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# P2 Functions/Iteration help watch

1. Edexcel P2 Jan 02
q.7

A curve is shown with equation f(x)=0.5ex - x2
Curve cuts y-axis at A and the is a minimum at point B

eqn of tangent to C at A is
y=0.5x+0.5

part (b)
the x-co-ord of B is approx 2.15. A more exact estimate is found using
xn+1=lng(xn)

show that a possible form for g(x) is g(x)=4x

how would I show this?
2. f(x) = 0.5e^x - x^2
f'(x) = 0.5e^x - 2x
at B, f'(x) = 0

0.5e^x - 2x = 0
e^x = 4x
ln (e^x) = ln (4x)
x = ln (4x)

COnvert to an interative formula, x_(n+1) = ln (4x_n)
so g(x) = 4x
3. (Original post by mik1w)
f(x) = 0.5e^x - x^2
f'(x) = 0.5e^x - 2x
at B, f'(x) = 0

0.5e^x - 2x = 0
e^x = 4x
ln (e^x) = ln (4x)
x = ln (4x)

COnvert to an interative formula, x_(n+1) = ln (4x_n)
so g(x) = 4x

I follow up to x=ln(4x) wouldnt lne^x be 1^x therefore 1 not x?

and WHAT is a interative formula?
4. (Original post by marisad_uk)
I follow up to x=ln(4x) wouldnt lne^x be 1^x therefore 1 not x?

and WHAT is a interative formula?
No. lne = 1 as its log in base e of e so its 1. Therefore it just leaves x as x multiplied by 1 is just x.
5. (Original post by !Laxy!)
No. lne = 1 as its log in base e of e so its 1. Therefore it just leaves x as x multiplied by 1 is just x.
sorry for sounding stupid but i am still confused

lne=logee=1

lnex =1x
1x =1
6. sorry but i gotta *bump*

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