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Inverse of f(x)=x^2+6x

  1. Feb 13, 2007 #1
    Hi
    I need to find the inverse of the function [tex]f\left( x \right) = x^2 + 6x\[/tex] with the domain of [tex]x \ge - 3[/tex] and [tex]x \le - 3[/tex].

    What i have done so far:
    [tex]
    \begin{array}{c}
    f\left( x \right) = x^2 + 6x \\
    x = y^2 + 6x \\
    \end{array}
    [/tex]

    and then from here I need to put the 2nd function in terms of y, however I do not know how to because of the function having the variable y twice. I am not asking for the answer of the problem, only suggestions on which way to approach the situation.

    Many thanks
    unique_pavadrin
     
  2. jcsd
  3. Feb 13, 2007 #2

    AlephZero

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    Shouldn't your last equation be
    [tex]y = x^2 + 6x[/tex]
    and then you want to get equation x = some function of y?

    [tex]x^2 + 6x - y = 0[/tex] is just a quadratic equation in x.
     
  4. Feb 13, 2007 #3

    Gib Z

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    No No you don't just swap the x and y in one case, it must be all the cases.

    so the [tex]f(x)=y=x^2 + 6x[/tex] Then [tex]x=y^2 + 6y[/tex] is your inverse.

    I can't see any obvious ways to get y on its own from there.
     
  5. Feb 13, 2007 #4

    dextercioby

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    ???????????????????????????
     
  6. Feb 13, 2007 #5

    Gib Z

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    >.< Well i thought what i said was correct, but if dexter says it isn't, then im wrong for sure :). BUt yea im pretty sure when you want the inverse, you change ALL the x's to ys and vice versa. in the OP the last bit of TEX was incorrect.
     
  7. Feb 13, 2007 #6

    HallsofIvy

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    No, Gib Z, dextercioby's ??????????????????????????? wasn't directed toward your saying that to find the inverse of y= x2+ 6x you start by changing it to x=y2+ 6y, it was toward "I can't see any obvious ways to get y on its own from there."

    That's a quadratic function of y. You can always solve it by using the ____________.
     
  8. Feb 14, 2007 #7

    Gib Z

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    Ahh I don't understand how the quadratic forumula would work >.<" Unless you mean take x to the other side and pretend its a constant? ....x=y(y+6)
     
  9. Feb 14, 2007 #8

    dextercioby

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    Exactly. y^2 +6y-x=0 is a quadratic and should be solved for y(x) which is the inverse of the initial function, once one takes care of the domain issues.
     
  10. Feb 14, 2007 #9

    Gib Z

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    ahh ok thanks for telling me that :) Didnt realise x could be treated like that. ty
     
  11. Feb 16, 2007 #10
    Thanks for all the post people, your help is greatly appreciated. Sorry, wasn't able to get online until now due to isp problems. In my 1st post, the 2nd equation was a typo, it should have read x=y^2+6y.

    So therefore the inverse of the function between the domain of [tex]x \ge - 3\,\,\,\,\,and\,\,\,\,\,x \le - 3[/tex] will be:

    [tex]f^{ - 1} \left( x \right) = - 3 \pm \sqrt {36 - 4x}[/tex]

    ?
    many thanks
     
  12. Feb 16, 2007 #11

    Gib Z

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    The quadratic equation yields:

    [tex]y(x)=\frac{-6 \pm \sqrt {36-4x}}{2}[/tex], which is definitely not what you have.
     
  13. Feb 16, 2007 #12

    cristo

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    What does this domain mean? Is there a typo here; as it stands, this is equal to R
     
  14. Feb 16, 2007 #13

    dextercioby

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    I get

    [tex] y(x)=-3\pm \sqrt{9+x} [/tex]
     
  15. Feb 16, 2007 #14
    ah i see where i have gone wrong.
    thanks for the help, and sorry for any inconvenience
    unique_pavadrin
     
  16. Feb 16, 2007 #15

    Gib Z

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    Well yea, thats after simplifications >.< I was still correct, technically :P.
     
  17. Feb 16, 2007 #16

    dextercioby

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    Up to the minus sign under the square root, yeah...
     
  18. Feb 16, 2007 #17

    Gib Z

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    >.< O jesus Christ shoot me
     
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