We have the model $X_{n+1} = 4\left(X_n - \dfrac{1}{2}\right)^2$ with a given $X_0$ on the domain $[0,1]$. We have the following question:
Use your graphing calculator to figure out if there are periodic solutions with period 2 and if yes; calculate for which $X_0$ this is the case.
So to find period 2 solutions I rewrote and got:
$$ x = 4\left(\,4±\left(\,x-\dfrac{1}{2}\right)^2 - \dfrac{1}{2}\right)^2$$
But I need some hints on how to continue with this problems. Thanks.
The equation is $x_{n+1} = f(x_n)$. A period-2 solution means $x_{n+2} = x_n$. So you can find these by solving $x = f(f(x))$. To do this, plot $x$ and $f(f(x))$ and see where they cross (they cross iff $x=f(f(x))$).
In your case, $f(x) = 4\left(x-\frac{1}{2}\right)^2$, so $f(f(x)) = 4\left( 4\left( x-\frac{1}{2}\right)^{2}-\frac{1}{2}\right)^{2}$.
Plot $x, 4\left( 4\left( x-\frac{1}{2}\right)^{2}-\frac{1}{2}\right)^{2}$ on $[0,1]$ and look for where they cross.
Four solutions are apparent (as expected, since you are solving a quartic).
This gives:
Also, you could note that the equation for the period-2 fixed points is $x=4\,( 4 ( x-\frac{1}{2})^{2}-\frac{1}{2} )^{2}$, which simplifies to $64x^4-128x^3+80x^2-17x+1=0$. This can be factored as $(x-1)(4x-1)(16 x^2-12 x+1) = 0$, so two solutions are obvious, and the other two straightforward to compute.
Addendum: Here is another graph that may be more illustrative: The two fixed points (period-1 solutions) at $\frac{1}{4},1$ are clear. (Period-1 solutions are period-2 solutions, of course). The period-2 orbit is also clear.