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

Nov 24, 2016

$- 1 , - 1 , - 1$

#### Explanation:

Applying the function, we find:

${x}_{1} = f \left({x}_{0}\right) = 3 {x}_{0}^{2} - 4 = 3 {\left(1\right)}^{2} - 4 = 3 - 4 = - 1$

${x}_{2} = f \left({x}_{1}\right) = 3 {x}_{1}^{2} - 4 = 3 {\left(- 1\right)}^{2} - 4 = 3 - 4 = - 1$

${x}_{3} = f \left({x}_{2}\right) = 3 {x}_{2}^{2} - 4 = 3 {\left(- 1\right)}^{2} - 4 = 3 - 4 = - 1$

In fact, for any $n \ge 1$ we have ${x}_{n} = - 1$, since:

$f \left(- 1\right) = 3 {\left(- 1\right)}^{2} - 4 = - 1$

is a fixed point.