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Maxima/Minima problem

  1. Apr 23, 2008 #1
    1. The problem statement, all variables and given/known data

    For the following function, find a point of maxima and a point of minima, if these exist

    [tex]
    f(x) = 12x^6 - 4x^4 + 15x^3 - 1
    [/tex]

    3. The attempt at a solution

    I first started by finding the 1st and 2nd derivatives:

    [tex]
    f'(x) = 72x^5 - 16x^3 + 45x^2
    [/tex]

    [tex]
    f''(x) = 288x^3 - 48x^2 + 50x
    [/tex]

    On equating, f'(x) with 0,

    [tex]
    x^2(18x^3 - 12x + 15) = 0
    [/tex]

    hence, i get x = 0. But, at x = 0, f''(x) = 0, and f'''(x) = 50 i.e. the first non-zero derivative is an odd derivative i.e. the third. Hence, clearly, the point x=0 is a point of inflection.

    Now, i need to solve for

    [tex]
    18x^3 - 12x + 15 = 0
    [/tex]

    I have no idea how to solve this. On using Mathematica, I am getting 3 solutions, out of which one is real:

    [tex]
    x = \frac{1}{3} \left(-\frac{2 2^{2/3}}{\left(45-\sqrt{1897}\right)^{1/3}}-\frac{\left(45-\sqrt{1897}\right)^{1/3}}{2^{2/3}}\right)
    [/tex]

    This value, upto 10 significant digits is:

    [tex]
    x = -1.173395267
    [/tex]

    And indeed, there is a minima at this point, which I am able to see when I plot the curve of the function:

    [​IMG]

    But, since a CAS or a calculator is not allowed on this assignment, I am just not able to understand how to tackle this problem without one. Even the value I got using Mathematica, is too complex for me to arrive at for this level of the assignment. Plus, we've not been taught the solution for cubic equations.

    Any help is appreciated.
    Thanks,
    rohan
     
  2. jcsd
  3. Apr 23, 2008 #2
    The first derivative tells you where the tangent line is zero when you set it equal to zero. The tangent line is zero at x=0 (as well as an inflection point). I'm not sure why you are looking for inflection points when the question was to find maxima and minima.

    Your first derivative is correct. I'm not seeing where the [tex](18x^3-12x+15)[/tex] came from.

    If you carefully look at your graph, the minimum isn't at -1.17 it looks like it's about -.9
     
  4. Apr 24, 2008 #3
    You really need to recheck your work there. Your f'(x) is correct as far as I can see, but not when you reduce it to [tex]x^2(18x^3-12x+15)[/tex]. Expand it out and you get a much different answer to your first one of f'(x).
    Also, your f''(x) is wrong. [tex]72x^5[/tex] differentiated doesn't equal [tex]288x^3[/tex] Nor does [tex]45x^2[/tex] differentiated = [tex]50x[/tex].

    First derivative tells you the x co-ordinates for maxima and minima (putting y'=0 and solving for x). Put these back into f(x) to find the y co-ordinates.
    Second derivative tells you which ones are maxima and which are minima. eg. say you find x=1 from the first derivative. if y''<0 for x<1 and y''>0 for x>1 then at x=1 it is a minima.
    2nd derivative also tells you the point(s) of inflection. Again put it to zero and solve for x.
     
    Last edited: Apr 24, 2008
  5. Apr 24, 2008 #4
    I took a '4' common as well while equating it with zero.

    ahh.. yes.. i can see that error. but still that doesn't make much difference in this problem since f''(x) is still 0 at x=0.

    Not necessarily. f'(x) = 0 even at points where a curve changes it's curvature.

    i happen to know that and have done the same thing already. You don't seem to have understood the question I was asking.

    But anyways.. i've just used Cardano's method to solve for the roots and did the question somehow.
     
  6. Apr 25, 2008 #5
    By that I assume you mean this: [tex]72x^3-16x+45)[/tex] = [tex]4x^2(18x^3-12x+15)[/tex]
    If so, expand back out. the two sides don't match up.

    That would indicate a flat spot then, not a max/min point. You can find that out by checking to see if the signs on f''(x) change at that point. If they don't then it isn't a max or min or even point of inflection.

    I did understand it. I just wasn't going to tell you the answer outright. This is a homework HELP forum, not a do-my-homework-for-me forum.
     
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