In Exercises 3 – 8, the roots of f(x) are known or are easily found. Use 5 iterations of Newton’s Method with the given initial approximation to approximate the root. Compare it to the known value of the root.

The root of this are 1.41421, -1.41421

\(f(x) = x^{2} - 2, x_{0} = 1.5\)

\(f^{1}(x) = 2x\)

\(x_{1} = 1 - \frac{f(1)}{f^{1}(1)} = 1 - \frac{1-2}{2}\) = 1.5

\(x_{2} = 1.5 - \frac{f(1.5)}{f^{1}(1.5)}\) = 1.4166

\(x_{3} = 1.4166 - \frac{f(1.4166)}{f^{1}(1.4166)}\) = 1.414241

\(x_{4} = 1.414241 - \frac{f(1.414241)}{f^{1}(1.414241)}\) = 1.41423

\(x_{5} = 1.41423 - \frac{f(1.41423)}{f^{1}(1.41423)}\) = 1.41421

After 5th iteration, we can see that it equals to one of the roots