Answers 491

2a x Point B Gradient of AB

0 (0,0) 2

1 (1,1) 3

1 : 5 (1: 5 , 2 :25) 3 : 5

1 : 9 (1: 9 , 3 :61) 3 : 9

1 : 99 (1: 99 , 3 :9601) 3 : 99

1 : 999 (1: 999 , 3 :996 001) 3 : 999

x Point B Gradient of AB

5 (5,25) 7

3 (3,9) 5

2 : 5 (2: 5 , 6 :25) 4 : 5

2 : 1 (2: 1 , 4 :41) 4 : 1

2 : 01 (2: 01 , 4 :0401) 4 : 01

2 : 001 (2: 001 , 4 :004 001) 4 : 001

b lim

x! 2

x^2 ¡ 4

x¡ 2

=4

The gradient of the tangent to y=x^2 at the point(2,4)

is 4.

EXERCISE 13C

1af(2) = 3 bf^0 (2) = 0

2af(0) = 4 bf^0 (0) =¡ 1 3 f(2) = 3,f^0 (2) = 1

EXERCISE 13D

1af^0 (x)=1 bf^0 (x)=0 c f^0 (x)=2

2a

dy

dx

=¡ 1 b

dy

dx

=2x¡ 3 c

dy

dx

=4x+1

3a 3 b¡ 12 c 9 d 10

EXERCISE 13E

1af^0 (x)=3x^2 bf^0 (x)=6x^2

cf^0 (x)=14x df^0 (x)=p^3

x

ef^0 (x)=p 31

x^2

ff^0 (x)=2x+1

gf^0 (x)=¡ 4 x hf^0 (x)=2x+3

i f^0 (x)=2x^3 ¡ 12 x j f^0 (x)=^6

x^2

kf^0 (x)=¡

2

x^2

+

6

x^3

l f^0 (x)=2x¡

5

x^2

mf^0 (x)=2x+

3

x^2

nf^0 (x)=¡

1

2 x

p

x

of^0 (x)=8x¡ 4 pf^0 (x)=3x^2 +12x+12

2a

dy

dx

=7: 5 x^2 ¡ 2 : 8 x b

dy

dx

=2¼x

c

dy

dx

=¡

2

5 x^3

d

dy

dx

= 100

e dy

dx

=10 f dy

dx

=12¼x^2

3a 6 b^3

p

x

2

c 2 x¡ 10 d 2 ¡ 9 x^2 e 2 x¡ 1

f¡

2

x^3

+

3

p

x

g4+

1

4 x^2

h 6 x^2 ¡ 6 x¡ 5

4a 4 b ¡ 72916 c ¡ 7 d^134 e^18 f ¡ 11

5 b=3, c=¡ 4

6af^0 (x)=p^2

x

+1 bf^0 (x)=^1

33

p

x^2

cf^0 (x)=^1

x

p

x

df^0 (x)=2¡^1

2

p

x

ef^0 (x)=¡^2

xpx

ff^0 (x)=6x¡^32 px

gf^0 (x)=

¡ 25

2 x^3

p

x

hf^0 (x)=2+

9

2 x^2

p

x

7a

dy

dx

=4+

3

x^2

,

dy

dx

is the gradient function ofy=4x¡

3

x

from which the gradient at any point can be found.

b

dS

dt

=4t+4ms¡^1 ,

dS

dt

is the instantaneous rate of

change in position at the timet, or the velocity function.

c dC

dx

=3+0: 004 x$ per toaster, dC

dx

is the instantaneous

rate of change in cost as the number of toasters changes.

EXERCISE 13F.1

1agf(x)=(2x+7)^2 bgf(x)=2x^2 +7

cgf(x)=

p

3 ¡ 4 x dgf(x)=3¡ 4

p

x

egf(x)=

2

x^2 +3

fgf(x)=

4

x^2

+3

2 Note: There may be other answers.

ag(x)=x^3 , f(x)=3x+10

bg(x)=

1

x

, f(x)=2x+4

cg(x)=

p

x, f(x)=x^2 ¡ 3 x

dg(x)=^10

x^3

, f(x)=3x¡x^2

EXERCISE 13F.2

1au¡^2 , u=2x¡ 1 bu

(^12)
, u=x^2 ¡ 3 x
c 2 u
¡^12
, u=2¡x^2 du
(^13)
, u=x^3 ¡x^2
e 4 u¡^3 , u=3¡x f 10 u¡^1 , u=x^2 ¡ 3
2ady
dx
= 8(4x¡5) b dy
dx
= 2(5¡ 2 x)¡^2
c
dy
dx
=^12 (3x¡x^2 )
¡^12
£(3¡ 2 x)
d
dy
dx
=¡12(1¡ 3 x)^3 e
dy
dx
=¡18(5¡x)^2
f
dy
dx
=^13 (2x^3 ¡x^2 )
¡^23
£(6x^2 ¡ 2 x)
g dy
dx
=¡60(5x¡4)¡^3
h dy
dx
=¡4(3x¡x^2 )¡^2 £(3¡ 2 x)
i dy
dx
=6
³
x^2 ¡^2
x
́ 2
£
³
2 x+^2
x^2
́
3a¡p^13 b ¡ 18 c ¡ 8 d¡ 4 e ¡ 323 f 0
4 a=3, b=1 5 a=2,b=1
6ady
dx
=3x^2 , dx
dy
=^13 y
¡^23
Hint: Substitutey=x^3
b
dy
dx
£
dx
dy

dy
dy
fchain ruleg =1
cyan magenta yellow black
(^05255075950525507595)
100 100
(^05255075950525507595)
100 100 IB HL OPT
Sets Relations Groups
Y:\HAESE\CAM4037\CamAdd_AN\491CamAdd_AN.cdr Tuesday, 8 April 2014 8:39:19 AM BRIAN