Input / Question:
Input / Question:
Find the interval in which the smallest positive root of the following equations lies :
(a) tan x + tanh x = 0
(b) x^{3 }– x – 4 = 0.
Determine the roots correct to two decimal places using the bisection method.
Verified
Output/Answer
(a) Let f (x) = tan x + tanh x.
Note that f (x) has no root in the first branch of y = tan x, that is, in the interval (0, π / 2).
The root is in the next branch of y = tan x, that is, in the interval (π / 2, 3π / 2).
We have :
f (1.6) = – 33.31, f (2.0) = – 1.22,
f (2.2) = – 0.40, f (2.3) = – 0.1391, f (2.4) = 0.0676.
Therefore, the root lies in the interval (2.3, 2.4).
| f(m_{k})f(a_{k-1}) | m_{k} | b_{k-1} | a_{k-1} | k |
| >0 | 2.35 | 2.4 | 2.3 | 1 |
| <0 | 2.375 | 2.4 | 2.35 | 2 |
| >0 | 2.3625 | 2.375 | 2.35 | 3 |
| <0 | 2.36875 | 2.375 | 2.3625 | 4 |
After four iterations, we find that the root lies in the interval (2.3625, 2.36875). Hence,
the approximate root is m = 2.365625. The root correct to two decimal places is 2.37.
b) For f (x) = x^{ 3} – x – 4, we find f (0) = – 4, f (1) = – 4, f (2) = 2.
Therefore, the root lies in the interval (1, 2).
| f(m_{k})f(a_{k-1}) | m_{k} | b_{k-1} | a_{k-1} | k |
| >0 | 1.5 | 2 | 1 | 1 |
| >0 | 1.75 | 2 | 1.5 | 2 |
| <0 | 1.875 | 2 | 1.75 | 3 |
| >0 | 1.8125 | 1.875 | 1.75 | 4 |
| >0 | 1.78125 | 1.8125 | 1.75 | 5 |
| <0 | 1.796875 | 1.8125 | 1.78125 | 6 |
| >0 | 1.789063 | 1.796875 | 1.78125 | 7 |
| >0 | 1.792969 | 1.796875 | 1.789063 | 8 |
| >0 | 1.794922 | 1.796875 | 1.792969 | 9 |
| >0 | 1.795898 | 1.796875 | 1.794922 | 10 |
After 10 iterations, we find that the root lies in the interval (1.795898, 1.796875).
Therefore, the approximate root is m = 1.796387. The root correct to two decimal places is 1.80.