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    Hello there,

    Could someone tell me if the inverse for the funtion:-

    f(x) = 1 - 2/(3x^2+2) is correct please.

    (x)(3y^2+2) = 1 - 2
    3y^2+2 = -1/x
    3y^2 = (-1/x) - 2
    y^2 = [(-1/x) - 2]/3 If this is correct so far can I simplify it any further before square rooting? Also how do I chose between the + or - answer for the inverse?

    Thanks! :0
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    For the +ve and -ve answers it would be both.
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    (Original post by apo1324)
    Hello there,

    Could someone tell me if the inverse for the funtion:-

    f(x) = 1 - 2/(3x^2+2) is correct please.

    (x)(3y^2+2) = 1 - 2
    3y^2+2 = -1/x
    3y^2 = (-1/x) - 2
    y^2 = [(-1/x) - 2]/3 If this is correct so far can I simplify it any further before square rooting? Also how do I chose between the + or - answer for the inverse?

    Thanks! :0
    If you think f(x)=1-\frac{2}{3x^2+2} then your work is wrong.
    Let f(x)=y you need f^{-1}(x)
    to get get the inverse function
    change x with y and will get y=f^(-1)(x)

    x=1-\frac{2}{3y^2+2}
    \frac{2}{3y^2+2}=1-x
    \frac{3y^2+2}{2}=\frac{1}{1-x}
    and so on
 
 
 
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