A hydraulic model suggests that the depth ddd, in metres, of water in a specific section of a reservoir is determined by the cubic equation d3−6d+2=0d^3 - 6d + 2 = 0d3−6d+2=0. It is known that there is a single solution d=αd = \alphad=α for the operating depth in the interval [2.2,2.3][2.2, 2.3][2.2,2.3].
By considering a suitable change of sign, show that α\alphaα lies between 2.2 and 2.3.
Show that the equation d3−6d+2=0d^3 - 6d + 2 = 0d3−6d+2=0 can be rearranged into the iterative form
d=6−2d d = \sqrt{6 - \frac{2}{d}} d=6−d2Use the iterative formula
dn+1=6−2dn d_{n+1} = \sqrt{6 - \frac{2}{d_n}} dn+1=6−dn2with d1=2.2d_1 = 2.2d1=2.2, to find the values of d2,d3d_2, d_3d2,d3 and d4d_4d4, giving your answers to four decimal places.
Hence, deduce an interval of width 0.001 in which α\alphaα lies.
