(-1.001) = -2$ ⁠, $f(-1) = -2$ ⁠, $f(-0.999) = 2(-0.999) = -1.998$ ⁠. Around $1$ ⁠: $f(0.999) = 1.998$ ⁠, $f(1) = 2$ ⁠, $f(1.001) = 2$ ⁠. No jumps. ✓

A second check by a different route: the same function can be written as $f(x) = |x + 1| - |x - 1|$ ⁠. For $x \le -1$ this is $-(x + 1) + (x - 1) = -2$ ⁠; for $-1 < x \le 1$ it is $(x + 1) + (x - 1) = 2x$ ⁠; for $x > 1$ it is $(x + 1) - (x - 1) = 2$ ⁠. Since $|x|$ is continuous (Example 7), $|x + 1|$ and $|x - 1|$ are continuous (Theorem 2, composition), and so is their difference (Theorem 1). Same conclusion. ✓

Common mistakes

Using $2x$ for $f(-1)$ ⁠. The middle condition is $-1 < x$ ⁠, strict, so $f(-1) = -2$ from the first piece. (Here $2(-1) = -2$ too, but read the conditions properly.)
Sign slips at $-1$ ⁠. $2(-1) = -2$ ⁠, not $2$ ⁠.
Testing only one junction. Three pieces give two junctions.
Calling the corners discontinuities. The graph bends at $(\pm 1, \pm 2)$ but does not break.

What to take away

Test every junction; a function can be continuous at all of them.
When neighbouring pieces give the same value at the junction, the graph joins up there.
Rewriting a piecewise function with absolute values is a good independent check.

Every fact used, and where it comes from

Fact	Used in	Source
Constant functions and polynomials are continuous	Steps 2–4	Examples 5 and 14, Section 5.2 (Lesson: continuity at a point)
A limit at $c$ depends only on values near $c$	Step 2	Class XI, Limits and Derivatives; Example 10, Section 5.2
Left-hand and right-hand limits	Steps 3, 4	Class XI, Limits and Derivatives; recalled after Definition 1, Section 5.2
Continuity at $c$ ⁠: left-hand limit $=$ right-hand limit $= f(c)$	Steps 3–5	Definition 1, Section 5.2
$\|x\|$ is continuous	Check	Example 7, Section 5.2
Sums, differences and composites of continuous functions are continuous	Check	Theorems 1 and 2, Section 5.2.1 (Lesson: algebra of continuous functions)

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Exercise 5.1 · Q16: Continuity of −2-2−2 (x≤−1x \le -1x≤−1⁠), 2x2x2x (−1<x≤1-1 < x \le 1−1<x≤1⁠), 222 (x>1x > 1x>1⁠)

Basics you need

Solution

Domain and junctions

Points strictly inside a piece

The junction x=−1x = -1x=−1

The junction x=1x = 1x=1

Conclude

Every fact used, and where it comes from

Exercise 5.1 · Q16: Continuity of $-2$ ( $x \le -1$ ⁠), $2x$ ( $-1 < x \le 1$ ⁠), $2$ ( $x > 1$ ⁠)

The junction $x = -1$

The junction $x = 1$