IB Analysis and Approaches HL · Functions · Section 11 of 15

Continuity of a function

3 worked examples · By Álvaro López

Key idea

A function is continuous when you can draw it without lifting the pencil. Limits state this precisely, and from it come three useful things: classifying the "jumps", finding the parameter that makes a piecewise function continuous, and guaranteeing that an equation has a solution. Continuity and its relation to the derivative are HL content and appear in both papers.

Continuity at a point

To draw the graph through $x = a$ without lifting the pencil, three things are needed: the function must exist at $a$ , it must approach a value as you arrive at $a$ from both sides, and that value must be exactly the value of the function at $a$ .

Almost all the functions you use are continuous at every point where they are defined: polynomials, exponential functions, sine and cosine, and those obtained by adding, multiplying, composing or dividing (with a non-zero denominator) continuous functions. Problems appear at the points where the function is defined piecewise or where a denominator is zero.

Types of discontinuity

Depending on which condition fails, there are three cases. (These names help to organise the ideas; in the IB exam it is enough to describe what happens.)

A function with three discontinuities: at x = a a hole with the point displaced (removable), at x = b a finite jump and at x = c a vertical asymptote (infinite jump)

In the figure: at $a$ the limit exists (the curve passes through the height of the hole), but the point is higher up; at $b$ the curve jumps from $2.5$ to about $3.4$ ; at $c$ it shoots off to infinity.

Continuity on an interval

Piecewise functions with a parameter

When a function is defined by different formulas on each side of a point, continuity at that point forces the two pieces to "match".

If the parameter appears in a function with a hole, for example $f (x) = \frac{x ^{2} - 9}{x - 3}$ for $x \neq = 3$ and $f (3) = k$ , the condition is $k = x \to 3 lim f (x)$ : you choose the value that fills the hole.

Three theorems about continuous functions on $[a, b]$

Being continuous on a closed interval has important consequences that do not depend on the formula.

Intuitively: to go from height $f (a)$ to height $f (b)$ without lifting the pencil, you must pass through all the heights in between.

It is the special case $k = 0$ of the previous theorem: the graph crosses the $x$ -axis at least once. It may cross it more times; the theorem only guarantees one.

Function continuous on the interval from a to b with f(a) negative and f(b) positive: its graph crosses the x-axis at least once at a point c between a and b

The conditions matter. Without a closed interval the result fails: $f (x) = x$ on $] 0, 1 [$ has neither a maximum nor a minimum; and without continuity too: $f (x) = \frac{1}{x}$ on $] 0, 1]$ has no maximum. The maximum or the minimum may be at an endpoint of the interval.

Worked examples

Example 1 Basic

Classifying discontinuities

Consider the functions

$f (x) = {2 x + 1 5 - x^{2} if x < 1 if x \geq 1 g (x) = ⎩ ⎨ ⎧ \frac{x ^{2} - 4}{x - 2} 1 if x \neq = 2 if x = 2 h (x) = \frac{2}{x - 3}$

State the type of discontinuity, or write "continuous" if there is none, at:

a) $f$ at $x = 1$ .

b) $g$ at $x = 2$ .

c) $h$ at $x = 3$ .

Solution

a) From the left, $x \to 1^{-} lim f (x) = 2 \cdot 1 + 1 = 3$ . From the right, $x \to 1^{+} lim f (x) = 5 - 1 = 4$ . The one-sided limits exist but are different: finite jump.

b) For $x \neq = 2$ , $g (x) = \frac{( x - 2 ) ( x + 2 )}{x - 2} = x + 2$ , so $x \to 2 lim g (x) = 4$ . But $g (2) = 1 \neq = 4$ . The limit exists and does not coincide with the value: removable discontinuity.

c) $h (3)$ does not exist and, from the right, $\frac{2}{x - 3} \to + \infty$ (from the left, $\to - \infty$ ): infinite jump.

Example 2 Medium

Parameters that make a function continuous

a) Find the value of $k$ so that the function $f (x) = \frac{x ^{2} - 9}{x - 3}$ for $x \neq = 3$ , $f (3) = k$ , is continuous at $x = 3$ .

b) Find the value of $a$ so that the function

$g (x) = {a x + 1 x^{2} - a if x < 2 if x \geq 2$

is continuous at $x = 2$ .

