I am an amateur when it comes to math. I am currently taking CAL 1 and have a question about one of my assignments. Any help is appreciated.

Let $g(x)=x^3-x^3-2x$ and $f(x) = \ln(g(x))$

I have to find the domain of $y = f(x)$ I've figured out that when I do $x^3-x^2-2x > 0$, I end up with $0, 2, -1$.

But I'm still not sure what the domain is.... maybe I'm close? maybe it's obvious? Any help much appreciated!!!

Thanks.

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When you define $g(x)$ you have the $x^3$ term appearing twice. I can't edit since it's only one character. –  ae0709 Dec 24 '11 at 21:14

You need to figure out under what conditions on $x$ is $x^3 - x^2 - 2x >0$. Factoring the lhs we have:

$x (x^2-x-2)$

which then simplifies to:

$x (x-2) (x+1)$.

Thus, you need to identify the set of $x$ for which:

$x (x-2) (x+1) > 0$

Can you take it from here?

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yes, I got there, and I end up with 3 numbers, but I still do not fully understand what the domain is... –  Sam Nov 24 '11 at 17:07
No, you end up with three numbers $0$, $2$ and $-1$ only if you insist that the $>$ is a $=$ sign. What do you know must be true about three numbers $a$, $b$, $c$ if the product of these three numbers is greater than $0$ (i.e., $a b c >0$)? –  tards Nov 24 '11 at 17:10
they must all be positive numbers? –  Sam Nov 24 '11 at 17:17
What about $-1 \cdot -2 \cdot 3$? That is also positive. So,... –  tards Nov 24 '11 at 17:24
hmmmm well either an even number of negatives and any number of positives, or all positives –  Sam Nov 24 '11 at 17:25

If we define $P(x)=x(x-2)(x+1)$ , then we can find solution for $P(x)>0$ as it is shown on picture below. So domain is :

$x\in (-1,0) \cup (2,+\infty)$

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