# Using Newton's method, a root of x^3+3x+7=0 correct to one decimal place is: A. x=1.6 B. x=1.5 C. x=1.4 D. x=1.3 Help please?

Oct 7, 2017

The answer is option $C$, i.e $= - 1.4$

#### Explanation:

The Newton's method : To obtain successive approximations of ${x}_{1} , {x}_{2} , {x}_{3} , \ldots \ldots$, start with a guess ${x}_{0}$ and use

${x}_{n + 1} = {x}_{n} - f \frac{{x}_{n}}{f ' \left({x}_{n}\right)}$

Here,

$f \left(x\right) = {x}^{3} + 3 x + 7$

$f ' \left(x\right) = 3 {x}^{2} + 3$

Let ${x}_{0} = 0$

Therefore,

${x}_{1} = 0 - f \frac{0}{f ' \left(0\right)} = - \frac{7}{3}$

${x}_{2} = - \frac{7}{3} - \frac{f \left(- \frac{7}{3}\right)}{f ' \left(- \frac{7}{3}\right)} = - 2.3333 + \frac{12.704}{19.333} = - 1.676$

${x}_{3} = - 1.676 + \frac{2.74}{11.4} = - 1.44$

${x}_{4} = - 1.44 + \frac{0.27}{0.19} = - 1.41$

${x}_{5} = - 1.41 + \frac{0.0038}{8.94} = - 1.41$

graph{x^3+3x+7 [-7.56, 4.92, -2.27, 3.97]}