IB Mathematics AI SL Concept of a function, domain, range and graph MAI Study Notes - New Syllabus

 FUNCTIONS IN GENERAL, DOMAIN, RANGE, GRAPH

DEFINITION

Let us formally introduce the notion of the function:

\( f: X \rightarrow Y \)

A function from a set \( X \) to a set \( Y \) assigns to each element \( x \) of \( X \)
a unique element \( y \) of \( Y \)

We write:

\( \begin{aligned}
f(x) &= y \\
f: x &\mapsto y
\end{aligned} \)

We say:

\( \begin{aligned}
f(x) &= y \\
f: x &\mapsto y
\end{aligned} \)

EXAMPLE 1
Let \( X = \{1, 2, 3\} \) and \( Y = \{a, b, c, d\} \). The following is a function \( f: X \rightarrow Y \)

Indeed, each element of \( X \) has a unique image in \( Y \).
We say
\( f \) maps 1 to \( a \) or \( a \) is the image of 1
2 to \( b \) \( b \) is the image of 2
3 to \( d \) \( d \) is the image of 3

We write

\( \begin{aligned}
f(1) &= a, \quad f(2) = b, \quad f(3) = d \\
\text{or } f: 1 &\mapsto a \quad f: 2 \mapsto b \quad f: 3 \mapsto d
\end{aligned} \)

EXAMPLE 2
Let \( X = \{1, 2, 3\} \) and \( Y = \{a, b, c, d\} \)

 The following is a function
(we do not mind if two elements of \( X \) have the same image)

Notice though that the following is not a function
(we said “each \( x \) of \( X \),” but here 3 has no image)

 Finally, the following is not a function
(we said “unique \( y \) of \( Y \),” but 2 has two images)

DOMAIN AND RANGE

For a function \( f: X \rightarrow Y \),
The set of all \( x \)’s involved is called DOMAIN
The set of all \( y \)’s involved (only the images) is called RANGE

Consider again the function \( f: X \rightarrow Y \) given by

Then
DOMAIN: \( x \in X = \{1, 2, 3\} \)
RANGE: \( y \in \{a, b, d\} \)

We usually denote the domain by \( D_f \) and the range by \( R_f \).

Here, the sets \( X \) and \( Y \) are subsets of \( \mathbb{R} \), the set of real numbers.
Our functions usually have a specific pattern. For example, consider the function \( f \) which maps

\( 1 \mapsto 2 \quad 2 \mapsto 4 \quad 3 \mapsto 6 \quad 4 \mapsto 8 \quad \text{and so on} \)

in other words \( f \) maps each value \( x \) to its double \( 2x \).
We say that the function \( f \) is given by

\( \begin{aligned}
f: x &\mapsto 2x \\
\text{or } f(x) &= 2x \\
\text{or } y &= 2x
\end{aligned} \)

Thus the formula of the function gives any possible result, e.g.

\( f(1) = 2, f(15) = 30, f(2.4) = 4.8 \quad \text{etc} \)

If we restrict the function \( f \) from \( \mathbb{R} \) to the interval \( X = [0, 10] \), we still have the function \( f: X \rightarrow \mathbb{R} \), given by

\( f(x) = 2x, \quad 0 \leq x \leq 10 \)

but now

\( \begin{aligned}
\text{DOMAIN} &: x \in [0, 10] \\
\text{R Alicia} &: y \in [0, 20] \text{ (why?) }
\end{aligned} \)

GRAPH

We know that the pairs \( (x, y) \) that satisfy the equation of the function \( y = f(x) \) can be represented as points \( (x, y) \) on the Cartesian plane and form the graph of the function.

The graph clearly shows the DOMAIN and the RANGE of the function. For example,
DOMAIN: Projection on the \( x \)-axis, i.e. \( D_f: x \in [1, 8] \)
RANGE: Projection on the \( y \)-axis, i.e. \( R_f: y \in [2, 6] \)

We may observe, for example, that the points
\( (1, 2), (5, 3), (7, 6), (8, 5) \) lie on the curve.

That implies

\( f(1) = 2 \quad f(5) = 3 \quad f(7) = 6 \quad f(8) = 5 \)

We have already studied the graphs of two families of functions; linear and quadratic functions. The graphs are straight lines and parabolas respectively.

