Answer:
[tex]x_2 \approx -1.769[/tex]
Step-by-step explanation:
Let [tex]f(x)=x^3+x+7[/tex]
So [tex]f'(x)=3x^2+1[/tex]
[tex]x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)}[/tex]
Let [tex]x_1=-2[/tex]
We are going to find [tex]x_2[/tex]
So we are evaluating [tex]-2-\frac{f(-2)}{f'(-2)}[/tex]
First step find f(-2)
Second step find f'(-2)
Third step plug in those values and apply PEMDAS!
[tex]f(-2)=(-2)^3+(-2)+7=-8-2+7=-10+7=-3[/tex]
[tex]f'(-2)=3(-2)^2+1=3(4)+1=12+1=13[/tex]
So
[tex]x_2=-2-\frac{-3}{13} \\\\ x_2=\frac{-26+3}{13} \\\\ x_2=\frac{-23}{13} \\\\ x_2 \approx -1.769[/tex]
