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Thread: Inverse of a a function

  1. #1
    Super Member sakonpure6's Avatar
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    Inverse of a a function

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

    Hello, sakonpure6!

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

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


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

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

    Solve for $\displaystyle y\!:$

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


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

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

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


    Therefore: .$\displaystyle f^{-1}(x) \;=\;\frac{-11x \pm \sqrt{81x^2+120x}}{4x}$
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    Super 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: $\displaystyle f(x) = \frac{15}{2x^2 + 11x +5}$ ?
    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).

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