Domain and signum of the following function

**dttah** · November 26th 2011, 12:58 AM

Hello, I was having some trouble with this function here...

$\frac {log{|x^2-x+2|}}{\sqrt x}$

Domain:
We have an abs in the logarithm so all we need to do is the following:

$x^2-x+2 \ne 0$
$x >= 0$
$\sqrt x \ne 0$

The last two can be merged in... $x > 0$
The first one has no solution, means that I have no problem up there, so the domain is:

$(0;+ \infty)$

Now I have to study the signum...

$\frac {log{|x^2-x+2|}}{\sqrt x} >= 0$

I was used to distinguish the two cases and study two separate functions when I have the absolute value. But the stuff inside the absolute value is always positive, so the function with or without the absolute value is pretty much the same thing? So studying the function WITHOUT the abs, is the same thing? Thanks !

**Ackbeet** · November 26th 2011, 03:19 AM

I agree with your domain. As for the signum, what you must study is the two following cases:

$0<|x^{2}-x+2|<1,$ and

$1\le|x^{2}-x+2|.$

The logarithm function is negative in the first case, and non-negative in the second. When do these cases occur? That is, for which values of $x$ do these cases occur (if they occur at all)?

**dttah** · November 26th 2011, 03:42 AM

Well, looks to me that it's never in between 0 and 1 and it is always >= 1.

**Ackbeet** · November 26th 2011, 04:04 AM

Originally Posted by dttah

Well, looks to me that it's never in between 0 and 1 and it is always >= 1.

Yes. In fact,

$x^{2}-x+2=x^{2}-x+\frac{1}{4}-\frac{1}{4}+2$

$=\left(x-\frac{1}{2}\right)^{\!\!2}+\frac{7}{4}$

$\ge\frac{7}{4} \quad \forall x\in\mathbb{R}.$

Conclusion?

**dttah** · November 26th 2011, 05:16 AM

That's always positive so the absolute value is well.. useless? :P

**Ackbeet** · November 26th 2011, 07:13 AM

Originally Posted by dttah

That's always positive so the absolute value is well.. useless? :P

Well, it is, but I was talking about what conclusion you can achieve towards solving the problem. What can you conclude about the signum of the function?

**dttah** · November 26th 2011, 10:45 AM

always positive!

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