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Math Help - function inverse

  1. #1
    Senior Member furor celtica's Avatar
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    function inverse

    f(x) = x^2 - 8x
    find f^-1 (x)
    i know this must be very simple but i have no idea what to do here. can somebody just show me how it is done? i am really at a loss of what to do, especially since i usually have no problem with inverse functions. i just use f(x)= y and solve for x, then replace y with x.
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  2. #2
    Moo
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    Hello,

    x=y^2-8y

    And you want y with respect to x.

    y^2-8y-x=0

    Solve this quadratic equation, treating x as a constant.

    Be aware that the range of x is particular, because if the discriminant is negative, there is no real solution.
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  3. #3
    Super Member Gamma's Avatar
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    Tricky

    y=x^2-8x \Rightarrow x^2-8x-y=0 know how to solve for x now?

    Spoiler:

    x=\frac{8\pm \sqrt{64+4y}}{2}=\frac{8\pm \sqrt{4(16+y}}{2}=4\pm \sqrt{16+y}

    Just need a function to work as the inverse, so might as well take the positive one and check it to make sure it works. The inverse is f^{-1}(x)=4+\sqrt{16+x}.

    f(4+\sqrt{16+x})=(4+\sqrt{16+x})^2-8(4+\sqrt{16+x})
    =16+8\sqrt{16+x}+16+x-32-8\sqrt{16+x}=x

    Check to make sure it is a two sided inverse
    f^{-1}(f(x))=4+\sqrt{16+(x^2-8x)}=4+\sqrt{(x-4)^2}=4+(x-4)=x

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  4. #4
    Senior Member furor celtica's Avatar
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    i forgot to mention that it stated x<=4 (for f(x))
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  5. #5
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    Quote Originally Posted by furor celtica View Post
    i forgot to mention that it stated x<=4 (for f(x))
    A very good thing to have said initially because the function you gave does not have an inverse! Solving the equation y= f(x) for x gives two different solutions depending on whether you take the positive or negative square root. But those two solutions "meet" at x= 4. That is, one sign gives x>4, the other <4.
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  6. #6
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    Quote Originally Posted by Moo View Post
    Hello,

    x=y^2-8y

    And you want y with respect to x.

    y^2-8y-x=0

    Solve this quadratic equation, treating x as a constant.

    Be aware that the range of x is particular, because if the discriminant is negative, there is no real solution.

    Can some1 explain to me how to solve this equation? As it seems there is 2 constants?
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  7. #7
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    Quote Originally Posted by furor celtica View Post
    f(x) = x^2 - 8x, find f^-1 (x)
    ... i forgot to mention that it stated x<=4 (for f(x))
    y = x^2-8x

    x = y^2-8y

    use the method of completing the square ...

    x+16 = y^2-8y+16

    x+16 = (y - 4)^2

    \pm \sqrt{x+16} = y-4

    y = 4 \pm \sqrt{x+16}

    since x \le 4 for f(x) , f^{-1}(x) \le 4 ...

    f^{-1}(x) = 4 - \sqrt{x+16}
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