A research drone's vertical displacement yyy (in meters) from its starting altitude is modeled by the equation
y=2x2ln(x)−5x,x>0.5 y = 2x^2 \ln(x) - 5x, \quad x > 0.5 y=2x2ln(x)−5x,x>0.5where xxx is the horizontal distance in kilometers from the launch point. The drone reaches a minimum altitude at point PPP.
Show, using calculus, that the xxx-coordinate of PPP is a solution of the equation
4xln(x)+2x−5=0 4x \ln(x) + 2x - 5 = 0 4xln(x)+2x−5=0Using the Newton-Raphson method with f(x)=4xln(x)+2x−5f(x) = 4x \ln(x) + 2x - 5f(x)=4xln(x)+2x−5 and a first approximation x1=2x_1 = 2x1=2, find the value of x2x_2x2 and the value of x3x_3x3, giving your answers to 3 decimal places.
