1.6 Function, SPM Practice (Long Question)


Question 1:
The function f and g is defined by
f:x2x3 g:x 2 x ;x0
Find the expression for each of the following functions
(a) ff,
(b) gf,
(c) f-1 ,
Calculate the value of x such that ff(x) = gf(x).

Solution:
(a)
ff( x )=f[ f( x ) ]          =f( 2x3 )          =2( 2x3 )3          =4x9 ff:x4x9

(b)
gf( x )=g[ f( x ) ]           =g( 2x3 )           = 2 2x3 gf:x 2 2x3

(c)
Let  f 1 ( x )=y, thus  f( y )=x       2y3=x               y= x+3 2    f 1 ( x )= x+3 2 f 1 :x x+3 2 When ff( x )=gf( x ), 4x9= 2 2x3 ( 4x9 )( 2x3 )=2 8 x 2 30x+27=2 8 x 2 30x+25=0 ( 4x5 )( 2x5 )=0 4x5=0       or       2x5=0 x= 5 4              or              x= 5 2


Question 2:
The function f and g is defined by
f( x )=3x2 g( x )= 3 x ,x0 Find (a)  f 1 ( 2 ), (b) gf( 3 ), (c) function h if hf( x )=3x+2, (d) function k if fk( x )=4x7.

Solution:
(a)
Let  f 1 ( 2 )=x, thus  f( x )=2       3x2=2            3x=4              x= 4 3 f 1 ( 2 )= 4 3

(b)
gf( 3 )=g[ 3( 3 )2 ]            =g( 11 )            = 3 11

(c)
h[ f( x ) ]=3x+2 h( 3x2 )=3x+2 Let y=3x2 thus     x= y+2 3      h( y )=3( y+2 3 )+2             =y+2+2             =y+4  h( x )=x+4

(d)
f[ k( x ) ]=4x7 3k( x )2=4x7 3k( x )=4x5 k( x )= 4x5 3

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