two functions at a time

Let us first solve the function $g(x)$ by differentiating it with respect to x and y respectively:

$\frac{1}{3} \cdot g'(\frac{x+2y}{3}) = \frac{1}{3}g'(x)$ (i)

$\frac{2}{3} \cdot g'(\frac{x+2y}{3}) = \frac{2}{3}g'(y)$ (ii)

which equating (i) with (ii) gives $g'(x) = g'(y) = A \Rightarrow g(x) = Ax + B$ , where $A,B \in \mathbb{R}.$ Plugging in the given boundary conditions ultimately results in $\boxed{g(x) = x + 2}$ .

Now, turning the to the quadratic inequality below we find that:

$f^{2}(g(x)) - 5f(g(x)) + 4 > 0 \Rightarrow [f(g(x)) - 1][f(g(x)) - 4] > 0 \Rightarrow 2arctan(x+2) < 1$ or $2arctan(x+2) > 4 \Rightarrow \boxed{arctan(x+2) \in (-\infty, \frac{1}{2}) \cup (2, \infty)}.$ . Since $-\frac{\pi}{2} < arctan(x+2) < \frac{\pi}{2}$ must be satisfied for all $x \in (-10,10)$ , we restrict the range to the interval $(-\frac{\pi}{2}, \frac{1}{2}).$ To find the allowable domain values, we require:

$-\frac{\pi}{2} < arctan(x+2) < \frac{1}{2} \Rightarrow tan(-\frac{\pi}{2}) < x+2 < tan(\frac{1}{2}) \Rightarrow -\infty < x < tan(\frac{1}{2}) - 2 \Rightarrow x \in (-\infty, -1.4537)$ . The final domain required of this inequality problem is $x \in (-10, -1.4537)$ , which includes $\boxed{EIGHT}$ integral values.