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Find the domain of the following function:

$$g(x) = \ln(x^2+3x+2)$$

Here's my approach:

$$g(x) = \ln(x^2+3x+2)$$

$$g(x) = \ln(x+2)(x+1)$$

$$g(x) = \ln(x+2)+\ln(x+1)$$

Therefore,

$$x+1>0$$

$$x>-1$$

And,

$$x+2 > 0$$ $$x>-2$$

Both values on both sides of both inequalities are correct, however, in the answer, the sign is flipped for the second inequality. It will become apparent that the sign must be flipped (in the second inequality) when some numbers are inputted into the original function, however, I cannot seem to spot an error in my working out.

Thanks in advance, any help will be greatly appreciated.

GoodChessPlayer
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    As a note, remember that you can only split logarithmic functions when you are sure that the function is always defined. – Gummy bears Aug 16 '15 at 13:12

2 Answers2

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The problem is, that you want $$ (x^2+3x+2)=(x+2)(x+1) > 0 $$ but this means, that either $(x+2)>0\wedge(x+1)>0$ or $(x+2)<0\wedge(x+1)<0$. Well and the second case $$ (x+2)<0\wedge(x+1)<0\Leftrightarrow x<-2 \tag 1 $$ and in the first case $$ (x+2)>0\wedge(x+1)>0\Leftrightarrow x >-1 \tag 2 $$ so you actually just missed to check the $(1)$ case.

user190080
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Wrong, the condition is $(x^2 + 3x + 2) > 0$. So, $x \in (-\infty, -2) \cup (-1, \infty)$.

Björn Friedrich
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