Another function question!

- anonymous

Another function question!

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- schrodinger

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- anonymous

so

- anonymous

goodness my internet is slow hold on sorry

- anonymous

|dw:1378204282177:dw|

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## More answers

- anonymous

goodness the equation button wasn't working so I had to do that. Ugh. Anyways, I'm getting suspicious that a lot of my answers are infinity and I feel like I'm not doing something right. With this one, I'm not sure hot to tackle it . I know the denominator can't equal zero.

- anonymous

what is your question..

- anonymous

Oh just what is the domain. Sorry about that.

- hartnn

so, what vales can x not take ?
(hint: denominator = 0)

- anonymous

|dw:1378204623747:dw|

- hartnn

you want this also in set builder notation ?

- anonymous

Yes, and I'm still trying to comprehend why and what I'm doing.

- hartnn

trying to find domain
means trying to find values, that x can take
for that, we find the values that 'x' cannot take.
here, x cannot take 3 and -3, got that part ?

- anonymous

OH okay Wow that was obvious! Sorry. Now, in set builder notation would I write both 3 and -3 cannot equal x?

- hartnn

yes, we just write it in a particular form
like
{ x| x \(\in\)R ^ x \(\ne 3\)^x \(\ne -3\)}
^ ---->AND
read as
domain is x, such that x belongs to Real number, AND x not =3, and x not= -3
makes sense ?

- anonymous

\[
D=R-\{-3,3\}=(-\infty,-3)\cup(-3,3)\cup(3,\infty)
\]

- hartnn

hmm, shorter version....is R-{3,-3} also considered as set builder notation?

- anonymous

normally it is used but I could not put \ sign between them|dw:1378205844148:dw|

- anonymous

I guess it is used for interval notation, I did not notice we should write it as set builder notation..

- hartnn

thats what i thought... :)

- hartnn

@shorty7096 doubts ?

- anonymous

hmmm No I'm just trying to apply all of this to my next problem I'm trying to tackle

- hartnn

glad to hear that! :)

- anonymous

\[f(x)\=\ sqrt{x \^\{\2\}\ \-\8x+\1\2\}\\]

- anonymous

|dw:1378206960853:dw|

- anonymous

I think I'm stuck at that point I'm not sure What to do?

- hartnn

\(x^2-8x+12 = (x-6)(x-2)\)
isn't it ?

- anonymous

Oh the bottom is how far I got. Sorry the top is the original problem. I think I got it though. So it means x has to be greater or equal to both 2 and 6?

- hartnn

the first step was to factorize x^2-8x+12
and then equate it to\(\ge0\)
i just pointed out that you factorized incorrectly....try again?
to get (x-6)(x-2)

- anonymous

ohhh I see I get it. Man these questions just keep escalating when I'm barely getting a grasp on the easiest ones! This is my last problem! Thanks so much for the help and this one is a doozie for me. :C \[f(x)= \sqrt[3]{\frac{ x=1 }{ x ^{3} -8}}\]

- anonymous

oops
the top is supposed to be x+1 sorry

- anonymous

is it just x cannot equal 2? is it just more complicated than it seems?

- hartnn

did u get domain for (x-6)(x-2) \(\ge \)0
?

- anonymous

yes I did thank you

- hartnn

in your last problem, remember that the quantity under \(\sqrt[3].\)
must be non-negative
so ,also solve for (x+1)/(x^3-8) >= 0

- anonymous

hmmm okay I think I got it now! Thanks very much for your help!

- hartnn

good!
welcome ^_^

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