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How can I find the max/min of this function?

  1. Jun 3, 2013 #1
    1. The problem statement, all variables and given/known data
    Find the critical point of f(x)=(-5x^2+3x)/(2x^2-5), and show whether this point is a max/min.

    2. Relevant equations

    f(x)=(-5x^2+3x)/(2x^2-5)

    3. The attempt at a solution
    When I tried solving for the derivative of; f(x)=(-5x^2+3x)/(2x^2-5), I got (-6x^2+50x-15)/2x^2-5)^2.

    then I set it equal to 0 and I got 0=-6x^2+50x-15

    Now from here do I use the quadratic formula to solve for x? I tried that and I am getting x=(25+sqrt535)/6 and x=(25-sqrt535)/6. Have I done it right so far? When I try to plug in the x-values into the original eq to get y my calc says error... What do I do now?
     
  2. jcsd
  3. Jun 3, 2013 #2
    Try again. Everything looks OK.
     
  4. Jun 3, 2013 #3

    SteamKing

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    Make sure you are using the right quadratic formula. Remember, it starts out with '-b', and your answers don't have the correct value for '-b'.
     
  5. Jun 3, 2013 #4

    MarneMath

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    The op's quadratic answer is correct.* If you're plugging in the exact formula into the calculator, I would simply instead just take an approximation of your answers (ie .311 and 8.02) and plug that in and then proceed with determining if the values are a max or a min.

    *(The -b is negated by the 2a, where a is negative.)
     
    Last edited: Jun 3, 2013
  6. Jun 3, 2013 #5

    Ray Vickson

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    Your answers are correct; it sounds like you need to buy a new calculator.
     
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