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

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Answer 1 · 2017-02-16T22:42:49

If we look at the graph of the function $y=f(x)$ :

graph{3x^2-3x+2 [-5, 5, -2, 5]}

We see that it is a quadratic that has no real roots. We can confirm this by looking at the discriminant;

$Delta = b^2-4ac = 9-24 lt 0$

Therefore $f(x)=0$ has no (real) solution and therefore an iterative approach will fail.

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