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Inverse of a binomial

  1. Dec 25, 2004 #1
    the inverse of 1/(2x+6) x cannot=-3

    is -3 + 1/2x ? Is this correct?
  2. jcsd
  3. Dec 25, 2004 #2
    oh, x =\= 0 ..
  4. Dec 25, 2004 #3
    can someone show me how to check my self my using composites

    g(f(x))=x and f(g(x))=x then both are inverse of each other.
  5. Dec 25, 2004 #4
  6. Dec 25, 2004 #5
    I need more than that I dont need a hint I need to see how its done because if my answer is right then how come I dont know how to write out the composite function so that f(g(x)) and g(f(x)) both equal x?

    Can someone please show me how its done?

    I know how to do it im able to do it for f(x)=x^2 and g(x)=x+1 but in my question the fractions are throwing me off I dont know how to write it out, someone plz help me!!!! PLEASE!!! :uhh:
    Last edited: Dec 25, 2004
  7. Dec 26, 2004 #6


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    Okay,i'll be a nice guy... :tongue2:
    [tex] y=\frac{1}{2x+6}\Rightarrow 2x+6=\frac{1}{y}\Rightarrow x=\frac{1}{2y}-3 [/tex]
    So the function and the inverses are:
    [tex] f(x)=\frac{1}{2x+6};f^{-1}(x)=\frac{1}{2x}-3 [/tex]
    U wan to compute 2 functions:
    [tex] f(f^{-1}(x))=...??;f^{-1}(f(x))=...??[/tex]
    I'll take the first and leave you with the second.
    [tex] f(f^{-1}(x))=\frac{1}{2f^{-1}(x)+6}=\frac{1}{2(\frac{1}{2x}-3)+6}=
    \frac{1}{\frac{1}{x}-6+6}=x [/tex]

    I hope u saw the pattern and you won't have any trouble with the second.

  8. Dec 26, 2004 #7
    Yeesh. I saw the title of this thread and was very confused for a second. lol
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