Correct. If you use a calculator, you can do it very rapidly and get a better answer of about 0.652704
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