Newton's method is being used to approximate the non-zero \(x\)-intercept of the function with the equation \(f(x)=\dfrac{x^3}{5}-\sqrt{x}\). An initial estimate of \(x_0=1\) is used.
Which one of the following gives the first estimate that would correctly approximate the intercept to three decimal places?
- \(x_6\)
- \(x_7\)
- \(x_8\)
- \(x_9\)
- The intercept cannot be correctly approximated using Newton's method.
| \(f(x)\) | \(=\dfrac{x^3}{5}-\sqrt{x}\) |
| \(f'(x)\) | \(=\dfrac{3x^2}{5}-\dfrac{1}{2\sqrt{x}}\) |
\(\text{For Newton’s Method: }\)
| \(x_o\) | \(=1\) |
| \(x_{n+1}\) | \(=x_n-\dfrac{f(x_n)}{f'(x_n)}\) |
\(\text{Check using CAS:}\)
\begin{array} {|c|c|}
\hline
\ \ \ n\ \ \ & x_n \\
\hline
\ 0 \ & 1\\
\hline
\ 1 \ & 1-\dfrac{\dfrac{1^3}{5}-\sqrt{1}}{\dfrac{3\times 1^2}{5}-\dfrac{1}{2\sqrt{1}}}=9\\
\hline
\ 2 \ & 9-\dfrac{\dfrac{9^3}{5}-\sqrt{9}}{\dfrac{3\times 9^2}{5}-\dfrac{1}{2\sqrt{9}}}\approx 6.05\\
\hline
\ 3 \ & 6.05-\dfrac{\dfrac{6.05^3}{5}-\sqrt{6.05}}{\dfrac{3\times 6.05^2}{5}-\dfrac{1}{2\sqrt{6.05}}}\approx 4.13\\
\hline
\ 4 \ & 4.13-\dfrac{\dfrac{4.13^3}{5}-\sqrt{4.13}}{\dfrac{3\times 4.13^2}{5}-\dfrac{1}{2\sqrt{4.13}}}\approx 2.92\\
\hline
\ 5 \ & 2.92-\dfrac{\dfrac{2.92^3}{5}-\sqrt{2.92}}{\dfrac{3\times 2.92^2}{5}-\dfrac{1}{2\sqrt{2.92}}}\approx 2.24\\
\hline
\ 6 \ & 2.24-\dfrac{\dfrac{2.24^3}{5}-\sqrt{2.24}}{\dfrac{3\times 2.24^2}{5}-\dfrac{1}{2\sqrt{2.24}}}\approx 1.96\\
\hline
\ 7 \ & 1.96-\dfrac{\dfrac{1.96^3}{5}-\sqrt{1.96}}{\dfrac{3\times 1.96^2}{5}-\dfrac{1}{2\sqrt{1.96}}}\approx 1.906\\
\hline
\ 8 \ & 1.906-\dfrac{\dfrac{1.906^3}{5}-\sqrt{1.906}}{\dfrac{3\times 1.906^2}{5}-\dfrac{1}{2\sqrt{1.906}}}\approx 1.90366\\
\hline
\end{array}
\(\Rightarrow C\)
