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Math Help - Find where f is increasing, it's minima...

  1. #1
    Member dokrbb's Avatar
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    Find where f is increasing, it's minima...

    I have such function f(x) = x^6(x-4)^3 for the interval [-11, 10],

    I need to evaluate on which intervals the function is increasing, on which one is positive and where is the minimum.

    I tied to find the derivative , so f'(x) = 6x^5(x-4)^3*3(x-4)^2*x = and wrote it as 18x^6(x-4)^3(x-4)^2

    even before calculating the critical points it seems that I complicate something,

    Could you help me by indicating one simpler way of factoring and differentiating this?
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  2. #2
    Junior Member Barioth's Avatar
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    Re: Find where f is increasing, it's minima...

    Hi Dokrbb

    If I did understand the question right,
    We are looking for the intervals where the function f(x) given is increassing.
    So as you (should) know (or remember now that I say it : ) f(x) is increasing in an interval if its derivative is Positive (so if f'(x) > 0) So I had try to find the value for f'(x) =0 then look beetween the value of x for f'(x) =0 and see if its positive.

    That should get you started.

    Hope I helped
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    Member dokrbb's Avatar
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    Re: Find where f is increasing, it's minima...

    well, your advice was helpful and not helpful in the same time:

    it's helpful since from what you said I am on the right track, but it's not really,( I'm joking, sorry) because I already tried to obtain those points and I obtained the critical points as for

    x=-1, x=0 and x=4, but it's something wrong with these points, that's why I posted the derivative I obtained (I suppose that's where I got wrong...) and would really appreciate your help in getting those critical points,
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  4. #4
    Junior Member Barioth's Avatar
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    Re: Find where f is increasing, it's minima...

    I had recheck the derivative you found!

    remeber the (u*v)' = u'v+uv'
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    Re: Find where f is increasing, it's minima...

    There is some thing wrong with your critical points
    The derivative f'(x) = 6x^5 (x-4)^3 + 3 x^6 (x-4)^2
    For critical points f'(x) = 0 gives
    3x^5 (x-4)^2 [ 2 ( x-4 ) + x ] = 0 That is 3x^5 (x-4)^2 ( 3x-8) = 0 and that gives x = 0, 4 and 8/3 = 2.67
    Now pick up any convenient in the intervals and evaluate f'(x) the interval where it is > 0 the function is increasing and where it is < 0 the function is decreasing.
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  6. #6
    Member dokrbb's Avatar
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    Re: Find where f is increasing, it's minima...

    Thank you guys, now I see where I was wrong...
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