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use Newton's Method to determine $x_{2}$ for the given function and given value of $x_{0}$.

$f(x)=x^{3}-7 x^{2}+8 x-3, x_{0}=5$
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Here is the derivative of the function since we'll need that.
$f^{\prime}(x)=3 x^{2}-14 x+8$
We just now need to run through the formula above twice.

The first iteration through the formula for $x_{1}$ is,
$x_{1}=x_{0}-\frac{f\left(x_{0}\right)}{f^{\prime}\left(x_{0}\right)}=5-\frac{f(5)}{f^{\prime}(5)}=5-\frac{-13}{13}=6$

The second iteration through the formula for $x_{2}$ is,
$x_{2}=x_{1}-\frac{f\left(x_{1}\right)}{f^{\prime}\left(x_{1}\right)}=6-\frac{f(6)}{f^{\prime}(6)}=6-\frac{9}{32}=5.71875$
So, the answer for this problem is $x_{2}=5.71875$
Although it was not asked for in the problem statement the actual root is $5.68577952608963$. Note as well that this did require some computational aid to get and it not something that you can, in general, get by hand.
by Platinum (102k points)
Find the value of $\dfrac{x^{3}+5 x^{2}+4 x+3}{x^{3}}$

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