00:01
In this problem we are provided with the function f of x which equals to x cubed plus 2 times x squared plus 1.
00:11
In subpart a we are asked to solve this equation that is we need to find the roots of this equation using the bisection method.
00:23
So here let us first find out the value of the function at negative 2 we get the answer to be 1 and when we find out.
00:31
Find the value of the function at negative 3, we get the answer to be negative 8.
00:37
Since we have both the values to be of opposite sign, it implies that there exists a root that lies in the interval negative 3 to negative 2.
00:51
So therefore we have x not to be equal to negative 3 plus negative 2 divided by 2 which equals to negative 2 .5.
01:01
F of negative 2 .5 equals to negative 2 .125 which is less than 0.
01:11
So in the second iteration, we have x1 to be equal to negative 2 .5 plus negative 2 divided by 2 and this equals to negative 2 .25.
01:30
So evaluating the function at this point we have f of negative 2 .25 which equals to negative 0 .2656 which is less than 0.
01:41
So next in the third iteration we have x2 to be equal to negative 2 .25 plus negative 2 divided by 2 to be equal to negative 2 .125 and f of x2 equals to 0 .4355 which is greater than 0.
02:06
So proceeding this way, we obtain the following table of values and from the table it can be seen that at the 11th iteration the value of x is obtained to be negative 2 .2056 and this is the required root of the function.
02:25
So this is the required answer for subpart a.
02:31
Next, in subpart b, we're asked to use the method of false position.
02:38
So here, again, we would have the value of x -0 to be negative 3 and the value of x -1 would be negative 2.
02:46
So let us perform the first iteration.
02:49
We have x -2 equals to x -not minus f -of -x -not times x -1 minus x -not divided by f -of -x -1, minus f of x not.
03:03
So substituting the values we have negative 3 minus negative 8 times negative 2 minus negative 3 divided by 1 minus of negative 8 and evaluating this we obtain the answer for x2 to be negative 2 .1111.
03:25
So now evaluating the function at this point we get 0 .504 .0 .0 .0 .0 .0.
03:27
So now evaluating the function at this point we get 0 .504.
03:31
Which is greater than 0.
03:33
So now we consider x not to be equal to negative 3 and we consider x1 to be equal to negative 2 .1111.
03:42
So now let us evaluate x3.
03:47
X3 equals to negative 3 minus of negative 8 times negative 2 .111 -1 minus of negative 3 divided by 0 .55 .5 .5 .5 .5 .5 .5 .5 .5 .5 .5.
04:02
048 minus of negative 8.
04:06
So evaluating this, we obtain the answer for x3 to be negative 2 .1639.
04:13
Next let us evaluate the function at this point.
04:19
So we get f of x3 to be equal to 0 .2327 which is greater than 0.
04:27
So for the next iteration we have x not equals to negative 3 and x1 equals to negative 2 .1639.
04:36
So using this we find x4 which is negative 3 minus of negative 8 times negative 2 .1639 minus of negative 3 divided by 0 .2327 minus of negative 8.
04:55
So evaluating this we get negative 2 .1875 and the value of the function at this point equals to 0 .1028 which is greater than 0...