Question

Use Newtons method to find the roots of the equation 1/x=1+x^3

Answer 1

f(x)=x^3-1/x + 1

guess, f(x)
1, 1 positive
2, 8.5 postive go the other way.
1/2, 1/8-2+1=-7/8 so root is between 1, and 1/2
try .75 as root.
Now is the time to use the google search window
.75^3-1/.75+1= (put that in google window), .089 postive, so go lower toward 1/2
.65^3-1/.65+1=-.26 go postive now
.70^3-1/.70+1=-.08 negative, but notice it is close to zero....
.73^3-1/.73+1=+.019, go back
.72^2-1/.72+1=-.0156 go back
.726^2-1/.726 + 1=.005 go back
.724^2-1/.724+1= you do it. You can get the roots as accurate as you wish. Calculator required.

Answer 2

multiply by x
1 = x + x^4
x^4 + x - 1 = 0


let f(x) = x^4 + x - 1
f'(x) = 4x^3 + 1

newton's formula for his method says
f(newx) = x - f(x)/f'(x) , where x is some intial guess
= x - (x^4 + x -1)/(4x^2 + 1)
= (3x^4 + 1)/(4x^3 + 1)

let's start with bobpursleys first guess of x = .75

x ---- newx
.75------ .72529
.72529 -- .724492755
.72449275-- .7244919

<b>x = .72449 correct to 5 decimals in 3 iterations</b>

there appears to be another root around x = -1.25
x = -1.25

x ---- newx
-1.25 ----- +1.2219..
-1.2219.. -- -1.2207459..
-1.220749.. -- -1.2207440..
-1.220744 --- -1.2207440..

<b> x = -1.220744 correct to 6 decimals in 4 iterations </b>

my rough sketch shows only 2 real roots,
since it is a 4th degree, there have to be 2 complex roots.

you could do two consecutive synthetic divisions using
x=.72449 followed by x = -1.220744
to reduce it down to a quadratic, then using the formula to find the imaginary roots.