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HI!!

is this an on line class or do you really have to use newton's method?

I really have to use newtons method...

better still \[f(x)=x^2-4x+4-\ln(x)\]

then you make a guess
you got a good guess in mind?

woah lot of algebra method
you can use that too, but it is sometimes easier not to

lets make a guess
i guess 3

\[f(3)=3^2-4\times 3+4-\ln(3)=-0.098612\]

pretty close to zero already

you gotta start somewhere right?

figure it is easier to start with an integer

for reference if needed... http://tutorial.math.lamar.edu/Classes/CalcI/NewtonsMethod.aspx

now for the second approxmition it is
\[3-\frac{f(3)}{f'(3)}\]

\[f'(x)=2x-4-\frac{1}{x}\] so \[f'(3)=2\times 3-4-\frac{1}{3}=\frac{5}{3}\]

so next guess is
\[3+\frac{-0.098612}{\frac{5}{3}}\] at this point you really need a calculator

this gets very very messy quick but you can do it with wolfram or a spreadsheet

i get \[x_2=2.9408328\]

now you have to repeat it so it is really going to be messy

Yea, didnt get it...

what did you not get?

why are you using 3? whats so special about it

one is very near 3, so i picked 3 as \(x_1\)

ouch! I think it should be:
\(\large 3\color{red}{-}\frac{-0.098612}{\frac{5}{3}}\)

yeah you are right, it should be a plus

so \[x_2=3.0591672\]

click on the numeric answer, then rewrite with it substituted where i wrote x = 3