Solving equations by iterative methods 79
1.05, correct to 3 significant figure. We therefore stop
the iterations here.
Thus, correct to 3 significant figures, the positive root
of 5x 2 + 11x − 17 = 0 is 1.05
Problem 2. Use the bisection method to determine the positive root of the equation x + 3 = e x ,
correct to 3 decimal places.
Let f (x) = x + 3 −e x
then, using functional notation:
f (0)= 0 + 3 − e 0 = +2
f (1)= 1 + 3 − e 1 = +1.2817 . . .
f (2)= 2 + 3 − e 2 = −2.3890 . . .
Since f (1) is positive and f (2) is negative, a root lies
between x = 1 and x = 2. A sketch of f (x) = x + 3 − e x ,
i.e. x + 3 =e x is shown in Fig. 9.3.
f(x)
1
2
3
4
0
21
22
1
2 x
f(x) 5 x 1 3
f(x) 5 e x
Figure 9.3
Bisecting the interval between x = 1 and x = 2 gives
1 +2
2
i.e. 1.5.
Hence
f (1.5) = 1.5 + 3 − e
1.5
= +0.01831. . .
Since f (1.5) is positive and f (2) is negative, a root lies
between x = 1.5 and x = 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. . .
Since f (1.75) is negative and f (1.5) is positive, a root
lies between x = 1.75 and x = 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. . .
Since f (1.625) is negative and f (1.5) is positive, a root
lies between x = 1.625 and x = 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 . . .
Since f (1.5625) is negative and f (1.5) is positive, a
root lies between x = 1.5625 and x = 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 . . .
Since f (1.53125) is negative and f (1.5) is positive, a
root lies between x = 1.53125 and x = 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 . . .
Since f (1.515625) is negative and f (1.5) is positive, a
root lies between x = 1.515625 and x = 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 . . .
1.05, correct to 3 significant figure. We therefore stop
the iterations here.
Thus, correct to 3 significant figures, the positive root
of 5x 2 + 11x − 17 = 0 is 1.05
Problem 2. Use the bisection method to determine the positive root of the equation x + 3 = e x ,
correct to 3 decimal places.
Let f (x) = x + 3 −e x
then, using functional notation:
f (0)= 0 + 3 − e 0 = +2
f (1)= 1 + 3 − e 1 = +1.2817 . . .
f (2)= 2 + 3 − e 2 = −2.3890 . . .
Since f (1) is positive and f (2) is negative, a root lies
between x = 1 and x = 2. A sketch of f (x) = x + 3 − e x ,
i.e. x + 3 =e x is shown in Fig. 9.3.
f(x)
1
2
3
4
0
21
22
1
2 x
f(x) 5 x 1 3
f(x) 5 e x
Figure 9.3
Bisecting the interval between x = 1 and x = 2 gives
1 +2
2
i.e. 1.5.
Hence
f (1.5) = 1.5 + 3 − e
1.5
= +0.01831. . .
Since f (1.5) is positive and f (2) is negative, a root lies
between x = 1.5 and x = 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. . .
Since f (1.75) is negative and f (1.5) is positive, a root
lies between x = 1.75 and x = 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. . .
Since f (1.625) is negative and f (1.5) is positive, a root
lies between x = 1.625 and x = 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 . . .
Since f (1.5625) is negative and f (1.5) is positive, a
root lies between x = 1.5625 and x = 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 . . .
Since f (1.53125) is negative and f (1.5) is positive, a
root lies between x = 1.53125 and x = 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 . . .
Since f (1.515625) is negative and f (1.5) is positive, a
root lies between x = 1.515625 and x = 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 . . .