Use Newton's method with initial approximation x1 = ˆ’1 to find x2 the second approximation to the root of the equation x^3 + x + 7 = 0. x2 = ?

Sunday, April 21, 2013

Use the Newton
iteration formula:

$\displaystyle{x}_{{2}}={x}_{{1}}-\frac{{f{{\left({x}_{{1}}\right)}}}}{{f{'}}}{\left({x}_{{1}}\right)}$

In
our case, $\displaystyle{f{{\left({x}_{{1}}\right)}}}=-{1}-{1}+{7}={5}$

$\displaystyle{f{'}}{\left({x}\right)}={3}{{x}}^{{2}}+{1}$

$\displaystyle{f{'}}{\left({x}_{{1}}\right)}={3}{{\left(-{1}\right)}}^{{2}}+{1}={4}$

$\displaystyle{x}_{{2}}=-{1}-\frac{{{5}}}{{4}}=-\frac{{9}}{{4}}$

Answer: $\displaystyle{x}_{{2}}=-\frac{{9}}{{4}}$

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