The equation x3−4x+1=0x^3 - 4x + 1 = 0x3−4x+1=0 has a single solution, x=αx = \alphax=α, in the interval [1.8,1.9][1.8, 1.9][1.8,1.9].
By considering a suitable change of sign, show that α\alphaα lies between 1.8 and 1.9.
Show that the equation x3−4x+1=0x^3 - 4x + 1 = 0x3−4x+1=0 can be rearranged into the form
x=4−1x x = \sqrt{4 - \frac{1}{x}} x=4−x1Use the iterative formula
xn+1=4−1xn x_{n+1} = \sqrt{4 - \frac{1}{x_n}} xn+1=4−xn1with x1=1.8x_1 = 1.8x1=1.8, to find x2,x3x_2, x_3x2,x3 and x4x_4x4, giving your answers to four decimal places.
Hence, deduce an interval of width 0.001 in which α\alphaα lies.
