Estimatio of Roots Q's (1 Viewer)

[puffynutty]

1.i) Show that f(x)=e^x-3x^2 has a root between x=3.7 and x= 3.8

ii) Starting with x=3.8, use one application of Newtons method to find a better approximation for this root.

2.The function f(x)=sinx-2x/3 has a zero near x=1.5. Taking x=1.5 as a first approximation, use one application of Newtons method to find a second approximation to zero.

 

[x3.eddayyeeee]

just use newton's method like it tells you?
whats wrong with that?

 

[puffynutty]

Dude if u read my last ones i've done them, i don't have the answers though so i need someone to state them for me, and besides estimation of roots is the worst topic in the history of topics, so if you dont have any answers don't post SIMPLE

 

[x3.eddayyeeee]

relax. im just saying.
sorry i cant help you.
but i dont know how to find derivatives for those equation. U__U"

if you can just use the x value that you given in this formula:

a - f(a)/f'(a) = closer approx.

in other words. given value minus (subed x value into original equation DIVIDED by subed x value into equations first derivative.

 

[dance2urownbeat]

1.i) Show that f(x)=e^x-3x^2 has a root between x=3.7 and x= 3.8

f (3.7) = e^3.7-3(3.7)^2
= -0.622...

f (3.8) = e^3.8-3(3.8)^2
= 1.381...

therefore, since f (3.7) and f (3.8) are opposite in sign, there is a root between x = 3.7 and x = 3.8

EDIT: estimation of roots was my favourite topic haha. but sorry, i can't be bothered doing the rest...

 

[tommykins]

1.i) Show that f(x)=e^x-3x^2 has a root between x=3.7 and x= 3.8

Find the value for f[3.7] and f[3.8], they will be opposite in sign hence showing it has a root between the two intervals.

ii) Starting with x=3.8, use one application of Newtons method to find a better approximation for this root.

f(x) = e^x - 3x²
f'(x) = e^x - 6x

New root = 3.8 - f(3.8)/f'(3.8)

2.The function f(x)=sinx-2x/3 has a zero near x=1.5. Taking x=1.5 as a first approximation, use one application of Newtons method to find a second approximation to zero.

f[x] = sinx - 2x/3
f'[x] = cosx - 2/3

new root = 1.5 - f[1.5]/f'[1.5]

 

[puffynutty]

thanks all

soz x3.eddayyeeee

 

[x3.eddayyeeee]

thanks all

soz x3.eddayyeeee

its okay.