anonymous
  • anonymous
Another function question!
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
anonymous
  • anonymous
so
anonymous
  • anonymous
goodness my internet is slow hold on sorry
anonymous
  • anonymous
|dw:1378204282177:dw|

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anonymous
  • 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
  • anonymous
what is your question..
anonymous
  • anonymous
Oh just what is the domain. Sorry about that.
hartnn
  • hartnn
so, what vales can x not take ? (hint: denominator = 0)
anonymous
  • anonymous
|dw:1378204623747:dw|
hartnn
  • hartnn
you want this also in set builder notation ?
anonymous
  • anonymous
Yes, and I'm still trying to comprehend why and what I'm doing.
hartnn
  • 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
  • anonymous
OH okay Wow that was obvious! Sorry. Now, in set builder notation would I write both 3 and -3 cannot equal x?
hartnn
  • 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
  • anonymous
\[ D=R-\{-3,3\}=(-\infty,-3)\cup(-3,3)\cup(3,\infty) \]
hartnn
  • hartnn
hmm, shorter version....is R-{3,-3} also considered as set builder notation?
anonymous
  • anonymous
normally it is used but I could not put \ sign between them|dw:1378205844148:dw|
anonymous
  • anonymous
I guess it is used for interval notation, I did not notice we should write it as set builder notation..
hartnn
  • hartnn
thats what i thought... :)
hartnn
  • hartnn
@shorty7096 doubts ?
anonymous
  • anonymous
hmmm No I'm just trying to apply all of this to my next problem I'm trying to tackle
hartnn
  • hartnn
glad to hear that! :)
anonymous
  • anonymous
\[f(x)\=\ sqrt{x \^\{\2\}\ \-\8x+\1\2\}\\]
anonymous
  • anonymous
|dw:1378206960853:dw|
anonymous
  • anonymous
I think I'm stuck at that point I'm not sure What to do?
hartnn
  • hartnn
\(x^2-8x+12 = (x-6)(x-2)\) isn't it ?
anonymous
  • 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
  • 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
  • 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
  • anonymous
oops the top is supposed to be x+1 sorry
anonymous
  • anonymous
is it just x cannot equal 2? is it just more complicated than it seems?
hartnn
  • hartnn
did u get domain for (x-6)(x-2) \(\ge \)0 ?
anonymous
  • anonymous
yes I did thank you
hartnn
  • 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
  • anonymous
hmmm okay I think I got it now! Thanks very much for your help!
hartnn
  • hartnn
good! welcome ^_^

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