Fanduekisses
  • Fanduekisses
*****WHY U NO HELP ME??!!
Mathematics
chestercat
  • chestercat
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
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Fanduekisses
  • Fanduekisses
\[f(x)= \frac{ 1 }{ x^4 -1}\] and \[g(x) = \sqrt[6]{x}\]
Fanduekisses
  • Fanduekisses
\[f(g(x)) = \frac{ 1 }{ (\sqrt[6]{x})^4 }-1\]
Fanduekisses
  • Fanduekisses
The radical and exponent part is confusing, how will I go about that?

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Fanduekisses
  • Fanduekisses
Nnesha
  • Nnesha
you should convert radical to an exponent form
Nnesha
  • Nnesha
\[\huge\rm \sqrt[n]{x^m}=x^\frac{ m }{ n }\]index become the denomiantor of the fraction
Nnesha
  • Nnesha
denominator *
Nnesha
  • Nnesha
\(\color{blue}{\text{Originally Posted by}}\) @Fanduekisses \[f(g(x)) = \frac{ 1 }{ (\sqrt[6]{x})^4 }-1\] \(\color{blue}{\text{End of Quote}}\) btw, -1 should be in the denominator
Nnesha
  • Nnesha
\[\huge\rm \frac{ 1 }{(\color{ReD}{ \sqrt[6]{x}} )^4} =\frac{ 1 }{ (\color{Red}{x^{??}})^4 }\]
Fanduekisses
  • Fanduekisses
\[\frac{ 1 }{ (x ^{1/6} )^4}\]
Nnesha
  • Nnesha
yes right now you can multiply 1/6 by 4 exponent rule \[\huge\rm (a^m)^x = a^{m \times x}\]
Fanduekisses
  • Fanduekisses
\[\frac{ 1 }{ x ^{2/3} }\]
Nnesha
  • Nnesha
don't forget the negative one
Fanduekisses
  • Fanduekisses
ok so the domain will be \[x \ge0 \] and
Fanduekisses
  • Fanduekisses
Wait how do I solve for x? \[x ^{2/3}= 1\]
Fanduekisses
  • Fanduekisses
The denominator part, to find the domain, what x cannot equal.
Nnesha
  • Nnesha
ugh sorry my laptop is acting up,slow net connection alright so whats the question ? find domain ?
Fanduekisses
  • Fanduekisses
yes of the composition f(g(x))
Nnesha
  • Nnesha
you can take cube both sides to get rid of the fraction
Nnesha
  • Nnesha
like we rewrite x^2/3 into radical form \[\huge\rm \sqrt[3]{x^2}=1\] to cancel out cube root we should take cube sides just like we take square root to cancel out square
Fanduekisses
  • Fanduekisses
wait be right back
Nnesha
  • Nnesha
btw, you can apply difference of squres....
mathmate
  • mathmate
\(\frac{ 1 }{ (\sqrt[6]{x})^4 }-1=\frac{1}{x^{2/3}}-1=x^{-2/3}-1=0\) It doesn't change the answer in this case. lol
Fanduekisses
  • Fanduekisses
So th\[x \neq1 ?\]e domain of f(g(x)) is \[x \ge0\] and
Fanduekisses
  • Fanduekisses
Sorry for the late reply, I had an urgency.
Nnesha
  • Nnesha
how many solutions would you get when u take square root of something ? and it's okay :=)
Fanduekisses
  • Fanduekisses
+- 1 hehe
Fanduekisses
  • Fanduekisses
:D got it now hehe thanks so much!
Fanduekisses
  • Fanduekisses
So the domain is x cannot equal +-1 and x has to be greater than or equal to 0 ?
Nnesha
  • Nnesha
to write in interval u can draw a number line |dw:1443037013961:dw|
Nnesha
  • Nnesha
|dw:1443037153659:dw|
Fanduekisses
  • Fanduekisses
\[(-\infty, -1) U (-1, 1)(1,\infty)\]
Nnesha
  • Nnesha
\[(-\infty, -1) U (-1, 1) \rm U(1,\infty)\]
Nnesha
  • Nnesha
there is another way to test each number if it's satisfy or not in the original equation i don't think that's necessary for that question hm
Fanduekisses
  • Fanduekisses
I forgot the union lol
Fanduekisses
  • Fanduekisses
thanks
Nnesha
  • Nnesha
ye
Nnesha
  • Nnesha
np here is good notes to practice on these stuff http://math.arizona.edu/~rwilliams/math112-spring2012/Notes/Domain_of_a_Composition.pdf np that was a good review for me actlly :/

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