For the function f(x) = x^3 + x + 4 , write Newton's formula as x_n+1 = F( x_n) = x_n − [f(x_n)/f'(x_n)] for solving f(x) = 0.

Question

x_n+1=

Answer 1

Take the first derivative of [x³ + x + 4] as [3x² + 1] and use a graphing calculator to

note the intersection of y = x³ + x + 4 with the x-axis as falling between x = -1 and x = -2.

A programmable graphing calculator will aid in repeated analysis of

x − [x³ + x + 4] ÷ [3x² + 1] beginning with x = -1 (with each result of

the expression in bold print being fed back into the expression as the

new value of x) to show results of

-1.5

-1.387096774

-1.378838948

-1.378796701

-1.3787967

-1.3787967

The repetition of -1.3787967 signals the limit of the calculator's accuracy

in obtaining a root of x³ + x + 4 = 0.

(-1.3787967)³ + (-1.3787967) + 4 is returned  

by my calculator as simply 0; results are 

often shown as tiny fractions barely above 

(or below) zero such as "4.7E-12" or "-1.6E-9".