Steady state of a Dynamical System

Question

Why here can we say that $x=x_s$ increases monotonically, when it depends on time?

Answer 1

If we define the functions $$f_1(x)=\frac{1}{x}\\ f_2(x)=\alpha\frac{x^m}{1+x^m}$$ the steady state points $x_s$ are the solutions of $$f_1(x)=f_2(x)$$ What the question wants to prove is that there is a unique such point. This is proved as follows:

If we allow $x$ to vary from $0$ to $\infty$ these two curves $\{(x,f_1(x))|x\in(0,\infty)\}$, $\{(x,f_2(x))|x\in(0,\infty)\}$ are created in the plane. The steady state points $x_s$ are the points were these two curves cross. The function $f_1(\cdot)$ is a strictly decreasing function of $x$ and the function $f_2(\cdot)$ is a strictly increasing function of $x$ since $$\frac{x^m}{1+x^m}=1-\frac{1}{1+x^m}$$ with $-1/(1+x^m)$ strictly increasing. Now since the one is strictly decreasing and the other is strictly increasing it is easy to prove that there is exactly one such point $x_s$ such that $f_1(x)=f_2(x)$.