Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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.

share|improve this question
    
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
add comment

2 Answers

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?

share|improve this answer
    
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
show 1 more comment

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 :

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

enter image description here

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.