Estimation of roots (1 Viewer)

[unLimitieDx]

2010 3b vi) (2)

For the second part how would you know whether to let
f(x) = x - e^-x2

OR f(x) = e^-x2 - x

2010 3b vi) (2)

 

[RealiseNothing]

Why not just use:

 

[unLimitieDx]

When we halve the interval and sub the value into that equation it will give us +0.005 rather than negative and ultimately will give us an answer of 0.6 which is not the case

 

[Trebla]

It shouldn't matter. In either case, the true root is the same so the approximation yields the same answer.

Consider f(x) = x - e^x2
f(0.6) ~ - 0.098
f(0.7) ~ 0.087
When you check 0.65 you get
f(0.65) ~ - 0.005406
So the true root is between 0.65 and 0.7, hence the true root rounds up 0.7 to one decimal place

Consider f(x) = e^x2 - x
f(0.6) ~ 0.098
f(0.7) ~ - 0.087
When you check 0.65 you get
f(0.65) ~ 0.005406
So the true root is between 0.65 and 0.7, hence the true root rounds up 0.7 to one decimal place

 

[Dedication_]

Very misleading title, I wanted tallies.

 

[RealiseNothing]

Very misleading title, I wanted tallies.

hahahahaha.

 

[unLimitieDx]

It shouldn't matter. In either case, the true root is the same so the approximation yields the same answer.

Consider f(x) = x - e^x2
f(0.6) ~ - 0.098
f(0.7) ~ 0.087
When you check 0.65 you get
f(0.65) ~ - 0.005406
So the true root is between 0.65 and 0.7, hence the true root rounds up 0.7 to one decimal place

Consider f(x) = e^x2 - x
f(0.6) ~ 0.098
f(0.7) ~ - 0.087
When you check 0.65 you get
f(0.65) ~ 0.005406
So the true root is between 0.65 and 0.7, hence the true root rounds up 0.7 to one decimal place

Ahh thats right! Thanks Trebla!