The equilibrium yield YYY of a specific chemical synthesis is modeled by the equation Y3−6Y+3=0Y^3 - 6Y + 3 = 0Y3−6Y+3=0. This equation has a single positive solution, Y=αY = \alphaY=α, in the interval [2.1,2.2][2.1, 2.2][2.1,2.2].
By considering a suitable change of sign, show that α\alphaα lies between 2.1 and 2.2.
Show that the equation Y3−6Y+3=0Y^3 - 6Y + 3 = 0Y3−6Y+3=0 can be rearranged into the form
Y=6−3Y Y = \sqrt{6 - \frac{3}{Y}} Y=6−Y3Use the iterative formula
Yn+1=6−3Yn Y_{n+1} = \sqrt{6 - \frac{3}{Y_n}} Yn+1=6−Yn3with Y1=2.1Y_1 = 2.1Y1=2.1, to determine the values of Y2,Y3Y_2, Y_3Y2,Y3 and Y4Y_4Y4, giving your answers to four decimal places.
Hence, deduce an interval of width 0.001 in which α\alphaα lies.
Practise AQA A Level Maths 1.12 I: Numerical methods (A-level only) with exam-style questions for A Level Maths. 107 questions covering 1.12.1 Locating roots by change of sign (A-level only), 1.12.2 Iterative methods and Newton-Raphson (A-level only), 1.12.3 Numerical integration (A-level only), and 1.12.4 Numerical methods in context (A-level only), matched to the AQA A Level Maths (7357) specification and written in Paper 1, Paper 2 and Paper 3 style. Every question includes a full worked solution and mark scheme, so you can see where marks are awarded rather than just whether you got the answer right.
