The equation x3−3x2+31=0x^3 - 3x^2 + 31 = 0x3−3x2+31=0 has exactly one real root. For this equation, the Newton-Raphson formula can be written
xn+1=2xn3−3xn2−313xn2−6xn x_{n+1} = \frac{2x_n^3 - 3x_n^2 - 31}{3x_n^2 - 6x_n} xn+1=3xn2−6xn2xn3−3xn2−31Using the formula given above with x1=−1x_1 = -1x1=−1, find the values of x2 x_2\,x2 and x3x_3x3
329 exam-style questions on Edexcel A Level Maths Numerical Methods, covering 10.1 Locating Roots, 10.2 Iteration, 10.3 The Newton-Raphson Method, and 10.4 Applications to Modelling. Each one has a worked solution and a mark scheme showing where the marks go.