2. This chapter is simple with 4 understanding
into it:
a) Function relation
b) Absolute function
c) Composite function
d) Inverse function

3. a) Function relation Set A is always > Object
> Domain
Example: Value of x depend on question
Square root of 9 = 3
Square root of 49 = 7
Square root of x = 9
So x is 81
9 3
49 7
x 9
The image that connect
Range to object only called
The images only. range.
{3,7,9}
Set A Set B
Object of 3 = 9 Image of 9 = 3 Set B is always > Image
Object of 7 = 49 Image of 49 = 7 > Codomain
Object of 9 = x Image of x = 9

4. b) Absolute function
Example:
The absolute value function f(x)= |7-x| is defined for the domain 0< x < 15
Find f(2)
This means you should substitute the
Modulus “| |” should be value of 2 into the equation.
removed. So your x=5
answers should end
with (+) sign.
Image of 15
|7-x|= 15
x= -8 Range of f(x)
So, x=8 corresponding to given
Find f(9) domain.
This means same substitution. You So it will be
will get x= -2. But since it is in 0 < f(x) < 8
modulus form. Your answer must be 2

5. c) Composite function
Find f(x)
Example: The “f” is located inside g:x.
So you must use
the method of SUBSTITUTION
Given g:x is 3x-2 and gf:x is 3x²+4.
Find g(2) Since gf:x is 3x²+4, so you must use g:x for
3x-2 Find g²x substitution. You must make the both
3(2)-2 It means gg(x). equation equal. The “x” in g:x should be
=4 So, you must substitute changed into f(x). In this way,
g:x into x of g:x. gf(x) = 3f(x)-2
3f(x)-2= 3x²+4
Find g(x)=4 3(3x-2)-2 3f(x)=3x²+4+2
3x-2=4 9x-6-2 3f(x)=3x²+6
3x=4+2 9x-8 f(x)=3x²+6
3x=6 g²x=9x-8 3
x=2 f(x)=x²+2

6. d) Inverse function
Example:
So, y 4= x
Given h:x is x 4 and kh:x is 2x-19
3
3
Then, y= x+4
3
Next, y=3x+12
Find hˉ¹(x)
This means you should inverse the h:x. hˉ¹(x) is 3x+12
When inverse always,
> make “x” as “y”
> write the normal equation
> after the equation is =x
Since kh:x is 2x-19, then you must use h:x
for inverse.You must do the usual inverse to
h:x same as hˉ¹(x). The inverse=3x+12 will
be substituted into x of kh:x.You
Find k(x) will get k(x) by expanding the equation like
below:
The “k” is located outside kh:x.
So, you must use the method of khˉ¹(x)= 2(3x+12)-19
INVERSE k(x)=6x+24-19
k(x)=6x+5

7. d) Inverse function
Example:
So, y 4= x
Given h:x is x 4 and kh:x is 2x-19
3
3
Then, y= x+4
3
Next, y=3x+12
Find hˉ¹(x)
This means you should inverse the h:x. hˉ¹(x) is 3x+12
When inverse always,
> make “x” as “y”
> write the normal equation
> after the equation is =x
Since kh:x is 2x-19, then you must use h:x
for inverse.You must do the usual inverse to
h:x same as hˉ¹(x). The inverse=3x+12 will
be substituted into x of kh:x.You
Find k(x) will get k(x) by expanding the equation like
below:
The “k” is located outside kh:x.
So, you must use the method of khˉ¹(x)= 2(3x+12)-19
INVERSE k(x)=6x+24-19
k(x)=6x+5