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$$\frac{x^2+8x+12}{x^2+5x+6}>0$$ First of all while solving inequalities I need to check domain so in this case $$x^2+5x+6\neq0$$ $$x\neq-2,\ x\neq-3$$ Later on $$\frac{(x+6)(x+2)}{(x+3)(x+2)}>0$$ Then get critical values draw number line and get $$x\in(-\infty;-6)\cup(-3;-2)\cup(-2;+\infty)$$ However according to wolframalpha $x=-2$ is included as an answer.

So am I wrong or wolframalpha is wrong?

Also I checked $\frac{x}{x}=1$ and wolframalpha also includes $x=0$ but once again I think it's incorrect?

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Well, there is a removable singularity at $x=-2$. –  copper.hat Jan 2 '13 at 1:47
    
Try solving $x/x > 0$ with Wolfram. However, $1/x > 0$ produces the correct answer. –  copper.hat Jan 2 '13 at 1:51
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I am inclined to agree with Wolfram Alpha. Your expression "really wants" to be $\frac{x+6}{x+3}$, and perhaps assumed its more convoluted form by accident. With trigonometric identities, one tends to take this poin of view. For example, there is a standard identity that expresses $\sum_{k=1}^n \cos k\theta$ as a fraction with $\sin(\theta/2)$ in the denominator. The identity is technically incorrect when $\sin(\theta/2)=0$, but it is standard not to worry about it. –  André Nicolas Jan 2 '13 at 1:58
    
@copper.hat ok so it's that way probably because wolfram simplifies expression at the beginning, but why it does so, if it's not always correct? why it doesnt check for a domain first? –  Templar Jan 2 '13 at 2:01
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I am a little surprised that Wolfram doesn't indicate that it has removed the removable singularities first. –  copper.hat Jan 2 '13 at 2:05

4 Answers 4

As it seems you are simply solving an inequality, and entered it as such, you are correct: the expression is not defined at $-2$.

Why Wolfram Alpha omitted $\,-3\,$ from the solutions, but included $\,-2\,$, seems inconsistent to me, as you note! It seems that if Wolfram is going to at least be consistent...either both values should be omitted, or both included.

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Wolfram alpha probably simplified the expression before it calculated the solutions. You are right. It is worth mentioning, however, that there is a hole at $x=-2$, and the limit as $x\rightarrow -2$ satisfies the inequality.

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so is it a bug? can't believe noone noticed it for such long time –  Templar Jan 2 '13 at 1:49
    
Yeah, I guess it must be. Perhaps alpha is programmed to include limit points in its solution sets automatically - your $x/x$ example suggests the latter. –  Alexander Gruber Jan 2 '13 at 1:54
    
@AlexanderGruber This is probably not the explanation. Try solving $1/x^2>0$. –  mrf Jan 2 '13 at 8:15
    
@Templar: can't believe noone noticed it for such long time... An awful lot of people did notice it, and many other features of W|A, for a very long time. In fact these are quite well known. The trouble begins when people consider W|A as something it is not. –  Did Jan 2 '13 at 10:03

The function as written is certainly not defined at $x= -2,$ but it does have a removable singularity there. The only way to extend the definition (note the word extend) of the function so that it becomes continuous at $x = -2$ is to define the value at $-2$ to be $4$. Then the new function is identically equal to $\frac{x+6}{x+3}$ except at $x = -3,$ where it is not defined (and worse, the singularity at $x = -3$ is not removable).

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No, your expression is undefined at $x = -2$. Are you taking a limit from the left or right???

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why wolframalpha gives -2 as an answer then? –  Templar Jan 2 '13 at 1:43
    
It's kinda hairsplittery. I think that the expression is undefined there. Wolfram is eliding the removable singularity. I think it's a matter of taste, but I say $x \not= -2$. –  ncmathsadist Jan 2 '13 at 1:45
    
Wolfram is wrong, but justifiably so. –  copper.hat Jan 2 '13 at 1:48
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Wolfram is not perfectt. –  copper.hat Jan 2 '13 at 1:53

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