Higher Engineering Mathematics

78 NUMBER AND ALGEBRA

Sincef(1) is positive andf(2) is negative, a root
lies betweenx=1 andx=2. A sketch off(x) =
x+ 3 −ex, i.e.x+ 3 =exis shown in Fig. 9.3.

f(x)

4

− 2

f(x) = ex

f(x) = x+ 3

3

2

0

1

−1 1 2 x

Figure 9.3

Bisecting the interval betweenx=1 andx=2gives
1 + 2
2

i.e. 1.5.

Hence

f(1.5)= 1. 5 + 3 −e^1.^5

=+0.01831...

Sincef(1.5) is positive andf(2) is negative, a root
lies betweenx= 1 .5 andx=2.

Bisecting this interval gives

1. 5 + 2

2

i.e. 1.75.

Hence

f(1.75)= 1. 75 + 3 −e^1.^75

=−1.00460...

Sincef(1.75) is negative andf(1.5) is positive, a root
lies betweenx= 1 .75 andx= 1 .5.

Bisecting this interval gives

1. 75 + 1. 5

2

i.e. 1.625.

Hence

f(1.625)= 1. 625 + 3 −e^1.^625

=−0.45341...

Sincef(1.625) is negative andf(1.5) is positive, a
root lies betweenx= 1 .625 andx= 1 .5.

Bisecting this interval gives

1. 625 + 1. 5

2

i.e. 1.5625.

Hence

f(1.5625)= 1. 5625 + 3 −e^1.^5625

=−0.20823...

Sincef(1.5625) is negative andf(1.5) is positive, a

root lies betweenx= 1 .5625 andx= 1 .5.

Bisecting this interval gives

1. 5625 + 1. 5

2

i.e. 1. 53125

Hence

f(1.53125)= 1. 53125 + 3 −e^1.^53125

=−0.09270...

Sincef(1.53125) is negative andf(1.5) is positive,

a root lies betweenx= 1 .53125 andx= 1 .5.

Bisecting this interval gives

1. 53125 + 1. 5

2

i.e. 1. 515625

Hence

f(1.515625)= 1. 515625 + 3 −e^1.^515625

=−0.03664...

Sincef(1.515625) is negative andf(1.5) is positive,

a root lies betweenx= 1 .515625 andx= 1 .5.

Bisecting this interval gives

1. 515625 + 1. 5

2

i.e. 1. 5078125

Hence

f(1.5078125)= 1. 5078125 + 3 −e^1.^5078125

=−0.009026...

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...

Sincef(1.50390625) is positive andf(1.5078125)

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

x= 1 .5078125.