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Published by weilyew0123, 2020-04-02 05:10:00

MT F4 Chapter 1 Functions_2

MT_F4_1 Functions_2

1.6 Absolute Value Functions

(i) Absolute value of x = |x|, read as ‘the modulus of x’.
(ii) |x|is the numerical value of x.

For example: |5|= 5 and |–5|= 5

Example:
1 Given the function f : x  4x 11 , find
(a) the image of 1,

f (x)  4x 11
f (1)  4(1) 11

 4 11
 7
7

The image of 1 is 7.

(b) the image of 5,

f (5)  4(5) 11
 20 11
9
9

The image of 5 is 9.

(c) the objects that have the image 13.

f (x)  13

4x 11  13

4x 11  13

4x 11  13 or 4x 11  13
4x  13 11
4x  13 11 4x  2
x  2 1
4x  24 42

x  24  6
4

The objects that have the image 13 are 6 and  1 .
2

2 Given the function f (x)  2  5x , find

(a) the image of 2,
(b) the objects that have the image 7.

1.7 Composite Functions

Example:

1 The functions f and g are defined by f : x  2x 1 and

g : x  x2  2 respectively. Find
(a) the value of fg(3) ,
(b) the value of gf (2) .

(a) fg(3)  f g3 object

 f 32  2 g(x)  x2  2
g(3)  32  2
 f 7
 27 1 object

 15 f (x)  2x 1

(b) gf (2)  g  f  2 f (7)  27 1
 g 2 21
 g  3
  32  2

7

2 Find the composite functions fg and gf for the following pairs of
functions f and g .

(i) f : x  x2 1 and g : x  3x 1 fg  fg(x)
gf  gf (x)

(ii) f :x 2x and g:x 1
x2  2

1. Given the function f x  2 , x  0, , find the composite function f 2.

x

2. The functions f and g are defined by f : x  x 12 and g : x  x  2

resprctively. Find the composite functions of

(a) fg
(b) gf

Can u find out the difference?

1 kx  x  4 4 hx  x 1

2 f x 1  3x  2 5 f x  x2  5

3 hx  x2  2x 1 6 g2x  x2  x  3

2 f  x 1   3 x  2 f  y   3y 1  2

object  3y  3 2
 3y1
2 f  x 1   3 x  2
f x   3 x  1
let x 1  y ,
x  y 1

f y  3y  1  2

 3y  3 2
 3y 1

 f x  3x 1

6 g2x  x2  x  3  g x  x 2  2 x  12
4
let 2x  y ,
x y
2

g2x  x2  x  3
gy   y 2  y  3

2 2
 y2  y  3

42
 y2  2 y 12

4

 gx  x2  2x 12

4


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