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  1. #1
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    maxima and minima

    Hi friends,

    I am trying to find the maxima and minima of

    f(x) = 12x^6 – 4x^4 + 15x^3 –1.

    but after calculating f'(x) I am unable to solve it. Please guide me...
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  2. #2
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    Quote Originally Posted by mailquark View Post
    Hi friends,

    I am trying to find the maxima and minima of

    f(x) = 12x^6 4x^4 + 15x^3 1.

    but after calculating f'(x) I am unable to solve it. Please guide me...
    The solutions are either x = 0 or solutions to 72x^3 - 16x + 45=0.

    The cubic does not have easily found solutions (although it can be solved). What makes you think exact solutions are required?
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  3. #3
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    The question is in my book. I tried a lot almost 30 mins to solve this cubic... but could not....
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  4. #4
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    Quote Originally Posted by mailquark View Post
    The question is in my book. I tried a lot almost 30 mins to solve this cubic... but could not....
    State the question exactly as it's worded in your book.
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    For the following function, find a point of maxima and a point of minima, if these exist
    f(x) = 12x^6 4x^4 + 15x^3 1

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  6. #6
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    Quote Originally Posted by mailquark View Post
    For the following function, find a point of maxima and a point of minima, if these exist
    f(x) = 12x^6 – 4x^4 + 15x^3 –1
    There's a stationary point of inflection at (0, -1).

    There's a minimum turning at x \approx -0.941. If you want to use the cubic formula to solve for x (at least it's a depressed cubic so that's a start) be my guest. There's no simple way of getting the exact value of x.

    Does the book have an answer at the back?
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    No its collection of previous year question papers... I am practicing as exams are coming in December... If you can guide on this question I will be thankful.
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  8. #8
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    Quote Originally Posted by mailquark View Post
    No its collection of previous year question papers... I am practicing as exams are coming in December... If you can guide on this question I will be thankful.
    Then you should ask your instructor for solutions or, at least, answers.

    If you want an exact value for the x-coordinate of the minimum turning point, solve the cubic equation by following the process given here: Cubic Formula -- from Wolfram MathWorld

    Good luck substituting this exact value into y = f(x) to get the y-coordinate.
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  9. #9
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    Thanks a lot for giving for time...

    actually I am studying part time through distant learning and there is no instructor to guide so i thought anyone here can help. hence i posted two questions which only i was not able to solve in last 4 question papers. one is this question and another is

    http://www.mathhelpforum.com/math-he...-equation.html

    regards
    Kr.

    PS: I also tried formula given here http://en.wikipedia.org/wiki/Cubic_equation
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