Calculus: Early Transcendentals, Chapter 4, 4.8, Section 4.8, Problem 26

cos(x^2-x) = x^4
Set the left side equal to zero.
0=x^4-cos(x^2-x)
To solve using Newton's method, apply the formula:
x_(n+1)=x_n - (f(x_n))/(f'(x_n))
Let the function of the given equation be:
f(x) =x^4-cos(x^2-x)
And its derivative is:
f'(x) = 4x^3 + (2x-1)sin(x^2-x)
Plug-in f(x) and f'(x) to the formula of Newton's method.
x_(n+1) = x_n - ((x_n)^4-cos((x_n)^2-x_n))/(4(x_n)^3+(2x_n-1)sin((x_n)^2-x_n))
To get the initial value of x, refer to the graph of f(x). (See figure.)
Notice that when f(x) =0, the values of x are near -0.8 and 1. Use these two values of x to solve for the roots of the function to eight decimal places.
For the first root, let the initial value be -0.8.
x_1=-0.8
x_2= x_1 - ((x_1)^4-cos((x_1)^2-x_1))/(4(x_1)^3+(2x_1-1)sin((x_1)^2-x_1))=-0.7396478896
x_3= x_2 - ((x_2)^4-cos((x_2)^2-x_2))/(4(x_2)^3+(2x_2-1)sin((x_2)^2-x_2))=-0.7348883415
x_4= x_3 - ((x_3)^4-cos((x_3)^2-x_3))/(4(x_3)^3+(2x_3-1)sin((x_3)^2-x_3))=-0.7348591049
x_5= x_4 - ((x_4)^4-cos((x_4)^2-x_4))/(4(x_4)^3+(2x_4-1)sin((x_4)^2-x_4))=-0.7348591038
Notice that the two approximates have the same eight decimal places. So we stop the iteration here. Thus, one of the roots of f(x) is x=-0.73485910 .
For the second root, let the initial value be 1.
x_1=1
x_2= x_1 - ((x_1)^4-cos((x_1)^2-x_1))/(4(x_1)^3+(2x_1-1)sin((x_1)^2-x_1))=1
Notice that the second root is an integer. It has an exact value which is x=1.

Therefore, the solution of the equation cos(x^2-x) =x^4 is x={-0.73485910, 1}.