Use Newton’s Method to find a fourth approximation to the positive real zero of the function $f( x) =x^{3}+x^{2}-x-2$.

Solution Since $f(1) =-1$ and $f(2) =8,$ we know from the Intermediate Value Theorem that $f$ has a zero in the interval $(1, 2)$. Also, since $f$ is a polynomial function, both $f$ and $f^\prime$ are differentiable functions. Now $$f(x) =x^{3}+x^{2}-x-2 \qquad \hbox{and}\qquad f^\prime ( x) =3x^{2}+2x-1$$

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We choose $c_{1}=1.5$ as the first approximation and use Newton’s Method. The second approximation to the zero is $$c_{2}=c_{1}-\dfrac{f( c_{1}) }{f^\prime ( c_{1}) }=1.5-\dfrac{f(1.5) }{f^\prime (1.5) }=1.5-\dfrac{2.125}{ 8.75}\approx 1.2571429$$

Use Newton’s Method again, with $c_{2}=1.2571429$. Then $f(1.2571429) \approx 0.3100644$ and $f^\prime (1.2571429)\approx 6.2555106$. The third approximation to the zero is $$c_{3}=c_{2}-\dfrac{f( c_{2}) }{f^\prime ( c_{2}) }=1.2571429-\dfrac{0.3100641}{6.2555102}\approx 1.2075763$$

Figure 13 $f(x) = x^{3} + x^{2} - x - 2$

The fourth approximation to the zero is $$c_{4}=c_{3}-\dfrac{f( c_{3}) }{f^\prime ( c_{3}) } =1.2075763-\dfrac{0.0116012}{5.7898745}\approx 1.2055726$$