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Math Help - Another beginners calculus problem

  1. #1
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    Another beginners calculus problem

    Find an M>0 so that every x>M fulfills \frac{6x^2-7\sin x}{3x+1} > 1000

    I've been brainstorming ideas, but it's a bit frustrating since I can't simplify the equation like I usually do.

    Here are my thoughts and questions:

    1) I'm thinking along the lines of proving something like \frac{6x^2-7\sin x}{3x+1} > \frac{9x^2-1}{3x+1} > 1000 because then I can simplify to 3x-1>1000. Is there any way I can make that happen?

    2) Can 6x^2-7\sin x ever be < 0. Because if not, we know that 3x+1 must also be  > 0 since the function must be > 1000.

    3) Or can I just say M = 1000000000000, and since sinx is always between 1 and -1, it doesn't effect us, and 6x^2 is always bigger than 3x+1. so a x>M is definitely bigger than 1000.

    Thanks!
    Last edited by jayshizwiz; October 28th 2010 at 05:52 AM.
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  2. #2
    A Plied Mathematician
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    You'd like a sequence of inequalities that look like this:

    \dfrac{6x^2-7\sin(x)}{3x+1} >...>...>1000.

    So, if you can make the expression smaller but still with a size that you can control, you're good.

    Since you're subtracting the sin function, and you want your function to get smaller, make the 7\sin(x) term as big as possible:

    \dfrac{6x^2-7\sin(x)}{3x+1}\ge \dfrac{6x^2-7}{3x+1}.

    Now work with this, and make it bigger than the desired amount.
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  3. #3
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    Quote Originally Posted by Ackbeet View Post
    You'd like a sequence of inequalities that look like this:

    \dfrac{6x^2-7\sin(x)}{3x+1} >...>...>1000.

    So, if you can make the expression smaller but still with a size that you can control, you're good.

    Since you're subtracting the sin function, and you want your function to get smaller, make the 7\sin(x) term as big as possible:

    \dfrac{6x^2-7\sin(x)}{3x+1}\ge \dfrac{6x^2-7}{3x+1}.

    Now work with this, and make it bigger than the desired amount.
    Thanks but I'm still stuck for some reason. Can you give me the next step?

    I tried making \dfrac{6x^2-7}{3x+1} > 1005 > 1000 but I got stuck. I tried taking away a 1 and a 7 but I'm not so sure what to do.
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  4. #4
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    Hm...maybe I should simply solve for \dfrac{6x^2-7}{3x+1} > 1000

    and I should get x>(something). Then I can just make M=(something) so I know x is always >M
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  5. #5
    A Plied Mathematician
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    I think that procedure should work for you.
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