Yes, I am pretty sure you are correct.
Solution :
Given
![$f(x)=x^3-x-1, x_1=1$](https://tex.z-dn.net/?f=%24f%28x%29%3Dx%5E3-x-1%2C%20x_1%3D1%24)
![$f'(x)=3x^2-1$](https://tex.z-dn.net/?f=%24f%27%28x%29%3D3x%5E2-1%24)
Let the initial approximation is ![$x_1 =1$](https://tex.z-dn.net/?f=%24x_1%20%3D1%24)
So by Newton's method, we get
![$x_{n+1}=x_n-\frac{f(x_n)}{f'(x_n)},n=1,2,...$](https://tex.z-dn.net/?f=%24x_%7Bn%2B1%7D%3Dx_n-%5Cfrac%7Bf%28x_n%29%7D%7Bf%27%28x_n%29%7D%2Cn%3D1%2C2%2C...%24)
![$x_2=x_1-\frac{f(x_1)}{f'(x_1)}=1-\frac{1^3-1-1}{3(1)^2-1}=1.5$](https://tex.z-dn.net/?f=%24x_2%3Dx_1-%5Cfrac%7Bf%28x_1%29%7D%7Bf%27%28x_1%29%7D%3D1-%5Cfrac%7B1%5E3-1-1%7D%7B3%281%29%5E2-1%7D%3D1.5%24)
![$x_3=1.5-\frac{(1.5)^3-1.5-1}{3(1.5)^2-1}=1.34782608$](https://tex.z-dn.net/?f=%24x_3%3D1.5-%5Cfrac%7B%281.5%29%5E3-1.5-1%7D%7B3%281.5%29%5E2-1%7D%3D1.34782608%24)
![$x_4=1.34782608-\frac{(1.34782608)^3-1.34782608-1}{3(1.34782608)^2-1}=1.32520039$](https://tex.z-dn.net/?f=%24x_4%3D1.34782608-%5Cfrac%7B%281.34782608%29%5E3-1.34782608-1%7D%7B3%281.34782608%29%5E2-1%7D%3D1.32520039%24)
![$x_5=1.32520039-\frac{(1.32520039)^3-1.32520039-1}{3(1.32520039)^2-1}=1.32471817$](https://tex.z-dn.net/?f=%24x_5%3D1.32520039-%5Cfrac%7B%281.32520039%29%5E3-1.32520039-1%7D%7B3%281.32520039%29%5E2-1%7D%3D1.32471817%24)
![$x_6=1.32471817-\frac{(1.32471817)^3-1.32471817-1}{3(1.32471817)^2-1}=1.32471795$](https://tex.z-dn.net/?f=%24x_6%3D1.32471817-%5Cfrac%7B%281.32471817%29%5E3-1.32471817-1%7D%7B3%281.32471817%29%5E2-1%7D%3D1.32471795%24)
are identical up to eight decimal places.
The approximate real root is x ≈ 1.32471795
∴ x = 1.32471795
For every 3 cups of peanuts there are 2 cups of chocolate
9/3 = 3 so 3*2 = 6
For 9 cups of peanuts there are 6 cups of choclate
The answer is C
Here is the process, hope it helps!
y+2=-3/2(x-2)
y+2=-3/2x+3
y=-3/2x+1