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?

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Answer 1 · 2017-10-07T07:31:49

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

The Newton's method : To obtain successive approximations of #x_1, x_2, x_3,......#, start with a guess #x_0# and use

#x_(n+1)=x_n-f(x_n)/(f'(x_n))#

Here,

#f(x)=x^3+3x+7#

#f'(x)=3x^2+3#

Let #x_0=0#

Therefore,

#x_1=0-f(0)/(f'(0))=-7/3#

#x_2=-7/3-(f(-7/3))/(f'(-7/3))=-2.3333+12.704/19.333=-1.676#

#x_3=-1.676+2.74/11.4=-1.44#

#x_4=-1.44+0.27/0.19=-1.41#

#x_5=-1.41+0.0038/8.94=-1.41#

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