# 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.

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I don't understand your question: why is it necessary to use a $>$ sign? What do you mean by sign of the function? – Aru Ray Feb 3 '12 at 15:37
If I understand correctly, unNaturhal is trying to determine the sign of y given x. – Tanner Swett Feb 3 '12 at 15:47
Ehm... When you set up a "false system" ( || x > 0), and you put the solutions on the numbers line, with a "+" where the condition is verified and "-" where is not. Something like this... I don't know the name in english :/ – Overflowh Feb 3 '12 at 15:50

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.

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Ok, I think to have understood. But is not clear why you have studied the first derived; I use to study it to define the monotonicity. However, since you said that "the algebraic sign of $\arctan u$ is the same as the sign of $u$", I supposed that I could study just the sign of the function without the $\arctan$, in this way: $\frac{x+1}{x-3}+\frac{x}{4}>0$, and the result is just positive when $x>3$. It's correct this method? P.S.: if you want, I could give you a photo of what I did. – Overflowh Feb 4 '12 at 11:57
@unNaturhal: No, you can’t study $\frac{x+1}{x-3}=\frac{x}4$ instead: the signs of $\frac{x+1}{x-3}$ and $\arctan\frac{x+1}{x-3}$ are the same, but the sizes aren’t, so the signs of $\frac{x+1}{x-3}=\frac{x}4$ and $\arctan\frac{x+1}{x-3}=\frac{x}4$ won’t necessarily be the same. I used the first derivative to tell where the function was increasing and where it was decreasing, because I know that if it’s negatice and decreasing, for instance, then it stays negative. – Brian M. Scott Feb 4 '12 at 17:21
Ok, understood. You had to use a trick to find the positivity. However, this image is what I made (but as you said I'm wrong): i39.tinypic.com/20afluq.jpg Thanks for the help, but I don't think that I'd be able to do it again. – Overflowh Feb 4 '12 at 18:09

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.

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How I can obtain the points where the equation is true? Studing the domain? I studied it, and the function is defined $\forall x \in\mathbb{R}-\left\{3\right\}$. Is this? – Overflowh Feb 3 '12 at 16:00