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Math Help - Inverse of a a function

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    Senior Member sakonpure6's Avatar
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    Inverse of a a function

    Hi, how do you find the inverse of this function: f(x) = \frac{15}{2x^2 + 11x +5} ?
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    Re: Inverse of a a function

    Hello, sakonpure6!

    How do you find the inverse of this function? . f(x) = \frac{15}{2x^2 + 11x +5}

    Use the prescribed procedure.
    [However, the result will not be a function.]


    We have: . y \;=\;\frac{15}{2x^2+11x+5}

    "Switch" variables: . x \;=\;\frac{15}{2y^2 + 11y + 5}

    Solve for y\!:

    . . x(2y^2+11y + 5) \:=\:15 \quad\Rightarrow\quad 2xy^2 + 11xy + 5x - 15 \:=\:0


    Quadratic Formula: . y \;=\;\dfrac{-11x \pm\sqrt{(11x)^2-4(2x)(5x-15)}}{2(2x)}

    . . . . . . . . . . . . . . . y \;=\;\frac{-11x \pm \sqrt{121x^2 - 40x^2 + 120x}}{4x}

    . . . . . . . . . . . . . . . y \;=\;\frac{-11x \pm\sqrt{81x^2 + 120x}}{4x}


    Therefore: . f^{-1}(x) \;=\;\frac{-11x \pm \sqrt{81x^2+120x}}{4x}
    Thanks from sakonpure6
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    Senior Member sakonpure6's Avatar
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    Re: Inverse of a a function

    Oh I see.. i got stuck on the part with the quadratic. Thank you!
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    Re: Inverse of a a function

    Quote Originally Posted by sakonpure6 View Post
    Hi, how do you find the inverse of this function: f(x) = \frac{15}{2x^2 + 11x +5} ?
    Did you plot it? You'll notice that f is not bijective over it's entire domain so in order to establish an inverse you will have to restrict the domain of f to an interval, or union of intervals, where it is.

    Once you've done that you can solve the quadratic equation y=f(x) for x in terms of y. This gives you x=g(y). Replace every occurrence of y in g(y) with x to obtain g(x).

    f^{-1}(x)=g(x) for x in the restricted domain you chose.
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