Estimation of roots (1 Viewer)

unLimitieDx

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2010 3b vi) (2)
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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)
 

unLimitieDx

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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

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

Consider f(x) = x - ex2
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) = ex2 - 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
 
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unLimitieDx

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

Consider f(x) = x - ex2
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) = ex2 - 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!
 

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