3 − −= 3 50 x x x) 3 f2 3 ( ) f 3 13 ( ) - pmt

© Pearson Education Ltd 2019. Copying permitted for purchasing institution only. This material is not copyright free. 1

Exercise 3C

1 a 3 3 5 0x x− − = Let ( ) 3f 3 5x x x= − −

( )f 2 3= −

( )f 3 13= There is a sign change between f (2) and f (3) so a root of the equation lies in the interval [2, 3]. b

By similar triangles:

1

1

3 132 3

xx−

=−

1 1

1

9 3 13 2616 35

x xx

− = −=

135162.1875

x =

=

( )f 2.1875 1.094...= − There is a sign change, therefore a root lies between x = 2.1875 and x = 3.



2

2

3 132.1875... 1.094...

xx

−=

−

( ) ( )2 2

2

2

1.094... 3 13 2.187514.084... 31.722...

2.250...

x xxx

− = −

==

( )f 2.250... 0.351...= − There is a sign change, therefore a root lies between x = 2.250… and x = 3. By similar triangles:

3

3

3 132.250... 0.351...

xx

−=

−

( ) ( )3 3

3

3

0.351... 3 13 2.25013.351... 30.313...

2.270...

x xxx

− = −

==

( )f 2.270... 0.108...= −

Two successive approximations give x = 2.3, accurate to 1 d.p. 2 a 3 25 8 1 0x x− + = Let ( ) 3 2f 5 8 1x x x= − +

( )f 1 2= −

( )f 2 9= There is a sign change, therefore a root lies between x = 1 and x = 2.


1

1

2 91 2x

x−

=−

1 14 2 9 9x x− = −

111 13x =


1

1311

1.18. .

x =

=

f 1.18 1.920.... . = −

There is a sign change, therefore a root lies between 1.18. .

x = and x = 2.

By similar triangles: 2

2

2 91.920...1.18

. .x

x

−=

−

( )2 2

2

2

1.920... 2 9 1.18

10.920... 14.476...1.325...

. .x x

xx

− = −

==

( )f 1.325... 1.410...= − There is a sign change, therefore a root lies between 1.325...x = and x = 2.



3

3

2 91.325... 1.410...

xx

−=

−

( ) ( )3 3

3

3

1.410... 2 9 1.325...10.410... 14.745...

1.416...

x xxx

− = −

==



4

4

2 91.416... 0.841...

xx

−=

−

( ) ( )4 4

4

4

0.841... 2 9 1.416...9.841... 14.426...

1.465...

x xxx

− = −

==



2 b By similar triangles:

5

5

2 91.465... 0.440...

xx

−=

−

( ) ( )5 5

5

5

0.440... 2 9 1.465...9.440... 14.066...

1.490...

x xxx

− = −

=

=

( )f 1.490... 0.220...= − Two successive approximations give x = 1.5, accurate to 1 d.p. 3 a 3 33 3 0x x

x x+ = ⇒ − − =

Let ( ) 3f 3x xx

= − −

( )f 3 1= −

( )f 4 0.25= There is a sign change, therefore a root lies between x = 3 and x = 4. b


1

1

4 0.253 1x

x−

=−

( )1 1

1

1

4 0.25 31.25 4.75

3.8

x xxx

− = −

==

( )f 3.8 0.010...= There is a sign change, therefore a root lies between x = 3 and x = 3.8.



1

1

3.8 0.010...3 1x

x−

=−

( )1 1

1

1

3.8 0.010... 31.01... 3.831...

3.793

x xxx

− = −

==

Two successive approximations give x = 3.8, accurate to 1 d.p. 4 a 2 cos 1 0x x − = Let ( )f 2 cos 1x x x= −

( )f 1 0.080...=

( )f 1.5 0.787...= − There is a sign change, therefore a root lies between x = 1 and x = 1.5. b



1

1

1.5 0.787...1 0.080...x

x−

=−

( ) ( )1 10.080... 1.5 0.787... 1x x− = −

1

1

0.867... 0.907...1.046...

xx==

( )f 1.046 0.048...= There is a sign change, therefore a root lies between x = 1.046… and x = 1.5.


2

2

1.5 0.787...1.046... 0.048...

xx

−=

−

( ) ( )2 20.048... 1.5 0.787... 1.046...x x− = −

20.835... 0.895...x =

2 1.071...x =

( )f 1.071 0.025...= There is a sign change, therefore a root lies between x = 1.071… and x = 1.5.



3

3

1.5 0.787...1.071... 0.025...

xx

−=

−

( ) ( )3 30.025... 1.5 0.787... 1.071...x x− = −

30.812... 0.881...x =

3 1.085...x = Two successive approximations give x = 1.1, accurate to 1 d.p. 5 a 3 22 3 0x x− − = ( ) 3 2f 2 3x x x= − −

( )f 2 3= −

( )f 3 6= There is a sign change between f (2) and f (3), therefore there is a root of the equation in the

interval [2, 3]. 2f ( ) 3 4 (3 4)x x x x x′ = − = −

For 43

x > , ( )f x′ >0 and the function is increasing.

Therefore the root which lies in the interval [2, 3] must be the largest possible root of the equation. b


1

1

3 62 3

xx−

=−

( ) ( )1 13 3 6 2x x− = −

19 21x =

1

.2.3x =

.

f 2.3 1.185... = −

There is a sign change, therefore a root lies between x = .

2.3 and x = 3.


By similar triangles: 2

2

3 6. 1.185...2.3

x

x

−=

−

( )2 2

.1.185... 3 6 2.3x x − = −

27.185... 17.555...x =

2 2.443...x =

( )f 2.443... 0.353...= − There is a sign change, therefore a root lies between x = 2.443... and x = 3.


3

3

3 62.443... 0.353...

xx

−=

−

( ) ( )3 30.353... 3 6 2.443...x x− = −

36.353... 15.717x =

3 2.474...x =

( )f 2.474... 0.098...= − There is a sign change, therefore a root lies between x = 2.474… and x = 3.



4

4

3 62.474... 0.098...

xx

−=

−

( ) ( )4 40.098... 3 6 2.474...x x− = −

46.098... 15.140...x =

4 2.482x = Two successive approximations give x = 2.5, accurate to 1 d.p. 6 ( )f 2 3 1xx x= − −

( )f 3 2= −

( )f 4 3=


1

1

4 33 2x

x−

=−

( ) ( )1 12 4 3 3x x− = −

15 17x =


1175

3.4

x =

=

3 − −= 3 50 x x x) 3 f2 3 ( ) f 3 13 ( ) - pmt

Documents