Sometimes Newton's method fails no matter what initial value x₀ is chosen (unless you are lucky enough to choose the root itself). Let f(x) = ³√x and choose x₀ arbitrarily (x₀ ≠ 0). 

a. Show that x_n+1 = −2x_n for n = 0, 1, 2, ... so that the successive guesses generated by Newton's method are x₀, −2x₀, 4x₀,.... 

b. Use your graphing utility to graph f(x), and use an appropriate utility to draw the tangent lines to the graph of y = f(x) at the points that correspond to x₀, −2x₀, 4x₀,... Why do these numbers fail to estimate a root of f(x) = 0?