The curve C has the equation y=(12−x)lnxy = (12 - x) \ln xy=(12−x)lnx, x>0x > 0x>0
Verify that dydx=1x(12−x)−lnx=12x−1−1−lnx\displaystyle \frac{dy}{dx} = \frac{1}{x}(12 - x) - \ln x = 12x^{-1} - 1 - \ln xdxdy=x1(12−x)−lnx=12x−1−1−lnx
C has a stationary point at P. Show that the x x\,x coordinate of P lies between 4.5 and 5
Show that the x x\,x coordinate of P is a solution of x=121+lnx\displaystyle x = \frac{12}{1 + \ln x}x=1+lnx12
Use the iteration formula xn+1=121+lnxn\displaystyle x_{n+1} = \frac{12}{1 + \ln x_n}xn+1=1+lnxn12 with x0=4.75x_0 = 4.75x0=4.75 to find, to 3 decimal places, the values of x1x_1x1, 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.