How do you find the first three iterate of the function #f(x)=3x^2-4# for the given initial value #x_0=1#?

1 Answer
Nov 24, 2016

#-1, -1, -1#

Explanation:

Applying the function, we find:

#x_1 = f(x_0) = 3x_0^2-4 = 3(1)^2-4 = 3-4 = -1#

#x_2 = f(x_1) = 3x_1^2-4 = 3(-1)^2-4 = 3-4 = -1#

#x_3 = f(x_2) = 3x_2^2-4 = 3(-1)^2-4 = 3-4 = -1#

In fact, for any #n >= 1# we have #x_n = -1#, since:

#f(-1) = 3(-1)^2-4 = -1#

is a fixed point.