# I'm stuck with this problem, who can help me please ?

• Jan 5th 2008, 02:16 PM
supercheddarcheese
I'm stuck with this problem, who can help me please ?
Consider the function f(x)= log1/2 ( x^2-2x-3) and determine :

( A ) Its domain.
( B ) The intervals where it is strictly increasing.
( C ) The intervals where it is strictly decreasing.

I think that the domain is (-infinity,-1) the parenthesis after -1 is supposed to be a bracket. I have some trouble finding the range...and the intervals?:S
• Jan 5th 2008, 03:40 PM
Jhevon
Quote:

Originally Posted by supercheddarcheese
Consider the function f(x)= log1/2 ( x^2-2x-3) and determine :

( A ) Its domain.
( B ) The intervals where it is strictly increasing.
( C ) The intervals where it is strictly decreasing.

I think that the domain is (-infinity,-1) the parenthesis after -1 is supposed to be a bracket. I have some trouble finding the range...and the intervals?:S

clarify: is it $\log \frac 12(x^2 - 2x - 3)$ ?

you were not asked for the range, why are you looking for it?

the function is strictly increasing where the derivative is positive, it is strictly decreasing where the derivative is negative
• Jan 5th 2008, 03:55 PM
mr fantastic
Quote:

Originally Posted by Jhevon
clarify: is it $\log \frac 12(x^2 - 2x - 3)$ ?

you were not asked for the range, why are you looking for it?

the function is strictly increasing where the derivative is positive, it is strictly decreasing where the derivative is negative

A fitting and ironic reply as it happens - see here
• Jan 5th 2008, 04:01 PM
Jhevon
Quote:

Originally Posted by mr fantastic
A fitting and ironic reply as it happens - see here

indeed.

Hint: the domain is all real x such that $x^2 - 2x - 3 > 0$. (because the domain of log(x) is all x > 0, consult the defintion of a logarithm to see why). can you continue?
• Jan 5th 2008, 04:04 PM
supercheddarcheese
Quote:

Originally Posted by mr fantastic
A fitting and ironic reply as it happens - see here

Ohhh I went to the link, I know the user who posted the problems, he's part of my class in which we were asked to research this problems online... I tried these problems, but don't bother answering rlarach, he's a lazy and annoying student...

I'm almost done answering these by myself, I'm just missing around 3 or 4problems. Most of them were fairly easy but the rest I just don't know what to do in, including one which , at least I think so, was misswritten by our teacher.

As for asking for the range, it was automatic from me because my teacher usually asks us to find the range.

Anyways, thanks a lot to those of you who bothered to look in my post and help me in trying to find it :p
• Jan 5th 2008, 04:10 PM
supercheddarcheese
Quote:

Originally Posted by Jhevon
indeed.

Hint: the domain is all real x such that $x^2 - 2x - 3 > 0$. (because the domain of log(x) is all x > 0, consult the defintion of a logarithm to see why). can you continue?

Ohhhh so its from (negative infinity, -1.24) U (3.24, infinity) ?
• Jan 5th 2008, 04:13 PM
wingless
For the logarithm to be valid, $x^2 - 2x -3 > 0$. Let's find the domain from this inequality.

$\underbrace{(x-3)}_{x=3}\underbrace{(x+1)}_{x=-1} > 0$

$\begin{array}{cccccc}&-1&&3&\\\hline+&|&-&|&+\end{array}$

Then, $x = (-\infty,-1) \cup (3,\infty)$

I tried to explain it as clearly as I can. Do you know how to solve inequalities like this?
• Jan 5th 2008, 04:20 PM
supercheddarcheese
Quote:

Originally Posted by wingless
For the logarithm to be valid, $x^2 - 2x -3 > 0$. Let's find the domain from this inequality.

$\underbrace{(x-3)}_{x=3}\underbrace{(x+1)}_{x=-1} > 0$

$\begin{array}{cccccc}&-1&&3&\\\hline+&|&-&|&+\end{array}$

Then, $x = (-\infty,-1) \cup (3,\infty)$

I tried to explain it as clearly as I can. Do you know how to solve inequalities like this?

Oh my bad I saw the greater than as a lessthan by mistake.. but what I did was graph the fnction on the TI-89 calculator and what confused me was the empty void betwen the points negative one and 3.

why does it have to be greater than 0 though? if it is 0, then to the power of 0 anything is 1...and if it is negative then it would be one over .5 to the power of the positive (previously negative) number.

Im lost
• Jan 5th 2008, 04:26 PM
Jhevon
Quote:

Originally Posted by supercheddarcheese
... but don't bother answering rlarach, he's a lazy and annoying student...

that's not nice

Quote:

Originally Posted by supercheddarcheese
why does it have to be greater than 0 though? if it is 0, then to the power of 0 anything is 1...and if it is negative then it would be one over .5 to the power of the positive (previously negative) number.

Im lost

the logarithm function is only defined if what is being logged is greater than zero. the log graph had a vertical asymptote at zero and does not exist at all for negative numbers. this is because, by definition, the logarithm of a number to a given base is the power to which the base most be raised to give the number. since we usually consider positive bases (greater than 2), we know that nothing less than or equal to zero can work, since there is no number you can raise a positive number to and get zero, or a negative number.
• Jan 5th 2008, 04:26 PM
wingless
Oh, you're confusing the logarithm and the exponent.

$log_{a}{0} = b$

$0 = a^b$

As $a > 0$ and $a \neq 1$, there's no $b$ value that can make the expression $0$.
• Jan 5th 2008, 04:31 PM
supercheddarcheese
Ohh so you are just considering the x2 minus 2x minus 3 and factoring it. then you restrict it where it is 0 and less because it is impossible to get that for a power equation? ok, I get it now. So the domain is from negative infinity to negative one with a bracket, union 3 with a bracket to infinity?
• Jan 5th 2008, 04:53 PM
wingless
Well, factoring is not a part of finding the domain, we just needed it to solve the inequality.

No, it's not a bracket, they're all parentheses. -1 and 3 are not in the domain. Try to plug them in the expression, they will make it 0. And it must be greater than 0.

You must learn solving linear and nonlinear inequalities as soon as possible. (Nod)