A curve has the equation
y=ln(3−2x)+x2−2xy = \ln(3 - 2x) + x^2 - 2xy=ln(3−2x)+x2−2x
The curve intersects the xxx-axis at exactly one point.
Show that the xxx-coordinate of this point lies between 0 and 1.
Use the Newton-Raphson method, with x0=1x_0 = 1x0=1, to find the xxx-coordinate of this point, giving your answer correct to five decimal places. Show the result of each iteration.