IB Analysis and Approaches HL · Functions · Section 1 of 15

Function, domain and range

3 worked examples · By Álvaro López

Key idea

A function is a rule that turns each input value $x$ into a single output value $f (x)$ . The allowed input values form the domain, and the output values that are actually obtained form the range. The whole of topic 9 rests on these two ideas, and the IB exam asks for the domain or the range in almost every function problem.

Function, notation and image

A relation is any set of ordered pairs $(x, y)$ . It becomes a function when each value of $x$ is paired with only one value of $y$ : the same input never produces two different outputs.

It is read " $f$ of $x$ ", and it means "the value that the function $f$ gives when the input is $x$ ". To calculate an image, substitute the input into the formula. If $f (x) = 2 x^{2} - 3$ , then $f (4) = 2 \cdot 16 - 3 = 29$ and $f (- 1) = 2 \cdot 1 - 3 = - 1$ . The letter used for the variable does not matter: $f (t) = 2 t^{2} - 3$ is the same function.

Be careful with the notation: $f (x + 1)$ means that the input is $x + 1$ , not that $1$ is added to the result. With the function above, $f (x + 1) = 2 (x + 1)^{2} - 3$ .

Domain and range

Think of a machine: the domain is the inputs it accepts, and the range is the outputs it produces. When the question does not state the domain, the largest possible one is taken, the natural domain: all the $x$ for which the formula makes sense.

Usual restrictions on the domain

In the functions of this course, only three things can go wrong when you substitute a number.

If several of these situations occur at the same time, all of the conditions must be met. A root inside a denominator, for example $\frac{1}{x - 1}$ , requires $x - 1 \geq 0$ and also $x - 1 \neq = 0$ , that is, $x > 1$ .

Interval notation

The domain and the range are written with intervals. In this course we use square brackets that point outwards for the endpoints that are not included:

$[a, b]$ : from $a$ to $b$ , both included ( $a \leq x \leq b$ ).
$] a, b [$ : from $a$ to $b$ , neither included ( $a < x < b$ ).
$[a, b [$ and $] a, b]$ : one included and the other not.
$] - \infty, a]$ and $[a, + \infty [$ : half-lines. Infinity is never included.
$\cup$ joins separate intervals, for example $] - \infty, - 3] \cup [3, + \infty [$ .
$R ∖ {2}$ means "all real numbers except $2$ ".

Worked examples

Example 1 Basic

Image and domain of a fraction

Let $g (x) = \frac{2 x - 1}{x + 3}$ .

a) Calculate $g (1)$ .

b) Find the natural domain of $g$ .

Solution

a) Substitute $x = 1$ : $g (1) = \frac{2 \cdot 1 - 1}{1 + 3} = \frac{1}{4}$

b) The fraction only fails when the denominator is zero: $x + 3 = 0 \Rightarrow x = - 3$ . Therefore the domain is all real numbers except $- 3$ : $R ∖ {- 3} =] - \infty, - 3 [\cup] - 3, + \infty [$

Example 2 Medium

Domain with a root and a logarithm, and a range

a) Find the domain of $f (x) = x^{2} - 16$ .

b) Find the domain of $g (x) = \frac{ln ( 5 - x )}{x - 1}$ .

c) Find the range of $h (x) = 3 + x - 2$ .

Solution

a) We need $x^{2} - 16 \geq 0$ , that is, $(x - 4) (x + 4) \geq 0$ . The parabola $x^{2} - 16$ is positive outside the roots $- 4$ and $4$ , and it is zero at them, so $] - \infty, - 4] \cup [4, + \infty [$

b) There are two conditions. The logarithm requires $5 - x > 0$ , that is, $x < 5$ . The denominator requires $x \neq = 1$ . Both hold at the same time on $] - \infty, 1 [\cup] 1, 5 [$

c) The domain of $h$ is $x \geq 2$ . At $x = 2$ the root is $0$ and $h (2) = 3$ ; as $x$ increases, the root grows without limit. Since a root is never negative, $h (x) \geq 3$ , and every one of these values is attained: $[3, + \infty [$

Example 3 Hard

A rectangle with a fixed perimeter

A rectangle has a perimeter of $40$ cm. One of its sides measures $x$ cm, and $A (x)$ is its area in cm $^{2}$ .

a) Write $A (x)$ as a function of $x$ .

b) Find the domain of $A$ in this context.

c) Find the range of $A$ .

d) Find the values of $x$ for which the area is $75$ cm $^{2}$ .

Solution

a) If one side measures $x$ , the two sides together add up to $20$ (half the perimeter), so the other side measures $20 - x$ . The area is $A (x) = x (20 - x)$

b) A side of a rectangle must be positive, and so must the other one: $x > 0$ and $20 - x > 0$ . The domain is not $R$ but $] 0, 20 [$

c) $A (x) = - x^{2} + 20 x$ is a parabola that opens downwards. Its zeros are $0$ and $20$ , so the vertex lies at $x = 10$ , and there $A (10) = 10 \cdot 10 = 100$ . Within the domain $] 0, 20 [$ the area takes every value greater than $0$ up to $100$ (the value $0$ is not reached, because $0$ and $20$ do not belong to the domain): $] 0, 100]$

d) Solve $x (20 - x) = 75$ : $x^{2} - 20 x + 75 = 0 ⟹ (x - 5) (x - 15) = 0 ⟹ x = 5 or x = 15$ Both values are in the domain and give the same rectangle, rotated.

In the exam: the domain of a context problem is marked separately. Write down the physical conditions ( $x > 0$ , $20 - x > 0$ ) and do not stop at the domain of the formula. And for the range, check whether the endpoint is reached or not, so that you choose the right bracket.

Common mistakes

Summary

A function gives each input $x$ a single output $f (x)$ ; it is written $f : x \mapsto f (x)$ .
Domain: the permitted values of $x$ . Range: the values of $y$ that are obtained.
The natural domain excludes zero denominators, negative radicands and logarithm arguments that are not positive.
Intervals are written with $] a, b [$ if the endpoint is not included and $[a, b]$ if it is; infinity is always open.
For the range, a sketch of the graph helps (in Paper 2, with the GDC).
In a context problem, the domain includes the physical conditions stated in the question.

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