A first approximation to the equation x^3 - 3x^2 + 1 = 0 is 0.5. Using one application of Newton's method, find a better approximation to two decimal places.

is this correct??

P(0) = 1 -pos

P(1) = -1 -neg

therefore there is a root between 0 and 1

x^3 - 3x^2 + 1 = 0

f(x) = x^3 - 3x^2 + 1

f ' (x) = 3x^2 - 6x

f(a) = a^3 - 3a^2 + 1

f ' (a) = 3a^2 - 6a

a_1 = a - f(a) / f ' (a)

= 0.5 - [(0.5)^3 - 3x(0.5)2 + 1] / [3x(0.5)^2 - 6x(0.5)]

=2/3

=0.67