# Confusion over Bifurcation Diagram [closed]

I have the function $f(x,a) = \frac{2}{x} + 0.75x + a$ and want to create a bifurcation diagram of this function as $a$ varies.

An $\textit{equilibrium}$ point (I thought) of the trajectory made by the function is a point $x_{*}$ such that $f(x_{*}) = x_{*}$. A bifurcation diagram plots equilibrium points on the $y$-axis with $a$ varying as the $x$-axis.

1. Why does this refer to roots of $f$ as equilibrium points? Am I supposed to be looking for points s.t. $f(x_*)=x_*$ or am I supposed to for points like in that pdf?
2. Can someone me analyze this function so I can better understand how a bifurcation diagram works?

## closed as unclear what you're asking by John B, Claude Leibovici, N. F. Taussig, SchrodingersCat, user223391 Feb 20 '16 at 16:41

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Well, I'm guessing $f(x;a)$ is the RHS of an ODE, and you are looking for equilibrium points of $$\frac{d x}{d t} = f(x;a).$$
If this is the case, then such points, $x_*$, are given by the relation $$\frac{2}{x_*} +\frac{x_*}{4} = -a,$$
which translates to $$x_*^2 + 4 a x_* + 8 = 0.$$
What remains is to analyze this equation as $a$ varies (i.e., number of solutions and value), and construct the diagram by plotting $x_*$ as function of $a$. Note that there are values of $a$ where there are two real solutions, a critical value where there is only one, and values where there is none. All of this should be reflected in your diagram.
• Why do you say equil points are given by the relation $\frac{2}{x_{*}} + \frac{x_{*}}{4} = -a$? Shouldn't they be points $x_*$ that satisfy $$\frac{2}{x} + \frac{x_*}{4} + a = x_*$$ ? – bifurcationLost Feb 18 '16 at 23:47