EXAMPLE 3

 \( f(x) = 2x \), or otherwise \( y = 2x \) is represented by the graph
Here \( D_f: x \in \mathbb{R} \)
\( R_f: y \in \mathbb{R} \)

\( f(x) = x^2 \), or otherwise \( y = x^2 \) is represented by the graph
Here \( D_f: x \in \mathbb{R} \)
\( R_f: y \in [0, +\infty) \)

EXAMPLE 4
Consider the function \( f(x) = \begin{cases} x^2, & -2 \leq x \leq 0 \\ x, & 1 \leq x \leq 5 \end{cases} \)

The graph is given below

Clearly, \( D_f: x \in [-2, 0] \cup [1, 5] \) and \( R_f: y \in [0, 5] \)

NOTICE:
The graph also shows if we have a function or not

This is not a function, since \( f(3) \) for example is not unique!
Vertical line test:
Any vertical line intersects the graph at most once.

AN “AGREEMENT” FOR THE DOMAIN

Usually, a function is simply given as a formula of the form \( y = f(x) \), where \( x \) and \( y \) are real variables.

If the domain of the function is not given, we agree that

\( D_f \text{ is } \mathbb{R} \)

or \( D_f \) is the largest possible subset of \( \mathbb{R} \)
For example,

 if \( f \) is given by \( f(x) = 2x \), we assume that \( x \in \mathbb{R} \)
 if \( f \) is given by \( f(x) = \frac{2}{x} \), we assume that \( x \in \mathbb{R} – \{0\} = \mathbb{R} \)
(we may also write \( D_f: x \neq 0 \))

We mainly deal with the following cases

1. \( f(x) \) is a function with no restrictions on \( x \),
for example a polynomial [say \( f(x) = 2x^3 + 3x^2 + 1 \)], then

\( D_f = \mathbb{R} \)

2. \( f(x) = \frac{A}{B} \), then \( B \) cannot be 0, thus

\( D_f = \mathbb{R} – \{\text{roots of the equation } B = 0\} \)

3. \( f(x) = \sqrt{A} \), then \( A \geq 0 \).

\( D_f = \text{the solution set of the inequality } A \geq 0 \)

4. \( f(x) = \log A \) or \( f(x) = \ln A \), then \( A > 0 \).¹

\( D_f = \text{the solution set of the inequality } A > 0 \)

5. \( f(x) = \) is a combination of all the above.
We find the subset of \( \mathbb{R} \) where all our restrictions hold.

The functions \( f(x) = \log x \) and \( f(x) = \ln x \) are not known yet. They will be introduced later on within this topic.

EXAMPLE 5
a) \( f(x) = 3x – 9 \). Clearly, \( D_f: x \in \mathbb{R} \)

b) \( f(x) = \frac{5}{3x – 9} \).
Restriction: \( 3x – 9 \neq 0 \)
Solve: \( 3x – 9 = 0 \Leftrightarrow 3x = 9 \Leftrightarrow x = 3 \)
Thus, \( D_f: x \in \mathbb{R} – \{3\} \). We may also write \( D_f: x \neq 3 \)

c) \( f(x) = \sqrt{3x – 9} \).
Restriction: \( 3x – 9 \geq 0 \)
Solve: \( 3x – 9 \geq 0 \Leftrightarrow 3x \geq 9 \Leftrightarrow x \geq 3 \)
Thus, \( D_f: x \in [3, +\infty) \). We may also write \( D_f: x \geq 3 \)

d) \( f(x) = \ln (3x – 9) \). Restriction: \( 3x – 9 > 0 \)
Solve: \( 3x – 9 > 0 \Leftrightarrow 3x > 9 \Leftrightarrow x > 3 \)
Thus, \( D_f: x \in (3, +\infty) \). We may also write \( D_f: x > 3 \)

e) \( f(x) = \frac{x + 2}{x^2 – 3x + 2} \)
Restriction: \( x^2 – 3x + 2 \neq 0 \)
Solve: \( x^2 – 3x + 2 = 0 \Leftrightarrow x = 1 \text{ or } x = 2 \)
Thus, \( D_f: x \in \mathbb{R} – \{1, 2\} \)

f) \( f(x) = \sqrt{x – 1} + \sqrt[3]{2 – x} \)
Restrictions: \( x – 1 \geq 0 \) and \( 2 – x \geq 0 \)
Solve: \( x – 1 \geq 0 \Leftrightarrow x \geq 1 \)
\( 2 – x \geq 0 \Leftrightarrow x \leq 2 \)
Thus, \( D_f: x \in [1, 2] \quad \text{We may also write } D_f: 1 \leq x \leq 2 \)

g) \( f(x) = \frac{\sqrt{1 – x^2}}{x} \)
Restrictions: \( 1 – x^2 \geq 0 \) and \( x \neq 0 \)
Solve: \( 1 – x^2 \geq 0 \Leftrightarrow x^2 \leq 1 \Leftrightarrow -1 \leq x \leq 1 \)
Thus, \( D_f: x \in [-1, 0) \cup (0, 1] \)

