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Fanduekisses

  • one year ago

*****WHY U NO HELP ME??!!

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  1. Fanduekisses
    • one year ago
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    \[f(x)= \frac{ 1 }{ x^4 -1}\] and \[g(x) = \sqrt[6]{x}\]

  2. Fanduekisses
    • one year ago
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    \[f(g(x)) = \frac{ 1 }{ (\sqrt[6]{x})^4 }-1\]

  3. Fanduekisses
    • one year ago
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    The radical and exponent part is confusing, how will I go about that?

  4. Fanduekisses
    • one year ago
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    @ganeshie8

  5. Nnesha
    • one year ago
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    you should convert radical to an exponent form

  6. Nnesha
    • one year ago
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    \[\huge\rm \sqrt[n]{x^m}=x^\frac{ m }{ n }\]index become the denomiantor of the fraction

  7. Nnesha
    • one year ago
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    denominator *

  8. Nnesha
    • one year ago
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    \(\color{blue}{\text{Originally Posted by}}\) @Fanduekisses \[f(g(x)) = \frac{ 1 }{ (\sqrt[6]{x})^4 }-1\] \(\color{blue}{\text{End of Quote}}\) btw, -1 should be in the denominator

  9. Nnesha
    • one year ago
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    \[\huge\rm \frac{ 1 }{(\color{ReD}{ \sqrt[6]{x}} )^4} =\frac{ 1 }{ (\color{Red}{x^{??}})^4 }\]

  10. Fanduekisses
    • one year ago
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    \[\frac{ 1 }{ (x ^{1/6} )^4}\]

  11. Nnesha
    • one year ago
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    yes right now you can multiply 1/6 by 4 exponent rule \[\huge\rm (a^m)^x = a^{m \times x}\]

  12. Fanduekisses
    • one year ago
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    \[\frac{ 1 }{ x ^{2/3} }\]

  13. Nnesha
    • one year ago
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    don't forget the negative one

  14. Fanduekisses
    • one year ago
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    ok so the domain will be \[x \ge0 \] and

  15. Fanduekisses
    • one year ago
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    Wait how do I solve for x? \[x ^{2/3}= 1\]

  16. Fanduekisses
    • one year ago
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    The denominator part, to find the domain, what x cannot equal.

  17. Nnesha
    • one year ago
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    ugh sorry my laptop is acting up,slow net connection alright so whats the question ? find domain ?

  18. Fanduekisses
    • one year ago
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    yes of the composition f(g(x))

  19. Nnesha
    • one year ago
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    you can take cube both sides to get rid of the fraction

  20. Nnesha
    • one year ago
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    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

  21. Fanduekisses
    • one year ago
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    wait be right back

  22. Nnesha
    • one year ago
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    btw, you can apply difference of squres....

  23. mathmate
    • one year ago
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    \(\frac{ 1 }{ (\sqrt[6]{x})^4 }-1=\frac{1}{x^{2/3}}-1=x^{-2/3}-1=0\) It doesn't change the answer in this case. lol

  24. Fanduekisses
    • one year ago
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    So th\[x \neq1 ?\]e domain of f(g(x)) is \[x \ge0\] and

  25. Fanduekisses
    • one year ago
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    Sorry for the late reply, I had an urgency.

  26. Nnesha
    • one year ago
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    how many solutions would you get when u take square root of something ? and it's okay :=)

  27. Fanduekisses
    • one year ago
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    +- 1 hehe

  28. Fanduekisses
    • one year ago
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    :D got it now hehe thanks so much!

  29. Fanduekisses
    • one year ago
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    So the domain is x cannot equal +-1 and x has to be greater than or equal to 0 ?

  30. Nnesha
    • one year ago
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    to write in interval u can draw a number line |dw:1443037013961:dw|

  31. Nnesha
    • one year ago
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    |dw:1443037153659:dw|

  32. Fanduekisses
    • one year ago
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    \[(-\infty, -1) U (-1, 1)(1,\infty)\]

  33. Nnesha
    • one year ago
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    \[(-\infty, -1) U (-1, 1) \rm U(1,\infty)\]

  34. Nnesha
    • one year ago
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    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

  35. Fanduekisses
    • one year ago
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    I forgot the union lol

  36. Fanduekisses
    • one year ago
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    thanks

  37. Nnesha
    • one year ago
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    ye

  38. Nnesha
    • one year ago
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    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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