How to define the sign of a function $$y=\arctan\frac{x+1}{x-3} + \frac{x}{4}$$
I know that is necessary to put the function $>$ than $0$, but then?
$$\arctan\frac{x+1}{x-3} + \frac{x}{4}>0$$
It's a sum, so I can't set up a "false system" putting the two factors $>0$. In this case which is the rule to study the sign of this function?



Note: These are not homeworks.
 A: You want to determine the sign of $y$ given $x$, correct? In other words, you want to figure out when $y > 0$, which means, as you said, that we have to solve the inequality
$$\arctan \frac{x + 1}{x - 3} + \frac{x}{4} > 0.$$
The first step in solving this inequality are finding its "edges"—where the inequality goes from true to false, or vice versa. It turns out that there are only two places where this can happen: where the equation
$$\arctan \frac{x + 1}{x - 3} + \frac{x}{4} = 0$$
is true, and where the expression
$$\arctan \frac{x + 1}{x - 3} + \frac{x}{4}$$
is undefined. So, begin by making a list of all the points where the equation is true, along with all the points where that expression is undefined. You will use this list to determine the sign of $y$ given $x$ at all points.
A: Let $$f(x)=\arctan\frac{x+1}{x-3}+\frac{x}4\;.$$
You know that the algebraic sign of $\arctan u$ is the same as the sign of $u$, so 
$$
\arctan\frac{x+1}{x-3}\text{ is }\begin{cases}
\text{positive}&\text{when }x>3\\
\text{negative}&\text{when }-1<x<3\\
0&\text{when }x=-1\\
\text{positive}&\text{when }x<-1\;.
\end{cases}$$
Without further work we can be sure that $f(x)>0$ when $x>3$, and $f(x)<0$ when $-1\le x\le 0$; it’s only the intervals $(\leftarrow,-1)$ and $(0,3)$ that we need to investigate in detail.
Now
$$\begin{align*}
f\,'(x)&=\frac14-\frac4{(x-3)^2+(x+1)^2}\\
&=\frac{x^2-2x-3}{2\big((x-3)^2+(x+1)^2\big)}\\
&=\frac{(x-3)(x+1)}{2\big((x-3)^2+(x+1)^2\big)}
\end{align*}$$
is negative throughout the interval $0<x<3$, so $f$ is decreasing on that interval; since $f(0)<0$, it must be the case that $f(x)<0$ for all $x\in(0,3)$. Similarly, $f\,'(x)>0$ when $x<-1$, so $f$ is increasing on the interval $(\leftarrow,-1)$, and $f(-1)=0$, so $f(x)<0$ for all $x<-1$. To summarize,
$$f(x)\text{ is }\begin{cases}
\text{positive}&\text{when }x>3\\
\text{negative}&\text{when }x<3\;.
\end{cases}$$
I don’t at the moment see any simple way to avoid using some calculus here. You can rewrite
$$\frac{x+1}{x-3}=1+\frac4{x-3}$$
to see easily that $\frac{x+1}{x-3}<1$ when $x<3$, so $0<\arctan\frac{x+1}{x-3}<\frac{\pi}4$, and therefore $f(x)<0$ when $x\le-\pi$, but that still leaves the intervals $(-\pi,-1)$ and $(0,3)$ to be dealt with.
