Cambridge International Mathematics

724 ANSWERS

d

e

f

EXERCISE 23A.2

1af(x)=2(x+ 1)(x¡2)(x¡3)

bf(x)=¡2(x+ 3)(x+ 2)(2x+1)

c f(x)=^14 (x+4)^2 (x¡3)

2af(x)=2x^3 ¡ 5 x^2 ¡ 6 x+9

bf(x)=3x^3 ¡ 16 x^2 +15x+18

3 b=4,c=¡ 10 4 b=2,d=¡ 8

EXERCISE 23B

1af¡^1 (x)=x+7 bf¡^1 (x)=x¡^2

3

c f¡^1 (x)=

3 ¡ 4 x

2

df¡^1 (x)=^3

p

x

ef¡^1 (x)=^3

q

x¡ 1

2

f f¡^1 (x)=^3 x+1

4

gf¡^1 (x)=x^2 ¡ 1 ,x> 0 hf¡^1 (x)=

x^2 +5

3

, x> 0

i f¡^1 (x)=

1

x

+2

2a if¡^1 (x)=8¡x ii f¡^1 (x)=^9

x

bBoth functions are their own inverses.

3aFor f(x)=mx+c, m 6 =0,

f¡^1 (x)=^1

m

x¡c

m

which is linear asm 6 =0

bThe gradients are reciprocals.

c If (a,b)lies ony=f(x) thenb=ma+c

and f¡^1 (b)=^1

m

b¡c

m

=

b¡c

m

=ma

m

=a asm 6 =0

dif¡^1 (x)=4x+2 ii f¡^1 (x)=¡^32 x¡ 1

4b idoes iidoes not iii does iv does not

5a

This horizontal line cuts the

graph more than once.

) f¡^1 (x)does not exist.

b

Likewise, f¡^1 (x) does not

exist.

c

Likewise, f¡^1 (x) does not

exist.

6a i

ii

iii

b ...... a reflection ofy=f(x) in the line y=x.

7 From 6 , y=f¡^1 (x) is a reflection ofy=f(x) in the line

y=x. So, when x=k is reflected in y=x to become

y=k,y=kmust be a horizontal asymptote ofy=f¡^1 (x).

8a if¡^1 (x)=^2

x

+3

ii

@=2!W(!-3)

y

3 x

O

@=-Qr_(!-2)W(!+1)

y

x

2

-1 -1

O

@ =-3(! + 1)W(! -We_)

y

-1 x

2

We_

O

y

x

yx¡=¡X

O

y

O x

y

x

4

-2 O

yx¡=¡

yx¡=¡2 ¡+¡1¡=¡¦()x

2

f-^1 =x-^1

y

x

1

1

O

O

y

x

y fx f x x

= ()= -^1 ()=^2

yx¡=¡

y

x

y= x=f(x),x> 0

f-^1 (x)=x^2 ,x> 0

O

y

x

3

()^2

fx=x-

y=f-^1 (x)

y¡=¡3

x¡=¡3

yx¡=¡

O

IB MYP_3 ANS

cyan magenta yellow black

(^05255075950525507595)
100 100
(^05255075950525507595)
100 100
Y:\HAESE\IGCSE01\IG01_an\724IB_IGC1_an.CDR Thursday, 20 November 2008 4:49:18 PM PETER