The equation x3−3x2+5=0x^3 - 3x^2 + 5 = 0x3−3x2+5=0 has exactly one real root.
Show that, for this equation, the Newton-Raphson formula can be written
xn+1=2xn3−3xn2−53xn2−6xn\displaystyle x_{n+1} = \frac{2x_n^3 - 3x_n^2 - 5}{3x_n^2 - 6x_n}xn+1=3xn2−6xn2xn3−3xn2−5
Using the formula in part (a) with x1=−1x_1 = -1x1=−1, find the values of x2 x_2\,x2 and x3x_3x3. Give your answers to four decimal places.
Explain why, for this equation, the Newton-Raphson method cannot be used with x1=0x_1 = 0x1=0.
137 exam-style questions on OCR (MEI) A Level Maths 1.10 Numerical Methods (A-level only), covering 1.10.1 Locate roots by change of sign (A-level only), 1.10.2 When change of sign methods fail (A-level only), 1.10.3 Fixed point iteration (A-level only), 1.10.4 Newton-Raphson method (A-level only), 1.10.5 Convergence of iterations (A-level only), 1.10.6 Trapezium rule (A-level only), 1.10.7 Upper and lower bounds using rectangles (A-level only), and 1.10.8 Numerical methods to solve problems (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.