. Find
(a) the value of gf(3),
(b) the value of fg(-1 ),
(c) the composite functions fg,
(d) the composite functions gf,
(e) the composite functions g²,
(f) the composite functions f².

Example 1

Posted on April 18, 2020 by user

If f : x → x + 5 and g : x → x² +2x + 3, find

the value of gf (2),
the value of fg (2 ),
the composite functions fg,
the composite functions gf,
the composite functions g² ,
the composite functions f².

correction for part (c)

$\begin{array}{l} f g (x) = f (x^{2} + 2 x + 3) \\ = (x^{2} + 2 x + 3) + 5 \\ = x^{2} + 2 x + 3 + 5 \\ = x^{2} + 2 x + 8 \end{array}$

1.3a Composite Function

Posted on April 18, 2020 by user

Composite Function

If function

f : X \mapsto Y

,
and function

g : Y \mapsto Z

,
hence, composite function

g f : X \mapsto Z

Example:
If,

f : x \mapsto 2 x + 5

and

g : x \mapsto x^{2} - 1

, find

g f (2)

Answer:
$\begin{array}{l} f (x) = 2 x + 5 \\ f (2) = 2 (2) + 5 = 9 \end{array}$

$g f (2) = g [f (2)] = g (9)$

$\begin{array}{l} g (x) = x^{2} - 1 \\ g f (2) = g (9) = 9^{2} - 1 = 80 \end{array}$

Online Video Lessons

Youtube Video Lesson 1
Youtube Video Lesson 2
Youtube Video Lesson 3

Example 4 and 5

Posted on April 18, 2020 by user

Example 4
Given the function

f : x \mapsto 3 x + 2

, find the value of
(a)

f (2)

(b)

f (- 5)

(c)

f (\frac{1}{3})

Example 5
If

f (x) = x^{2} + 3 x + 2

, express each of the following in terms of x:
(a)

f (2 x)

(b)

f (3 x + 1)

(c)

f (x^{2})

Example 2 and 3

Posted on April 18, 2020 by user

Example 2

Function f is defined as

f : x \to \frac{5}{2 x - 1}, x \neq k

Find the value of k.

Example 3

Given

g : x \to \frac{3 x - 5}{2 x + 7}

Function g is defined for all values of x except x = a. Find the value of a.

Domain, Range, Objects, Images and Absolute Value Functions

Posted on April 18, 2020 by user

(B) Domain, Range, Objects and Images of a Function

Example:

The arrow diagram above represents the function f : x → 2x² – 5. State
(a) the domain,
(b) the range,
(c) the image of –2,
(d) the objects of
(i) –3,
(ii) –5.

Solution:
(a) Domain = {–2, –1, 0, 1, 2}.
(b) Range = {–5, –3, 3}.
(c) The image of –2 is 3.
(d) (i) The objects of –3 are 1 and –1.
(d) (ii) The objects of –5 is 0.

(C) Absolute Value Functions

1. Symbol | | is read as ‘the modulus’ of a number. In general, the modulus of x, that is | x |, is defined as

| x | = {\begin{cases} x if x \geq 0 \\ - x if x < 0 \end{cases}

2. In other words, modulus of a number always positive.
3. The absolute value function | f(x) | is defined by

| f (x) | = {\begin{cases} f (x) if f (x) \geq 0 \\ - f (x) if f (x) < 0 \end{cases}

Example:
Given function f : x → |x + 2|.
(a) Find the image of –4, –3, 0 and 2.
(b) Sketch the graph of f (x) for the domain –4 ≤ x ≤ 2.
Hence, state the range of values of f (x) based on the given domain.

Solution:
(a)
Given f (x) = |x + 2|
Image of –4 is f(–4) = | –4 + 2| = | –2| = 2
Image of –3 is f(–3) = | –3 + 2| = | –1| = 1
Image of 0 is f(0) = | 0 + 2| = | 2 | = 2
Image of 2 is f(2) = | 2 + 2| = | 4 | = 4

(b)
From (a),
f(–4) = 2
f(–3) = 1
f(0) = 2
f(2) = 4

Determine the point where the graph touches the x-axis.
At x-axis, f (x) = 0
|x + 2| = 0
x + 2 = 0
x = –2

Therefore, range of values of f (x) is 0 ≤ f (x) ≤ 4.

Notation of Function

Posted on April 18, 2020 by user

As shown in figure above, for a function

f : X \to Y

, each element x in the domain X has a unique image y in the codomain Y.

The function can be written as:

\begin{array}{l} y = f (x) \\ o r \\ f : x \mapsto f (x) \end{array}

For $y = f (x)$ , we say y is a function of x.
f(x) is also called the value of the function f at x.
f(x) is read as "f of x".

Example:
Given the function

f : x \mapsto 5 x + 1

, find the value of
a.

f (2)

f (- 3)

f (\frac{2}{5})

Answer:
(a)

\begin{array}{l} f (x) = 5 x + 1 \\ f (2) = 5 (2) + 1 = 11 \end{array}

(b)

\begin{array}{l} f (x) = 5 x + 1 \\ f (- 3) = 5 (- 3) + 1 = - 14 \end{array}

(c)

\begin{array}{l} f (x) = 5 x + 1 \\ f (\frac{2}{5}) = 5 (\frac{2}{5}) + 1 = 3 \end{array}