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A little inverse help cube root

  1. Dec 25, 2004 #1
    find the inverse of f(x)=3sqrt(x+2) -7 x>=-2

    I got this far (x+7)=3sqrt(y+2)

    I dont know how to get rid of the cube root someone help plz :cry:
  2. jcsd
  3. Dec 25, 2004 #2
    just cube both sides of the equation....
  4. Dec 25, 2004 #3
    therefore is the inverse of my equation f^-1(x)=(x+7)^(3)-2 ? :yuck:
  5. Dec 25, 2004 #4
  6. Dec 25, 2004 #5
    it doesnt look anything like the inverse when plugged into the graphing calculator into y= and compared to DrawInv
  7. Dec 25, 2004 #6
    I tried again now my answer is

    [(x+7)^3] -2

    Is this correct? OR totally wrong? :uhh:
  8. Dec 25, 2004 #7
    Is that 3 * cube root of (x + 2)

    or just cube root of (x + 2)??
  9. Dec 26, 2004 #8


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    This is really curious.She's posted 196 times,yet she hasn't had the time to learn how to edit the formulas in 'tex'.
    If your initial function was:
    [tex] f_{1}(x)=\sqrt[3]{x+2}-7 [/tex]
    ,then the inverse is found simply:
    [tex] f_{1}^{-1}(x)=(x+7)^{3}-2 [/tex].
    If your initial function was:
    [tex] f_{2}(x)=3\sqrt[3]{x+2}-7 [/tex]
    ,then the inverse is easily found to be
    [tex] f_{2}^{-1}(x)=\frac{(x+7)^{3}}{27}-2[/tex]
    The curious part is that i don't have any idea "which is which",as you posted both answers as 'correct'.Obviously only one is.
    In the end i'd like to ask you in a very polite way to sacrifice some of your spare time and read the .pdf document which explains you how you can edit your mathematical formulas in 'tex'. :smile:


    PS.Isn't it a little weird to you as well that on Christmas Evening you chose mathematics as a 'date'?? :tongue2: :wink:
  10. Dec 26, 2004 #9


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    By the way- the notation "3 sqrt(x)" to mean "cube root" really steams me. It makes it look like you don't know the difference between "cube root" and "square root". Even if you refuse to use [itex]^3\sqrt{x}[/itex] or 3√(x), you could at least write "cbrt(x)" or "3rdrt(x)".
  11. Dec 27, 2004 #10
    Ur right I really dont have time to read the txt, but when I do have I time I will for sure. The question is 3 * the square root of (x+2) subtract 7
    (7 is not under the square root). And yes I do know the difference between cube root and square root.

    I got the previous answer because I was told to cube both sides of the equation to get rid of the square root, but I think I was supposed to square both sides of the equation to get rid of the square root.

    So is the answer
    (x+7)^(2) -2
    Last edited: Dec 27, 2004
  12. Dec 28, 2004 #11


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    Yes,this time it's the good answer. :smile: One request,though,pleeaeaeaeaeaeaeaeaeaeaeaeaeaeeaeaeaeaeeaeaeaeaeaeaseread that file which explains the tex editing language.U might be able to have an adjustable fraction line,not a bunch of minuses... :tongue2:

  13. Dec 28, 2004 #12


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    What's wrong with that? I never get any dates, either. Of course, in my case, the reason is obvious.
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