Can you put up what you've done next time...

skipping forward a bit...

Let

Have,

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?

Definition...

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.

=> Stable

=> Unstable