THESE FIRST FEW CELLS (WHICH LOOK EMPTY) CONTAIN THE CODE FOR THE COMMANDS USED BELOW.  I WOULDN'T RECOMMEND DELETING THEM.  YOU ARE WELCOME TO TAKE A LOOK THOUGH!

iterate[functmm_,startmm_, itmm_,skipmm_]:= 
MatrixForm[N[NestList[functmm,Nest[functmm,N[startmm],skipmm],itmm],10]]

iterateexact[functmm_,startmm_, itmm_,skipmm_]:= 
MatrixForm[NestList[functmm,Nest[functmm,startmm,skipmm],itmm]]

f[x_] := 4 x(1-x);  (*This is the function we will be examining in detail.*)

f[x_]:=x^2

f[x_]:=Sin[x]

f[x_]:=Cos[x]

f[x_]:=ArcTan[x]

f[x_]:=Sqrt[x]

a=4;  
Plot[{f[x],x, 1}, {x,0,1}, PlotRange->All]

You can also solve the equation f(x)=x, or use FindRoot if the Solve command can't handle it.

a=4;
Solve[f[x]==x,x]

A periodic point of period n is a point p satisfying (f^n)(p)=p, and the point is said to be of least period n if (f^k)(p)p for all 0<k<n. The methods described above can also work for points of higher period.  I have included an example of a search for period 3 points.  To find points of period 3, we set f(f(f(x)))=x and solve for x.  The Mathematica command for (f^3)(x) is Nest[f,x,3].  The graph below shows the line y=x, y=f(x), and y=(f^3)(x).

a=4;
Plot[{Nest[f,x,4],f[x],x}, {x,0,1}, PlotRange->{0,1}]

a=4;
iterate[f,.123,10,100]

Sometime we want to iterate exact values.  Don't use this command unless you know that's what you want to do.

a=4;
iterateexact[f,6/50,5,0]

Why not use iterateexact all the time?  That’s obvious from the last line.  Note that if we left the input as 0.12 Mathematica would use decimal approximations.

a=4;
iterateexact[f,6/50.,5,0]