Can you put up what you've done next time...
skipping forward a bit...
If , 1 fixed point .
If , 3 fixed points at and .
If , 2 fixed points at , 1/2.
We consider what happens when a=4.
Now without doing any calculations I can tell you it's a Saddle node bifurcation. Why?
For , has no fixed points in the interval .
(ii) At , has one fixed point in at and .
(iii) For , has two fixed points in , one attracting and one repelling.
Where the c denoted 'critical' value. (a=4 and x=1/2 in this case)
So looking at that.
Condition i) is satisfied since if a<4 we get no real fixed points near x=1/2.
Condition ii) is satisfied since we get a single fixed point at x=1/2 when a=4.
Condition iii) is satisfied since when a>4 two fixed points appear. I haven't calculated stability yet but you can do that to check.
So lets us find the derivative at x=1/2 to see what we get just to be sure...
Evaluate this at and and we get...
Hence saddle node bifurcation.
Use the absolute value of the derivative to prove stability/unstability at the fixed points when if you wish to double check that it is in fact, a saddle-node bifurcation.