Example 2

Functions f and g are defined by $f:x\mapsto x-1$   and $g:x\mapsto \frac{3-x}{x+4}$  .  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.

Composite Function

Example:
If, $f:x\mapsto <!--\; \mapsto \; -->2x+5$ and $g:x\mapsto <!--\; \mapsto \; -->x^2-<!--\; -\; -->1$ , find $gf(2)$

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

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

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

Online Video Lessons

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

Example 4
Given the function $f:x\mapsto 3x+2$  , find the value of
(a) $f(2)$
(b) $f(-5)$
(c) $f\left(\frac{1}{3}\right)$

Example 5
If $f(x)=x^2+3x+2$  , express each of the following in terms of x:
(a) $f(2x)$
(b) $f(3x+1)$
(c) $f(x^2)$

Example 2

Function f is defined as $f:x\to \frac{5}{2x-1},x\ne k$
Find the value of k.

Example 3

(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

$\left|x\right|=\{\begin{array}{l}x\text{ if }x\ge 0\\ -x\text{ if }x<0\end{array}$

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

$\left|f(x)\right|=\{\begin{array}{l}f(x)\text{ if }f(x)\ge 0\\ -f(x)\text{ if }f(x)<0\end{array}$

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.

Example:
Given the function $f:x\mapsto <!--\; \mapsto \; -->5x+1$ , find the value of
a. $f(2)$
b. $f(-<!--\; -\; -->3)$
c. $f(\frac{2}{5})$

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

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

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

1. A function is a relation in which every element in the domain has a unique image (exactly one) in the codomain.
2. One-to-one relation and many-to-one relation are examples of a special kind of relation which we call function.

1. Relation can be classified into 4 types:
1. One-to-one relation
2. Many-to-one relation
3. One-to-many relation
4. Many-to-many relation

Domain and Codomain

1. In the relation between one set and another, the first set is known as the domain and the second set is known as the codomain.
2. Elements in the domain is called objects, whereas elements in the codomain mapped to the objects is called the image.
3. Elements in the codomain not mapped to the objects are not the image.
4. All images in codomain can be written as a set known as range.

Example:

Domain = {3, 4, 5}
Codomain = {7, 9, 12, 15}
Range = {9, 12, 15} [Note: 7 is not an image because it is not mapped to any object.]

3 is the object of 9, 12 and 15.
4 is the object of 12.
5 is the object of 15.

9, 12 and 15 are the images of 3.
12 is the image of 4.
15 is the image of 5.

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