Saddle-node bifurcation Taylor expansion I am working through the Taylor expansion of $\dot{x}=f(x,r)$ at $x=x^*$ and $r=r_c$ in Nonlinear Dynamics and Chaos (Strogatz) [Saddle-Node Bifurcations]:
\begin{align}
\dot{x} & = f(x,r) \\
 & = f(x^{*},r_c)+(x-x^{*}) \frac{\partial{f}}{\partial{x}}\Bigg|_{(x^{*},r_c)} +(r-r_c)\frac{\partial{f}}{\partial{r}}\Bigg|_{(x^{*},r_c)}+\frac{1}{2}(x-x^{*})^2\frac{\partial^2{f}}{\partial{x^2}}\Bigg|_{(x^{*},r_c)}+\ldots
\end{align}
The book says

$f(x^{*},r_c)=0$ since $x^{*}$ is a fixed point, and
  $\frac{\partial{f}}{\partial{x}}\Big|_{(x^{*},r_c)}=0$ by the tangency
  condition of a saddle-node bifurcation.

I understand $\dot{x}=f(x^{*},r_c)=0$ because it's at a fixed point.
What I don't understand is why $\frac{\partial{f}}{\partial{x}}\Big|_{(x^{*},r_c)}=0$. Isn't that the stability of the fixed point which can be $f'(x^{*})<0$, $f'(x^{*})>0$, or $f'(x^{*})=0$?
 A: Look at $\textbf{Example $3.1.2$}$ on page $47$ in your book $\textit{Nonlinear Dynamics and Chaos}$ by Steven H. Strogatz. 
The example goes as follows: consider $\dot{x}=f(x,r)$, where your $f(x,r)=r-x-e^{-x}$. First, you want to find fixed points $x^*$, which occur when $f(x,r)=0$. 
It is easier to plot $g(x,r)=r-x$ and $h(x)=e^{-x}$ on a graph and then visually look for the points of intersection, which are precisely the fixed points $x^*$. The $x$-axis on the graph remains the same while the $y$-axis represents $\dot{x}$, and we view $r$ as a constant for the time-being. There are two points of intersection as you can see below (photo credit: Figure $3.1.6 (a)$ on page $48$ in Strogatz): 

Now, when you decrease the parameter $r$, the line corresponding to the function $g(x,r)$ will move down. When this line is tangent to the curve defined by $h(x)$, you get one fixed point and this fixed point is defined to be a $\textbf{saddle-node bifurcation}$, and such $r=r_c$ is defined to be a $\textbf{bifurcation point}$. Note that if you continue to decrease $r$, then the line will no longer intersect the curve and thus, there aren't any fixed points. 
Now, returning to the saddle-node bifurcation, we obtain the bifurcation point $r_c$ precisely when: 


*

*the graphs $g(x,r)$ and $h(x)$ intersect, and 

*the tangent line of $g(x,r)$ is the tangent line of $h(x)$ at $x^*$. 


Mathematically, the conditions are given by 
$$
r_c - x^*=e^{-{x^*}} \hspace{4mm}\mbox{ and } \hspace{4mm}
\frac{\partial}{\partial x}(r-x)\Bigg|_{(x^*,r_c)} = \frac{d}{dx}(e^{-x})\Bigg|_{(x^*,r_c)}, 
$$
and we can rewrite the equalities as: 
$$
r_c-x^*-e^{-{x^*}}=0\hspace{4mm} \mbox{ and }\hspace{4mm}  \frac{\partial}{\partial x}(r-x-e^{-x})\Bigg|_{(x^*,r_c)}=0. 
$$
These are equivalent to the conditions 
$$
f(x^*, r_c)=0 \hspace{4mm} \mbox{ and }\hspace{4mm} \frac{\partial}{\partial x}f(x,r)\Bigg|_{(x^*,r_c)}=0. 
$$
