00:01
In this problem, we have been asked to use the bisection algorithm method to find the root of the given function to an interval of length less than or equal to 0 .1.
00:10
So the function we have been given is f of x is equal to x squared minus 2, and our interval is the interval from 0 to 3.
00:20
Now, first of all, let us determine how many iterations we need.
00:24
So what will the length of the interval be on the nth iteration? so we will have b minus a divided by 2 to the power of n.
00:33
Here, the interval is from 0 to b, and n will be the number of iterations.
00:39
And we want this to be less than or equal to 0 .1.
00:43
So we end up with 3 minus 0 divided by 2 to the power of n, that is less than or equal to 0 .1.
00:49
So that just means that we have 3 divided by 2 to the power of n is less than or equal to 0 .1.
00:56
And so that will mean that 2 to the power of n divided by 3 is greater than or equal to 1 divided by 0 .1, that will be 10.
01:05
So 2 to the power of n will be greater than or equal to 30, and n will be greater than or equal to the logarithm of 30, where the base is 2.
01:16
This is approximately equal to 4 .9.
01:19
And so that means that n will be 5 in this case.
01:22
We need a natural number.
01:23
So n will be 5.
01:24
We need 5 iterations.
01:26
So let us consider these iterations.
01:29
So let's consider the first iteration.
01:32
So we have x1, that will be 0 plus 3 divided by 2.
01:37
The interval is from 0 to 3, so 0 plus 3 divided by 2.
01:41
This will be 1 .5.
01:44
Now, let us consider the value of f of 0.
01:48
0 squared minus 2, that will be negative 2.
01:51
This is less than 0.
01:52
Let us consider f of 3.
01:54
That is 3 squared minus 2.
01:55
That's 9 minus 2, which is 7.
01:57
That is greater than 0.
01:58
Let us consider f of 1 .5.
02:00
That is 1 .5 squared minus 2, and 1 .5 squared minus 2 is 0 .25.
02:08
So that's greater than 0.
02:10
So we consider f of 0, which is less than 0, and f of 1 .5, which is greater than 0...