Defining #f(x) = 1/x-(1+x^3)# if #x_0# is such that
#f(x_0) approx 0# then
#f(x_1) = f(x_0) + f'(x_0)(x_1-x_0)+O_2(abs(x_1-x_0))#
where #O_2(abs(x_1-x_0))# is something that goes to #0# with the order of #abs(x_1-x_0)^2#. In those circumstances we can assume that #f(x_1) approx 0# so
#0=f(x_0)+f'(x_0)(x_1-x_0)# which is an iterative procedure used for root computation. So
#x_(k+1)=x_k-f(x_k)/(f'(x_k))#
In our case we have
#f(x_k) = 1/x_k-(1+x_k^3)# and
#f'(x_k) = -(1/x^2 + 3 x^2)#
With this procedure we will obtain all real roots.
Now beginning with #x_0 = -1# we have
#((k,x_k),(1,-1.0),(2,-1.25),(3,-1.22126),(4,-1.22074),(5,-1.22074))#
Trying now #x_0=1#
#((k,x_k),(1,1.0),(2,0.75),(3,0.724449),(4,0.724492),(5,0.724492))#
Attached a plot of #y=1/x# in blue and #y = 1+x^3# in red.
