Transcritical bifurcation

Remarks

Under these conditions, besides $x = 0$, the equilibria near $(0, 0)$ form a curve $x = c(\mu)$ with $c(0) = 0$ and $c'(0) = -2 f_{x\mu}(0, 0) / f_{xx}(0, 0) \ne 0$. So this second branch crosses the branch $x = 0$ at $\mu = 0$. Since $f_x(0, \mu)$ changes sign at $\mu = 0$, the equilibrium $0$ changes stability there, and the other branch changes stability the opposite way.

Example: $f(x, \mu) = \mu x - x^2$. Then $f(0, \mu) = 0$, $f_x(0, 0) = 0$, $f_{x\mu} = 1$ and $f_{xx} = -2$. The second branch is $x = \mu$, with $c'(0) = 1 = -2 \cdot 1 / (-2)$.

Non-example: $f(x, \mu) = \mu x - x^3$. Here $f_{xx}(0, 0) = 0$, so clause 3 fails; this family has a pitchfork bifurcation instead.