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Use newton's method to approximate the solution to x^5-2=3x+1; find the third iteration using x1=1 as the initial guess.
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\[x_2=x_1-\frac{f\left(x_1\right)}{f'\left(x_1\right)}\]
Rewrite function as
\[x^5-3x-3=0\]
\[1-\frac{\left(x^5-3x-3\right)}{5x^4-3}\]
Continueplz
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Lets' plug in 1 \[1-\frac{\left(1^5-3(1)-3\right)}{5(1)^4-3}\] =7/2
Now we have to do whole thing again with 7/2 instead of 1
So is 2.8 my final ans?
I got 2.28
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