00:01
Hello everyone in this problem we are given with the function f of x which is given as 15 sign of x into e power minus 0 .5 x minus 2.
00:16
Now we need to determine the upper positive root for the given function using secret method and also it is given that the relative error must be less than 0.
00:35
0 .1 percentage it is given that the value of x i minus 1 value as 1 .5 and x i value as 2 .0.
00:46
So now let us take this value as x0 to be equal to 1 .5 and x1 to be equal to 2 so now using second method so f of 1 .5 will be equal to 5 .068 that is substituting the value of x as 1 .5 in the given f of x we get this value similarly by substituting the value of x sorry value of x as 2 in the given f of x so we will get this value as 3 .018 so now we can calculate x2 so the x2 can be given by the formula x2 which is equal to x0 minus f of x0 into x1 minus x0 divided by f of x1 minus f of x not now substituting the values so we get x2 to be equal to x not x not is 1 .5 minus f of x not as 5 .068 into x1 as 2 minus 1 .5 divided by 3 .0 1 .0 18 minus 5 .0668.
02:15
So simplifying this we get the value of x2 as 2 .736.
02:28
Now let us find the value of x3 to be equal to it will be x1 minus f of x1 into x2 minus x1 divided by f of x2 minus x1 divided by f of x2 minus f of x1.
02:50
Now substituting the corresponding values, so we get it as 2 minus 3 .018 into x2 is 2 .736 minus 2 divided by minus of 0 .493 minus 3 .08.
03:11
So simplifying this, we get the value of x3 to be 2 .633.
03:18
So from this from the value of x3 and x2 we can find absolute value of x3 minus x2 is greater than the given relative error.
03:31
So now we need to continue in order to in order to obtain the required accuracy.
03:41
So f of x3 value is minus 0 .040.
03:49
Now we can find the value of x3.
03:51
4 it will be x2 it is 2 .736 minus f of x2 values minus 0 .493 into 2 .633 that is x3 minus x2 value is 2 .736 the whole divided by x3 values minus 0 .040 plus f of x2 value is 0 .040 plus f of x2 value is 0 .4 so simplifying this we get this value as x4 to be equal to 2 .623...