How do you use Newton's method to find the approximate solution to the equation #x^4=x+1,x>0#?

We have:

# x^4 = x+1 => x^4-x-1=0#

Let:

# f(x) = x^4-x-1 #

Our aim is to solve #f(x)=0#. First let us look at the graph:
graph{x^4-x-1 [-4, 4, -5, 10]}

We can see that there are two solutions; one solution in the interval #-1 lt x lt 0#, one in #0 lt x lt 2#. The root we find will depend upon our initial approximation #x_0#, and this value will require a little trial and error.

To find the solution numerically, using Newton-Rhapson method we use the following iterative sequence

# { (x_1,=x_0), ( x_(n+1), = x_n - f(x_n)/(f'(x_n)) ) :} #

Therefore we need the derivative:

# \ \ \ \ \ \ \f(x) = x^4-x-1 #
# :. f'(x) = 4x^3-1 #

Then using excel working to 10dp we can tabulate the iterations as follows:

Initial Value #x_0=1#

Incidentally, we can also find the other solution #(x lt 0)#

Initial Value #x_0=0#

We could equally use a modern scientific graphing calculator as most new calculators have an " Ans " button that allows the last calculated result to be used as the input of an iterated expression.

And we conclude that the solution (to 10dp) are:

# x_1 = 1.2207440846 #
# x_2 = -0.7244919590 #