Use Newton's method to estimate the two zeros of the function $f(x)=x^{4}-x-12$. Start with $x_{0}=-1$ for the left-hand zero and with $x_{0}=1$ for the zero on the right. Then, in each case, find $x_{2}$
Determine $x_{2}$ when $x_{0}=-1$
Answer
Final Answer: The value of \(x_{2}\) when \(x_{0}=-1\) is \(\boxed{-\frac{255}{109}}\).
Step 1 ：Given the function \(f(x)=x^{4}-x-12\) and its derivative \(f'(x)=4x^{3}-1\).
Step 2 ：We are asked to use Newton's method to estimate the zeros of the function. The formula for Newton's method is \(x_{n+1} = x_n - \frac{f(x_n)}{f'(x_n)}\).
Step 3 ：We are given that \(x_{0}=-1\). We need to find \(x_{2}\), which means we need to apply Newton's method twice.
Step 4 ：First, we calculate \(x_{1}\) using the formula: \(x_{1} = x_{0} - \frac{f(x_{0})}{f'(x_{0})} = -1 - \frac{(-1)^{4}-(-1)-12}{4*(-1)^{3}-1} = -3\).
Step 5 ：Then, we calculate \(x_{2}\) using the formula: \(x_{2} = x_{1} - \frac{f(x_{1})}{f'(x_{1})} = -3 - \frac{(-3)^{4}-(-3)-12}{4*(-3)^{3}-1} = -\frac{255}{109}\).
Step 6 ：Final Answer: The value of \(x_{2}\) when \(x_{0}=-1\) is \(\boxed{-\frac{255}{109}}\).
