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Math Help - incresing/decreasing functions

  1. #1
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    incresing/decreasing functions

    show that the function f(x)=\frac{-x^3}{3} + x^2 -x +14 is a decreasing function



    I just cant prove that f'(x) < 0
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  2. #2
    Super Member flyingsquirrel's Avatar
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    Hello,
    Quote Originally Posted by Musab View Post
    I just cant prove that f'(x) < 0
    In this case f' is a quadratic function. If you manage to factor it it'll be easy to show that f'(x)<0.
    Last edited by flyingsquirrel; September 5th 2008 at 12:53 PM. Reason: added missing word
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  3. #3
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    yes,

    but I still cant solve it
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  4. #4
    Super Member Matt Westwood's Avatar
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    Why, what have you got?
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  5. #5
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    Decreasing function

    It's easy bro. Once you've gotten  f'(x) , just put it  = 0 to get critical points. Then show that  f'(x) < 0 on either side of the critical point.

    I hope that helps.

    me07.
    Last edited by ILoveMaths07; September 5th 2008 at 01:43 PM. Reason: Umm... I'm new... This is my second post... Wanted to try on Latex...
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  6. #6
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    Quote Originally Posted by Musab View Post
    show that the function f(x)=\frac{-x^3}{3} + x^2 -x +14 is a decreasing function



    I just cant prove that f'(x) < 0
    function f(x)=\frac{-x^3}{3} + x^2 -x +14

    differentiate the function f'(x)=-x^2+2x-1

    f'(x)=-(x^2-2x+1)

    f'(x)=-(x-1)^2

    f'(x)=-(x-1)^2<0;\;always

    since (x-1)^2\geq 0, \text {being a perfect square}

    So, f(x) is a decreasing function.
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  7. #7
    Super Member 11rdc11's Avatar
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    Quote Originally Posted by Shyam View Post
    function f(x)=\frac{-x^3}{3} + x^2 -x +14

    differentiate the function f'(x)=-x^2+2x-1

    f'(x)=-(x^2-2x+1)

    f'(x)=-(x-1)^2

    f'(x)=-(x-1)^2<0;\;always

    since (x-1)^2\geq 0, \text {being a perfect square}

    So, f(x) is a decreasing function.
    f'(x)=-(x-1)^2<0;\;always

    Just wondering I don't think you can say always because when x = 1 this happens

    0 < 0

    which is false

    so it should be

    f'(x)=-(x-1)^2<0;\; \text {always, except when x = 1}

    Also at

    f'(1)= 0

    so this is niether decreasing or increasing.
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  8. #8
    Super Member Matt Westwood's Avatar
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    A *strictly* decreasing function has its derivative less than zero everywhere.

    A function which is just "decreasing" has its derivative less than or equal to zero everywhere.

    That is, its gradient is allowed to be zero at certain points, just never positive.

    You've been asked to show that this function is decreasing, not that it is strictly decreasing.

    It's a subtle point but one worth taking on board.
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