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Math Help - Reciprocal of a Quadratic Function

  1. #1
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    Reciprocal of a Quadratic Function

    For the function f(x)=1/(x^2-2x-15), determine the range. I got yER|y>0 or y<=-0.0625. However, the book's answer is different from mine. Is mine correct?
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  2. #2
    Super Member Quacky's Avatar
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    Re: Reciprocal of a Quadratic Function

    What is the book's answer? As far as I can see, y\leq{\frac{-1}{16} which is what you have.
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  3. #3
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    Re: Reciprocal of a Quadratic Function

    The textbook says yER|y is not equal to 0.
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  4. #4
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    Re: Reciprocal of a Quadratic Function

    Quote Originally Posted by Dragon08 View Post
    The textbook says yER|y is not equal to 0.
    Oh, of course! My mistake entirely. See the graph, it will help. I basically ignored the sections either side of the asymptotes. Consider, for example, x=6 or x=-4

    There is a maximum point at x=\frac{-1}{16}, but the only impossible value for the range is when y=0 because you can never obtain such a value from the function.

    Edit: Correction.
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  5. #5
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    Re: Reciprocal of a Quadratic Function

    oh okay, so yER|y>o or y<=-0.0625 is correct?
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  6. #6
    Super Member Quacky's Avatar
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    Re: Reciprocal of a Quadratic Function

    Quote Originally Posted by Dragon08 View Post
    oh okay, so yER|y>o or y<=-0.0625 is correct?
    No, Look at my example of x=-4

    h(-4)=\frac{1}{(-4)^2-2(-4)-15}
    =\frac{1}{16+8-15}
    =\frac{1}{9}

    You have found a maximum point in the graph. If you use the link I provided above, and find the point, you can clearly see that it is a maximum point. What I mean by "maximum point" is not "this is the highest possible value the function can obtain", rather "at this point, the function stops increasing and starts decreasing." This stops, however, at the asymptote at x=5 At this point, the function exceeds the maximum point previously calculated; it has nothing to do therefore with the range of the function.

    I'm waffling superfluously and confusingly; the core point is this:

    To calculate the possible range of this function, you need to consider the values h(x) cannot possibly take. In this case, there is just one such value: where h(x)=0. The maximum value is not necessary to find the range in this scenario.

    Edit:correction.
    Last edited by Quacky; October 14th 2011 at 07:19 PM.
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  7. #7
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    Re: Reciprocal of a Quadratic Function

    But looking at the graph, the h(x) cannot be any values above about -0.06 and below 0 as well :S
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  8. #8
    Super Member Quacky's Avatar
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    Re: Reciprocal of a Quadratic Function

    Actually that's true and stumped me. You were right from the start, I think I completely misunderstood what you were doing.

    It's clear that h(x)>0 or h(x)\leq{\frac{-1}{16}}

    I'm wondering if this hole in the graph hasn't been considered because of a mistake. Otherwise, I have no clue. Someone else will hopefully be able to provide a more resolute answer.
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  9. #9
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    Re: Reciprocal of a Quadratic Function

    Oh okay, thankss for you help
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