Answer:
[tex]x_{2}\approx 2.59, x_{4}\approx 4.3[/tex]
Step-by-step explanation:
1) Given that the formula for the Newton Method is:
[tex]x_{n+1}=x_{n}-\frac{f(x_{n})}{f(x_{n+1})}[/tex]
2) The Function and its 1st derivative
[tex]f(x)=3x^{7}+2x^{4}+2\\f'(x)=21x^{6}+8x^{3}[/tex]
3) Let's work with the first approximation. Plugging in in the function:
[tex]f(x)=3x^{7}+2x^{4}+2\Rightarrow f(3)=3(3)^{7}+2(3)^{3}+2\Rightarrow f(3)=6617\\f'(x)=21x^{6}+8x^{3}\Rightarrow f'(3)=21(3)^{6}+8(3)^{4}\Rightarrow f'(3)=15957\\\\x_{n+1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}\Rightarrow x_{2}=x_{1}-\frac{f(x_{1})}{f'(x_{1})}\\x_{2}=3-\frac{6617}{15957}\Rightarrow x_{2}\approx 2.59\\\\[/tex]
4) To approximate the other root x
[tex]f(x)=3x^{7}+2x^{4}+2\Rightarrow f(5)=3(5)^{7}+2(5)^{3}+2\Rightarrow f(5)=234627\\f'(x)=21x^{6}+8x^{3}\Rightarrow f'(5)=21(5)^{6}+8(5)^{4}\Rightarrow f'(5)=333125\\\\x_{n+1}=x_{n}-\frac{f(x_{n})}{f'(x_{n})}\Rightarrow x_{4}=x_{3}-\frac{f(x_{1})}{f'(x_{1})}\\x_{4}=5-\frac{234627}{333125}\Rightarrow x_{4}\approx 4.3[/tex]
