Chapter 5 · Continuity and Differentiability

Exercise 5.1 · Q11: Points of discontinuity: $x^3 - 3$ for $x \le 2$ ⁠, $x^2 + 1$ for $x > 2$

Exercise 5.1Continuity of piecewise functions · Section 5.2

The question

Find all points of discontinuity of $f$ ⁠, where $f$ is defined by

f(x) = \begin{cases} x^3 - 3, & \text{if } x \le 2 \\ x^2 + 1, & \text{if } x > 2 \end{cases}

Exam answer

For $x < 2$ ⁠, $f(x) = x^3 - 3$ ⁠, and for $x > 2$ ⁠, $f(x) = x^2 + 1$ ⁠. Both are polynomials, so $f$ is continuous at every $x \ne 2$ ⁠.

At $x = 2$ ⁠:

\begin{aligned} \text{LHL} &= \lim_{h\to0} f(2 - h) \\ &= \lim_{h\to0}\bigl((2 - h)^3 - 3\bigr) = 5 \\ \text{RHL} &= \lim_{h\to0} f(2 + h) \\ &= \lim_{h\to0}\bigl((2 + h)^2 + 1\bigr) = 5 \end{aligned}

and $f(2) = 2^3 - 3 = 5$ ⁠. So $\text{LHL} = \text{RHL} = f(2)$ ⁠: $f$ is continuous at $x = 2$ ⁠.

Hence $f$ is continuous at every real number: there is no point of discontinuity.

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

The one idea this problem is built on

Away from $x = 2$ ⁠, $f$ is a single polynomial near each point, and polynomials are continuous. So everything depends on the junction $x = 2$ ⁠. There the cubic gives $2^3 - 3 = 5$ and the quadratic gives $2^2 + 1 = 5$ ⁠: the two very different formulas happen to meet at the same height. Left-hand limit, right-hand limit and $f(2)$ are all $5$ ⁠, so there is no point of discontinuity.

Given, to find, and the plan

Given: $f(x) = x^3 - 3$ for $x \le 2$ and $f(x) = x^2 + 1$ for $x > 2$ ⁠.
To find: Every point of the domain at which $f$ is not continuous.
Plan: (1) Domain and junction. (2) Continuity at every $c < 2$ ⁠. (3) Continuity at every $c > 2$ ⁠. (4) At $c = 2$ ⁠: left-hand limit, right-hand limit and $f(2)$ ⁠, each separately. (5) Conclude.

Basics you need

1Reading a piecewise function

The point $x = 2$ satisfies $x \le 2$ ⁠, so $f(2)$ uses $x^3 - 3$ ⁠. The formula $x^2 + 1$ is used only for $x$ strictly greater than $2$ ⁠. Basics: piecewise functions

2A limit only looks near the point

If $f$ equals a polynomial $p$ on an interval around $c$ ⁠, then $\lim_{x\to c} f(x) = \lim_{x\to c} p(x)$ ⁠. Basics: what a limit is

3Polynomials are continuous

$\lim_{x\to c} p(x) = p(c)$ for every polynomial $p$ and every real $c$ (Example 14).

4Left-hand and right-hand limits

$x \to 2^-$ uses only $x < 2$ ⁠; $x \to 2^+$ uses only $x > 2$ ⁠. Basics: one-sided limits

5Continuity at a point (Definition 1)

At a point $c$ of the domain: continuous when left-hand limit $=$ right-hand limit $= f(c)$ ⁠.

Solution

Domain and junction

Each real $x$ satisfies exactly one of $x \le 2$ and $x > 2$ ⁠, and both formulas are polynomials, defined everywhere. So the domain of $f$ is $\R$ ⁠, and the only junction is $x = 2$ ⁠.

Continuity is examined at points of the domain, and the junction is the one place where the definition of $f$ switches formula, so it needs its own test.

The conditions split the line into $(-\infty, 2]$ and $(2, \infty)$ ⁠. Basics: piecewise functions
Every point $c < 2$

Let $c < 2$ ⁠, so $f(c) = c^3 - 3$ ⁠. Every $x$ within distance $2 - c$ of $c$ is also less than $2$ ⁠, so there $f(x) = x^3 - 3$ ⁠. Hence
$\lim_{x\to c} f(x) = \lim_{x\to c} (x^3 - 3) = c^3 - 3 = f(c),$
and $f$ is continuous at every $c < 2$ ⁠.

Near a point strictly inside the left region, only the cubic formula is ever used, so $f$ behaves exactly like the polynomial $x^3 - 3$ there. The limit only looks near $c$ ⁠, so it is the polynomial's limit, which is its value.

$x^3 - 3$ is a polynomial, continuous by Example 14. Lesson: continuity at a point
Every point $c > 2$

Let $c > 2$ ⁠, so $f(c) = c^2 + 1$ ⁠. Every $x$ within distance $c - 2$ of $c$ is also greater than $2$ ⁠, so there $f(x) = x^2 + 1$ ⁠. Hence
$\lim_{x\to c} f(x) = \lim_{x\to c} (x^2 + 1) = c^2 + 1 = f(c),$
and $f$ is continuous at every $c > 2$ ⁠.

