formal proof: Intersection of two function Given are the following two functions:
$
g(x,\theta)=1-\frac{\left(  1-\theta\right)  }{\theta}\left(  \frac{(x-1)R}
{(1-(1-\pi)i)R +\left(  1-\pi\right) i x}\right) \tag{NAG}
$
and
$f(x)=\frac{\pi xR}{\left[  \left(  1-i\right)  R+ix\right]  }
\cdot\frac{(1-i)}{\left(  1-(1-\pi)(1-i)\right)  } \tag{CMP}$
What I want to show is that the intersection point of $g(x,\theta)$ and $f(x)$ increases with $\theta$. I mean that an increase in $\theta$ leads to a new intersection which is characterized by both higher x and higher function value.
This can be shown graphically but what I need is a formal proof. The problem is that solving for the equilibrium is a solution of a complex (but just) quadratic equation. 
Note that the CMP is no function of $\theta$ but an increasing function of x $\forall x \in  \mathbb{R}_{\leq0}$, thus a change in $\theta$ moves the intersection along CMP. Additionally, NAC is a decreasing function of $x$ $(\frac{\partial g(x,\theta)}{\partial p_{n}}<0$) but the slope of the tangent increases with $\theta$ $(\frac{\partial^2 g(x,\theta)}{\partial p_{n} \partial \theta}>0$). Since the NAC passes independently of the parameter setting through $(1,1)$, an increase in $\theta$, from $\theta^{**}$ to $\theta^{*} $leads to a raise of the angel $\alpha$ moving a low intersection point $(x^{**},y^{**})$ to a higher $(x^{*},y^{*})$. 
To summarize, what I need is a formal proof that shows that the intersection point increases with an increase in $\theta$. I hope someone can help me to show this with a without the need of solving for the intersection point. (I actually do not know how to start). Thx for your help!!!
 A: Okay, you can do this in three steps (BTW, you need to assume also that $\theta\in (0,1)$; I take it that assumption is okay since you call $\theta$ a probability).
Step 1: As you noted, $f(x)$ is an increasing function of $x$ when $x > 0$. In fact, $f(x)$ is strictly increasing: its slope is always positive (using that $\pi,i\in (0,1)$ and $R \geq 1$ by assumption). This implies

If the intersection point increases in $x$, the intersection value $f(x)$ must also increases. 

Step 2:  As you also noted $g(x,\theta)$, for fixed value of $\theta$, is a decreasing function of $x$. In particular, it is monotonic. Which leads us to

$f(x)$ and $g(x,\theta)$ only intersect once. (Where $x\geq 0$.)

and hence

Consider $\theta_1$ and $\theta_2$. Let $x_1$ be the value such that $f(x_1) = g(x_1,\theta_1)$. Suppose that $g(x_1,\theta_2) > g(x_1,\theta_1)$, then necessarily the intersection point $x_2$ which solve $f(x_2) = g(x_2,\theta_2)$ must satisfy $x_2 > x_1$. 

Proof: by monotonicity, for every $0 \leq y \leq x_1$ we have $f(y) \leq f(x_1)$. Similarly we also have $g(y,\theta_2) \geq g(x_1,\theta_2)$. So if $g(x_1,\theta_2) > f(x_1)$, it is impossible that $f(y) = g(y,\theta_2)$. Since the two curves must intersect, they have to intersect somewhere $y > x_1$. q.e.d.
Note that similarly if $g(x_1,\theta_2) < g(x_1,\theta_1)$, the intersection point $x_2$ must be less than $x_1$. 
Step 3: So all you need to show now is that if $(x,\theta)$ is such that $f(x) = g(x,\theta)$, for any $\theta' > \theta$, you have that $g(x,\theta') > g(x,\theta)$. Looking at the equation for $g(x,\theta)$, you see that $g(x,\theta)$, for fixed $x > 1$, is an increasing function of $\theta$ (just compute $\partial g / \partial \theta$). So you'd be done if you can show that the intersection point of $f$ and $g$ cannot be a point where $x \leq 1$. 
Using the monotonicity again, it suffices to compute $g(1,\theta)$ and $f(1)$. If $g(1,\theta)> f(1)$, then from the same argument as in step 2, we know that the intersection point must be $x > 1. 
You see immediately that $g(1,\theta) = 1$. On the other hand it is a simple algebraic computation to see that $f(1) < 1$: you just need to show that $\pi (1-i) R < [ (1-i)R + i ][ 1- (1-\pi)(1-i)]$. If you multiply out the right hand side, and subtract from it the left hand side, you get that you just need the inequality $0 < (1-\pi)(1-i)i R + (1- (1-\pi)(1-i))i$ to hold, which follows from the assumed range of $\pi,i$ and that $R$ is positive. 
