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Math Help - Need help finding an inverse function

  1. #1
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    Need help finding an inverse function

    I have a function that describes a ball being thrown at an initial velocity of 192 ft/s, given by:

    f(t) = 192t - 16t^2

    I need to find an inverse function that yields the time 't' when the ball is at a height 'h' as the object travels upward:

    t = 6 - [sqrt(576 - h) / 4]

    Also, I need to find an inverse function that yields the time 't' when the ball is at height 'h' as the object travels downward:

    t = 6 + [sqrt(576 - h) / 4]

    The problem is that I can't figure out how to find these inverse functions from f(t) = 192t - 16t^2. I tried factoring, but didn't have any luck. Also, since this is a second order function, it's not one-to-one, which means that it shouldn't have an inverse, right (since it will fail the horizontal line test)? The two answers that were given above were what was listed, but I can't figure out how to get to the answer.


    Help would be greatly appreciated! Thanks!
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  2. #2
    MHF Contributor Siron's Avatar
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    Re: Need help finding an inverse function

    When the ball is at height h that means f(t)=h so the equation becomes:
    h=192t-16t^2
    \Leftrightarrow 16t^2-192t+h=0
    Calculating the discriminant gives:
    D=(-192)^2-4\cdot 16 \cdot h = 36 864 - 64h=64(576-h)

    So the two solutions are ...
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  3. #3
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    Re: Need help finding an inverse function

    Siron is referring to the "quadratic formula".

    Solutions to ax^2+ bx+ c= 0 are given by
    \frac{-b\pm\sqrt{b^2- 4ac}}{2a}.

    Your function, h= 192t- 16t^2 is the same as 16t^2- 192t+ h= 0 so a= 16, b= -192, c= h.
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  4. #4
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    Re: Need help finding an inverse function

    Or you can use the "completing the square" method.

    First make 16t^2-192t the subject:
    16t^2-192t = -h

    Factorise to make it easier:
    16(t^2-12t) = -h

    Completing the square gives:
    16((t-6)^2-36) = -h

    You should be able to take it from here...

    It's the same as HallofIvy's only different approach...
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