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Math Help - Finding stationary points.

  1. #1
    B33
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    Finding stationary points.

    Can somebody help me find the stationary points of this:

    \displaystyle{Q(x) = \frac{4x^2 + 20x + 25}{e^x}}
    and if the above doesnt display right it is:
    4x^2+20x+25/e^x

    and give the value of the points?
    Also say whether points are a local max or min and explain why to me?

    Thank you in advance!
    Last edited by mr fantastic; October 14th 2010 at 04:16 PM. Reason: Re-titled.
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  2. #2
    A Plied Mathematician
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    What work have you done so far?
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  3. #3
    B33
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    I did the quotient rule and got it down to a quadratic over e^x

    -4x^2-12x-5/e^2

    what do I do from there? or is that right?
    lol
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  4. #4
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    You need to be more careful with your parentheses. Technically, you wrote

    -4x^2-12x-(5/e^2).

    Also, I think you forgot the 2x in the exponent. Anyway, aside from notation issues, I think you have the general idea:

    \displaystyle Q'(x)=-\frac{4x^{2}+12x+5}{e^{2x}}.

    What now? How do you find stationary points?
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  5. #5
    B33
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    e^2x on the bottom? are you sure?

    Well you have to find out which values make the equation equal zero after the first derivative I'm fairly certain.
    Then you put that point back into the equation to see what value you get for that point.
    THEN get the second derivative and and judging by what THAT point is you can tell if you have a local max or min....
    but I def need some help lol
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  6. #6
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    e^2x on the bottom? are you sure?
    Quotient rule: low dee high minus high dee low over the square of what's below, or

    \displaystyle\left(\frac{f}{g}\right)'=\frac{gf'-fg'}{g^{2}}.

    Since you have an e^{x} in the denominator of the original fraction, you'd better have an (e^{x})^{2}=e^{2x} in the denominator afterwards, right? You're not going to lose the variable x, at any rate.

    So what do you get when you set the derivative equal to zero?
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  7. #7
    B33
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    Do I set the -4X^2-12x-5 to zero and forget about the bottom right now?
    In that case I get x=-5 and x=-3?
    -4x^2-12x-5=0
    -4x^2-12x=5
    -x(4x-12)=5
    x=-5 & x=-3?
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  8. #8
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    You are correct in forgetting about the denominator, since it will never contribute towards a zero. However, you are not solving the quadratic correctly.

    You cannot assume that if ab = c, where c is not zero, that a=c, or b=c. Just take 2 x 3 = 6. It is not true that 2 = 6 or 3 = 6.

    Complete the square, or use the quadratic formula. What do you get?
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  9. #9
    B33
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    Okay. So I used quadratic formula instead.
    I got x=-2.5 and x=-0.5
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  10. #10
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    So those are the x-coordinates for your stationary points. You need to find the y-coordinates, I think. What else are you going to need to do, do you think?
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  11. #11
    B33
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    Okay, so now I put the x values back into the function to get the y values. which should tell me if I have a local max or min?
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  12. #12
    B33
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    So I put the values back into the function and ignored the e^x or e^2x.... right thing to do?
    I got x=-62.5 and x=14
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