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Math Help - Critical points of a function

  1. #1
    Member sinewave85's Avatar
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    Critical points of a function

    Using the first and second order derivatives to find the critical points of this function has left me very confused -- they don't mesh at all with the critical points on the graph of the function (as displayed by a graphing calculator).

    f(x) = 3x^{4} - 2x^{3} - 12u^{2} + 18x -5

    f^{\prime}(x) = 12x^{3} - 6x^{2} - 24x + 18

    f^{\prime}(x) = 6(2x-2)(x-1)(x+1.5)

    f^{\prime \prime}(x) = 36x^{2} - 12x -24

    f^{\prime \prime}(x) = 12(3x + 2)(x -1)

    So, that give potential critical points of x= 1, and -3/2 for relative extremas and x=-2/3 for a point of inflection.

    The calculator displays critical points (min, max, min, respectively) of x= 3, 0.6861393, and -2.186138.

    Why the discrepancy?
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  2. #2
    Moo
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    Hello,

    why is it 18-5 ? Is there any typo here ? Because it's strange it hasn't been simplified into 13
    And hence it seems not logical that the derivative has 18 in it...

    Also, what is u ? If it doesn't depend on x, then its derivative is 0 and you shouldn't have any u left in the derivative of f.
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  3. #3
    Member sinewave85's Avatar
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    Quote Originally Posted by Moo View Post
    Hello,

    why is it 18-5 ? Is there any typo here ? Because it's strange it hasn't been simplified into 13
    And hence it seems not logical that the derivative has 18 in it...

    Also, what is u ? If it doesn't depend on x, then its derivative is 0 and you shouldn't have any u left in the derivative of f.
    Ok, I must not have my thinking cap on today. The 18 is missing its variable -- supposed to be 18x. And the u's are just a typo, too. The book uses u for the variable, and I was changing it to x as I typed (just because that is the standard variable) and missed those. I fixed the op.
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  4. #4
    Member javax's Avatar
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    hey there
    you'll have

    y' = 6(2x^3-x^2-4x+3) = 6(2x+3)(x-1)^2
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  5. #5
    Member sinewave85's Avatar
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    Quote Originally Posted by javax View Post
    hey there
    you'll have

    y' = 6(2x^3-x^2-4x+3) = 6(2x+3)(x-1)^2
    Hi, javax. The thing is, that gives the same solutions as the way I have it written out, right:

    y' = 6(2x+3)(x-1)^2

    0=(2x+3)(x-1)(x-1)

    x=\frac{-3}{2}, 1

    So we are still facing the same question.
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  6. #6
    Member javax's Avatar
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    If we're talking about y = 3x^4-2x^3-12x^2+18x-5
    then it's derivative is
    y' = 6(2x+3)(x-1)^2 which is negative in the interval (-\infty, -\frac{3}{2}) thus the function decreasing in this interval and it will be positive in the interval (-\frac{3}{2}, \infty) which means the function is increasing in this interval. I'm sure this is true for this function.
    The same procedure with y''...
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  7. #7
    Member sinewave85's Avatar
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    Quote Originally Posted by javax View Post
    If we're talking about y = 3x^4-2x^3-12x^2+18x-5
    then it's derivative is
    y' = 6(2x+3)(x-1)^2 which is negative in the interval (-\infty, -\frac{3}{2}) thus the function decreasing in this interval and it will be positive in the interval (-\frac{3}{2}, \infty) which means the function is increasing in this interval. I'm sure this is true for this function.
    The same procedure with y''...
    Ok, thanks. I think something was set incorrectly in my calculator and it was creating an inaccurate graph. Thanks for the help.
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