Given the equation:
[tex]4x^7+2x^4+2=0[/tex]You need to remember that Newton's method to approximate a root of the equation provides this formula:
[tex]x_{n+1}=x_n-\frac{f(x_n)}{f^{\prime}(x_n)}[/tex]In this case:
[tex]f(x_n)=4x^7+2x^4+2[/tex]Then, you need to derivate it, in order to find:
[tex]f^{\prime}(x_n)[/tex]Use these Derivative Rules:
[tex]\begin{gathered} \frac{d}{dx}(x^n)=nx^{n-1} \\ \\ \frac{d}{dx}(k)=0 \end{gathered}[/tex]Where "k" is a constant.
You get:
[tex]f^{\prime}(x_n)=(4)(7)x^6+(2)(4)x^3+0[/tex][tex]f^{\prime}(x_n)=28x^6+8x^3[/tex]• Knowing that:
[tex]x_1=2[/tex]You can set up that:
[tex]x_2=2-\frac{4x^7+2x^4+2}{28x^6+8x^3}[/tex]Substitute the given value of "x" and evaluate, in order to find the second approximation:
[tex]\begin{gathered} x_2=2-\frac{4(2)^7+2(2)^4+2}{28(2)^6+8(2)^3} \\ \\ x_2=\frac{1583}{928} \end{gathered}[/tex]• Now you can set up that:
[tex]x_3=\frac{1583}{928}-\frac{4x^7+2x^4+2}{28x^6+8x^3}[/tex]Then, substituting the corresponding x-value and evaluating, you get:
[tex]x_3=\frac{1583}{928}-\frac{4(\frac{1583}{928})^7+2(\frac{1583}{928})^4+2}{28(\frac{1583}{928})^6+8(\frac{1583}{928})^3}[/tex][tex]x_3\approx1.45[/tex]Hence, the answers are:
• Second approximation:
[tex]x_2=\frac{1583}{928}[/tex]• Third approximation:
[tex]x_3\approx1.45[/tex]