Find the values of r at which bifurcation occus and classify those as saddle node, transcritical, or pitchfork bifurcation $$f(x)= rx-\frac{x}{1+x}$$
Find the values of $r$ at which bifurcation occus and classify those as saddle node, transcritical, or pitchfork bifurcation.
I found the fixed points as $x^*=0,\frac{1}{r}-1$.
I am not sure how to identify what TYPE of bifurcation it is out of the three types. Is there a way to test which one? 
 A: $f(x)=rx-{x\over1+x}$. Fixed points where $f(x)=x$, so $$x=rx-{x\over1+x}$$ $x^*=0$ or $$1=r-{1\over1+x},\quad {1\over1+x}=r-1,\quad 1+x={1\over r-1},$$ $$x^*={1\over r-1}-1={2-r\over r-1}$$ I don't know how you got $x^*={1\over r}-1$. 
Bifurcations occur when $f'(x^*)=\pm1$. When $f'(x^*)=-1$, you may get a period-doubling bifurcation, but that's not among the ones you listed, so I'll ignore that. $$f'(x)=r-{1\over(x+1)^2}$$ so we need $r-{1\over(x+1)^2}=1$. When $x^*=0$, this gives $r=2$. The other fixed point, $x^*={2-r\over r-1}$, gives nothing (but check my algebra here). 
So, bifurcation occurs at $$(r^*,x^*)=(2,0)$$ Just from drawing the graphs of $x=0$ and $x={2-r\over r-1}$ and seeing how they cross at $(2,0)$ is, I think, enough to see that it's a transcritical bifurcation. But you can always calculate $f'(x)$ for fixed points nearby $(2,0)$ to see where you have stable fixed points and where unstable, and analyze the bifurcation that way. Or, you can use the formulas in Elaydi, for example, if $${\partial f\over\partial x}(r^*,x^*)=1,\quad {\partial f\over\partial r}(r^*,x^*)=0,\quad {\rm and}\quad {\partial^2 f\over\partial x^2}(r^*,x^*)\ne0$$ it's transcritical. 
