Numerical Analysis By Shanker Rao

24 NUMERICAL ANALYSIS
In the 4th step a n
, b n
and x n + 1
are equal up to two decimal places. Thus, x = 1.51 is the root of f(x) = 0,
correct up to two decimal places.
Example 2.4 Find the root of tan x + x = 0 up to two decimal places, which lies between 2 and 2.1.
Solution Let f(x) = tan x + x
Here f(2) = –0.18, f(2.1) = 0.39
Thus, the root lies between 2.0 and 2.1
∴ a 0
= 2, b 0
= 2.1
n a n
b n
x
n+ 1 =
F
HG
a
n
+ b
2
n
I K J
f(x n + 1
)
0 2.0 2.1 2.05 0.12
1 2.0 2.05 2.025 –0.023
2 2.025 2.05 2.0375 0.053
3 2.025 2.0375 2.03125 –0.0152
4 2.025 2.03125 2.02812 –0.0039
5 2.02812 2.03125 2.02968 0.0056
check 6 2.02813 2.02968 2.02890 0.00087
In the 5th step, a n
, b n
and x n + 1
are equal up to two decimal places.
∴ x = 2.03 is a root of f(x) = 0, correct up to two decimal places.
Exercise 2.2
1. Find a root of the equation x 3 – 4x – 9 = 0 correct to three decimal places by using bisection method.
2. Find the positive roots of the equation x 3 – 3x + 1.06 = 0, by method of bisection, correct to three
decimal places.
3. Compute one positive root of 2x – 3 sin x – 5 = 0, by bisection method, correct to three significant
figures.
4. Compute one root of x + log x – 2 = 0 correct to two decimal places which lies between 1 and 2.
5. Compute one root of sin x = 10(x – 1) correct to three significant figures.
6. Compute the root of log x = cos x correct to two decimal places.
7. Find the interval in which the smallest positive root of the following equation lies. Also find the root
correct to two decimal places. Use bisection method
(a) tan x + tan hx = 0
(b) x 3 – x – 4 = 0
8. Find the root of the equation x 3 – x – 11 = 0, using bisection method correct to three decimal places
(which lies between 2 and 3).
9. Find the root of the equation x 4 – x – 10 = 0, using bisection method.