 AN INFORMAL IDEA OF THE INVERSE FUNCTION: \( f^{-1} \)

Consider the functions \( f(x) = x + 10 \) and \( g(x) = x – 10 \)
Notice that

\( \begin{aligned}
f(0) &= 10 & g(10) &= 0 \\
f(1) &= 11 & g(11) &= 1 \\
f(2) &= 12 & g(12) &= 2 \\
f(3) &= 13 & g(13) &= 3
\end{aligned} \)

In simple words,
the function \( f \) adds 10 to any value
the function \( g \) does the inverse process, i.e. it subtracts 10.

The inverse function of \( f \), that is \( g \), will be denoted by \( f^{-1} \)

\( \begin{aligned}
f(x) &= x + 10 \\
f^{-1}(x) &= x – 10
\end{aligned} \)

Mathematically

\( \text{If } f(x) = y \text{ then } f^{-1}(y) = x . \)

In fact, \( f \) and \( f^{-1} \) are inverse to each other.
Along the same lines

Notice. In fact, the inverse function of \( f(x) = e^x \) is \( f^{-1}(x) = \ln x \) (known as logarithmic) which we are going to see in a while.

Guess the inverse function of

\( f(x) = 3x + 7 \)

Here
I give you \( x \), you multiply by 3, you add 7, you find \( y \)

The inverse process is
you give me \( y \), I subtract 7, I divide by 3, I obtain \( x \)

So the inverse function is

\( x = \frac{y – 7}{3} \)

But as we usually express functions in terms of \( x \),

\( f^{-1}(x) = \frac{x – 7}{3} \)

In a similar way, we can deduce that the inverse function of

\( f(x) = x^3 + 5 \)

is

\( f^{-1}(x) = \sqrt[3]{x – 5} \)

GRAPH OF \( f^{-1} \)

The graph of \( f^{-1} \) is a reflection of \( f \) about the line \( y = x \)

EXAMPLE 1
If \( f(x) = x^2 \), for \( x \geq 0 \), then \( f^{-1}(x) = \sqrt{x} \). Their graphs are

Notice:

 \( f \) and \( f^{-1} \) intersect on the line \( y = x \).
 The points of intersection are \( (0, 0) \) and \( (1, 1) \)
(they are on the line \( y = x \))
\( f(2) = 4 \) and thus \( f^{-1}(4) = 2 \).

THE INVERSE FUNCTION IN REAL LIFE PROBLEMS

In real life problems instead of \( x \) and \( y \) we may have other parameters.
For example, a square of side \( a \)

has
Perimeter \( P = 4a \)
Area \( A = a^2 \)

(it is the function \( y = 4x \))
(it is the function \( y = x^2 \))

If they give us the perimeter \( P \), then

\( P = 4a \Leftrightarrow a = \frac{P}{4} \)

Now the side is given in terms of the perimeter. This is in fact the inverse function of \( P = 4a \)

If they give us the area \( A \), then

\( A = a^2 \Leftrightarrow a = \sqrt{A} \)

Now the side is given in terms of the area. This is in fact the inverse function of \( A = a^2 \).

EXAMPLE 2
Let
\( F \) denote the temperature in Fahrenheit degrees
\( C \) denote the temperature in Celsius degrees

The conversion from Celsius to Fahrenheit is given by

\( F = 1.8C + 32 \)

\( F(30) = 86 \) implies that \( 30^\circ \) Celsius is equal to \( 86^\circ \) Fahrenheit

If we solve for \( C \) then

\( C = \frac{F – 32}{1.8} \)

\( C(86) = 30 \) implies that \( 86^\circ \) Fahrenheit is equal to \( 30^\circ \) Celsius

We can say that

\( \begin{aligned}
F(30) &= 86 \quad \text{(Celsius to Fahrenheit)} \\
\text{and} \\
F^{-1}(86) &= 30 \quad \text{(Fahrenheit to Celsius)}
\end{aligned} \)