Higher Engineering Mathematics, Sixth Edition

80 Higher Engineering Mathematics

Sincef( 1. 5078125 )is negative andf( 1. 5 )is positive,

a root lies betweenx= 1 .5078125 andx= 1 .5.

Bisecting this interval gives

1. 5078125 + 1. 5

2

i.e. 1. 50390625.

Hence

f(1.50390625)= 1. 50390625 + 3 −e^1.^50390625

=+0.004676...

Since f( 1. 50390625 )is positive and f( 1. 5078125 )

is negative, a root lies betweenx= 1 .50390625 and

x= 1 .5078125.

Bisecting this interval gives

1. 50390625 + 1. 5078125

2

i.e. 1. 505859375.

Hence

f(1.505859375)= 1. 505859375 + 3 −e^1.^505859375

=−0.0021666...

Since f( 1. 50589375 )is negative and f( 1. 50390625 )

is positive, a root lies betweenx= 1 .50589375 and

x= 1 .50390625.

Bisecting this interval gives

1. 505859375 + 1. 50390625

2

i.e. 1. 504882813.

Hence

f(1.504882813)= 1. 504882813 + 3 −e^1.^504882813

=+0.001256...

Since f(1.504882813) is positive and

f(1.505859375) is negative,

a root lies between x= 1 .504882813 and x=

1 .505859375.

Bisecting this interval gives

1. 504882813 + 1. 50589375

2

i.e.1.505388282.

The last two values of x are 1.504882813 and

1.505388282, i.e. both are equal to 1.505, correct to

3 decimal places.

Hence the root ofx+ 3 =exisx=1.505, correct to 3

decimal places.

The above is a lengthy procedure and it is probably

easier to present the data in a table as shown in the

table.

x 1 x 2 x 3 =

x 1 +x 2

2

f(x 3 )

0 + 2

1 +1.2817...

2 −2.3890...

1 2 1.5 +0.0183...

1.5 2 1.75 −1.0046...

1.5 1.75 1.625 −0.4534...

1.5 1.625 1.5625 −0.2082...

1.5 1.5625 1.53125 −0.0927...

1.5 1.53125 1.515625 −0.0366...

1.5 1.515625 1.5078125 −0.0090...

1.5 1.5078125 1.50390625 +0.0046...

1.50390625 1.5078125 1.505859375 −0.0021...

1.50390625 1.505859375 1.504882813 +0.0012...

1.504882813 1.505859375 1.505388282

Problem 3. Solve, correct to 2 decimal places,

the equation 2lnx+x=2 using the method of

bisection.

Let f(x)=2lnx+x− 2

f( 0. 1 )=2ln( 0. 1 )+ 0. 1 − 2 =− 6. 5051 ...

(Note that ln0 is infinite that is why

x=0 was not chosen)

f( 1 )=2ln1+ 1 − 2 =− 1

f( 2 )=2ln2+ 2 − 2 =+ 1. 3862 ...

A change of sign indicates a root lies betweenx=1and

x=2.

Since 2lnx+x=2then2lnx=−x+2; sketches of

2lnxand−x+2areshowninFig.9.4.