The same argument on the right of the junction, with the quadratic formula.

$x^2 + 1$ is a polynomial, continuous everywhere.
Test the junction $x = 2$

Value. $2 \le 2$ ⁠, so $f(2) = 2^3 - 3 = 8 - 3 = 5$ ⁠.

Left-hand limit. For $x < 2$ ⁠, $f(x) = x^3 - 3$ ⁠:
$\lim_{x\to 2^-} f(x) = \lim_{x\to 2^-} (x^3 - 3) = 8 - 3 = 5.$
Right-hand limit. For $x > 2$ ⁠, $f(x) = x^2 + 1$ ⁠:
$\lim_{x\to 2^+} f(x) = \lim_{x\to 2^+} (x^2 + 1) = 4 + 1 = 5.$
Compare. Left-hand limit $= 5$ ⁠, right-hand limit $= 5$ ⁠, $f(2) = 5$ ⁠. All equal, so $f$ is continuous at $x = 2$ .

Each one-sided limit is a limit of a polynomial, so it is found by substituting $x = 2$ into the formula that is in force on that side. The cubic and the quadratic are very different formulas, but at $x = 2$ both give $5$ ⁠, and that agreement of values (together with $f(2) = 5$ ⁠) is all that continuity at the junction requires.

One-sided limits: compute each with the formula valid on that side only. Basics: one-sided limits
Conclude

Steps 2, 3 and 4 cover every real number: $c < 2$ ⁠, $c > 2$ and $c = 2$ ⁠. At each, $f$ is continuous. So $f$ has no point of discontinuity.

As in Question 10, the graph has a corner at the junction (the cubic arrives with slope $3 \cdot 2^2 = 12$ ⁠, the parabola leaves with slope $2 \cdot 2 = 4$ ⁠), but no break. Corners concern differentiability (Section 5.3), not continuity.

Blue:

y = x^3 - 3

for

x \le 2

⁠, ending at the filled point

(2, 5)

⁠. Gold:

y = x^2 + 1

for

x > 2

⁠, starting from the same point. No gap. Horizontal scale: 1 unit = 80 px; vertical: 1 unit = 18 px.

Result

Both pieces are polynomials, so $f$ is continuous at every $x \ne 2$ ⁠. At $x = 2$ ⁠: left-hand limit $5$ ⁠, right-hand limit $5$ ⁠, $f(2) = 5$ ⁠. $f$ is continuous everywhere: there is no point of discontinuity. (Textbook answer: "No point of discontinuity".)

Check it with numbers

$f(1.999) = 1.999^3 - 3$ $= 7.988005999 - 3$ $= 4.988005999$ ⁠. $f(2.001) = 2.001^2 + 1$ $= 4.004001 + 1 = 5.004001$ ⁠. $f(2) = 5$ ⁠. Both neighbours are within about $0.012$ of $5$ ⁠, and halving the distance to $2$ roughly halves the gap ( $f(1.9995) \approx 4.994$ ⁠, $f(2.0005) \approx 5.002$ ⁠). ✓

Common mistakes

Using $x^2 + 1$ for $f(2)$ ⁠. The condition for the second formula is $x > 2$ ⁠, strict, so $f(2)$ comes from $x^3 - 3$ ⁠. (Both give 5 here, but the reason must be right.)
Arithmetic slips. $2^3 - 3 = 5$ ⁠, not $3$ ⁠; $2^2 + 1 = 5$ ⁠.
Thinking different formulas mean a break. A change of formula is only a candidate for a break; the one-sided limits decide.

What to take away

Polynomial pieces are continuous inside their own regions; test only the junctions.
At a junction, substitute into each side's formula for the one-sided limits, and into the correct formula for the value.
Equal one-sided limits that also equal the value mean continuity, even if the formulas look nothing alike.

Every fact used, and where it comes from

Fact	Used in	Source
Every polynomial is continuous	Steps 2–4	Example 14, Section 5.2 (Lesson: continuity at a point)
A limit at $c$ depends only on values near $c$	Steps 2, 3	Class XI, Limits and Derivatives; used in Example 10, Section 5.2
Left-hand and right-hand limits	Step 4	Class XI, Limits and Derivatives; recalled in Section 5.2
Continuity at $c$ ⁠: left-hand limit $=$ right-hand limit $= f(c)$	Steps 4, 5	Definition 1, Section 5.2
Slopes $3x^2$ and $2x$ (corner remark only)	Step 5	Class XI derivatives of $x^n$ ⁠; Section 5.3

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Exercise 5.1 · Q11: Points of discontinuity: x3−3x^3 - 3x3−3 for x≤2x \le 2x≤2⁠, x2+1x^2 + 1x2+1 for x>2x > 2x>2

Basics you need

Solution

Domain and junction

Every point c<2c < 2c<2

Every point c>2c > 2c>2

Test the junction x=2x = 2x=2

Conclude

Every fact used, and where it comes from

Exercise 5.1 · Q11: Points of discontinuity: $x^3 - 3$ for $x \le 2$ ⁠, $x^2 + 1$ for $x > 2$

Every point $c < 2$

Every point $c > 2$

Test the junction $x = 2$