00:01
In this question, they asked us to find out the root of the equation f of x given to x a cube minus 2x minus 5 using regular falsie method.
00:20
Okay? so they did not give degree of accuracy.
00:24
I would assume it to be 0 .01.
00:27
Fine.
00:29
So now i'm going to compute using the regular false.
00:33
Method so now we know that f of see we know that f of 1 okay f of 1 is equal to 2 minus 2 minus 5 which is nothing but minus 5 f of 2 is nothing but 2 power 4 minus 2 2 minus 5 which is nothing but 16 minus 4 minus 5 which is essentially 7 so basically for the first iteration, a is equal to 1, b is equal to 2.
01:18
The solution would be, see, f of 1 multiplied by f of 2 is less than 0.
01:29
So the solution lies between 1 and 2, correct? so a is equal to 1, b is equal to 2, f of a is equal to minus 5, which is less than 0, f of b is equal to 7, which is greater than equal to see what we have to remember is these two signs should be opposite so for the regular falsie method the second step is to okay the second step is to find out what is x x is equal to nothing but a minus a minus one minute b minus a times f of x minus f of a divided by f of v minus f of it okay this is what it is but then we know that we have to make this 0 correct so what do we get what do we get x x is equal to 1 minus p minus this is 1 and we would get minus minus 5 divided by 12 which essentially is 1.
03:12
1 .42, okay, nearly equal to.
03:20
Now, f of 1 .42 is equal to minus 2 .15, something around this, which is less than 0.
03:32
Okay, so we have to continue again.
03:36
So i'm scrolling down.
03:39
Now here, a would be 1 .42, b would be 2, and f of a would be minus 2 .15.
03:50
F of b would be 7.
03:53
So we have to compute x now.
03:55
What do we get? we get around 1 .25.
04:00
Okay.
04:00
So for this particular 1 .25 we would get something around minus 0 .61 which is less than 0.
04:16
Okay.
04:20
So since this is less than 0 and is not equal to 0, now the a becomes 1 .55.
04:29
B is 2...