Solution

a) For $x \neq = 3$ , $f (x) = \frac{( x - 3 ) ( x + 3 )}{x - 3} = x + 3$ , so $x \to 3 lim f (x) = 6$ . For continuity we need $f (3) = 6$ , that is, $k = 6$ .

b) From the left, $x \to 2^{-} lim g (x) = 2 a + 1$ . From the right, $x \to 2^{+} lim g (x) = 4 - a$ , which is also the value $g (2) = 4 - a$ . Equate them:

$2 a + 1 = 4 - a ⟹ 3 a = 3 ⟹ a = 1$

Check: with $a = 1$ , the left-hand limit is $3$ and $g (2) = 4 - 1 = 3$ .

Example 3 Hard

Two parameters and a guaranteed solution

Consider the function

$f (x) = ⎩ ⎨ ⎧ x + 2 a x + b 10 - x if x < 1 if 1 \leq x \leq 3 if x > 3$

and the function $h (x) = x^{3} + x - 3$ .

a) Find the values of $a$ and $b$ so that $f$ is continuous on $R$ . Give the pair $(a, b)$ .

b) Calculate $h (1)$ and $h (2)$ . Give the pair $(h (1), h (2))$ .

c) Justify that the equation $h (x) = 0$ has a solution in $] 1, 2 [$ , and find its value to two decimal places using the calculator.

d) Find the minimum and the maximum values of $h$ on $[0, 2]$ , knowing that $h$ is increasing. Give the pair (minimum, maximum).

Solution

a) There are two join points: $x = 1$ and $x = 3$ .

At $x = 1$ : from the left the limit is $1 + 2 = 3$ ; the middle piece gives $f (1) = a + b$ . It must be $a + b = 3$ .

At $x = 3$ : from the right the limit is $10 - 3 = 7$ ; the middle piece gives $f (3) = 3 a + b$ . It must be $3 a + b = 7$ .

Subtracting the two equations: $2 a = 4$ , so $a = 2$ and $b = 1$ . Check: $f (1) = 3$ and $f (3) = 7$ , and the one-sided limits coincide at both points.

b) $h (1) = 1 + 1 - 3 = - 1$ and $h (2) = 8 + 2 - 3 = 7$ .

c) $h$ is a polynomial, so it is continuous on $[1, 2]$ . As $h (1) = - 1 < 0$ and $h (2) = 7 > 0$ , Bolzano's theorem guarantees a $c \in] 1, 2 [$ with $h (c) = 0$ . With the GDC, the graph crosses the $x$ -axis at $c \approx 1.21$ .

d) As $h$ is continuous on the closed interval $[0, 2]$ , Weierstrass's theorem guarantees that it attains a maximum and a minimum. Since it is increasing, the minimum is at the left endpoint and the maximum at the right one: $h (0) = - 3$ and $h (2) = 7$ .

In the exam: in a justification using Bolzano you must write all three things: " $h$ is continuous on $[1, 2]$ ", " $h (1) < 0$ and $h (2) > 0$ " and the conclusion. If the continuity is missing, the part does not earn full marks.

Common mistakes

Summary

$f$ is continuous at $a$ when $f (a)$ exists, $x \to a lim f (x)$ exists and both coincide.
Discontinuities: removable (the limit exists but does not equal $f (a)$ ), finite jump (different one-sided limits) or infinite jump (an infinite one-sided limit).
With a parameter in a piecewise function: equate the one-sided limits and the value of the function at each join point.
Intermediate value: a function continuous on $[a, b]$ takes every value between $f (a)$ and $f (b)$ . Bolzano: if also $f (a) f (b) < 0$ , there is a root in $] a, b [$ .
Weierstrass: continuous on a closed $[a, b]$ implies an absolute maximum and minimum. In Paper 2, the GDC locates the solution.

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Continuity of a function

Continuity at a point

Types of discontinuity

Continuity on an interval

Piecewise functions with a parameter

Three theorems about continuous functions on [a,b]

Worked examples

Common mistakes

Summary

Want to work on it together?

Three theorems about continuous functions on $[a, b]$