80 Higher Engineering Mathematics
Since f (1.5078125) is negative and f (1.5) is positive,
a root lies between x = 1.5078125 and x = 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 between x = 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 between x = 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 of x + 3 =e
x is x = 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 2 ln x + x = 2 using the method of
bisection.
Let f (x) = 2 ln x + x − 2
f (0.1) = 2 ln(0.1) + 0.1 − 2 = −6.5051 ...
(Note that ln 0 is infinite that is why
x = 0 was not chosen)
f (1) = 2 ln1 + 1 − 2 =−1
f (2) = 2 ln2 + 2 − 2 =+1.3862 ...
A change of sign indicates a root lies between x = 1 and
x = 2.
Since 2 ln x + x = 2 then 2lnx =−x + 2; sketches of
2 ln x and −x + 2 are shown in Fig. 9.4.
Since f (1.5078125) is negative and f (1.5) is positive,
a root lies between x = 1.5078125 and x = 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 between x = 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 between x = 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 of x + 3 =e
x is x = 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 2 ln x + x = 2 using the method of
bisection.
Let f (x) = 2 ln x + x − 2
f (0.1) = 2 ln(0.1) + 0.1 − 2 = −6.5051 ...
(Note that ln 0 is infinite that is why
x = 0 was not chosen)
f (1) = 2 ln1 + 1 − 2 =−1
f (2) = 2 ln2 + 2 − 2 =+1.3862 ...
A change of sign indicates a root lies between x = 1 and
x = 2.
Since 2 ln x + x = 2 then 2lnx =−x + 2; sketches of
2 ln x and −x + 2 are shown in Fig. 9.4.