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    Heres the question
    http://img694.imageshack.us/img694/4826/aaaaaaof.jpg
    I got up to the point x>+/-2
    When you draw the graph I don't get why it's the part below the X axis at -2<x<2
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    Values of x for which f(x) is an increasing function means values of x for which f'(x)>0.
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    An increasing function means \displaystyle f(x+1) &gt; f(x).

    If you find \displaystyle x such that \displaystyle f'(x) = 0, then you're finding the turning points of \displaystyle f(x), and due to the sign of the coefficient of the variable with highest power, we know the function overall is generally decreasing (as it's a cubic), but there'll be a part of it which is increasing, and it's increasing between the turning points.
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    Try doing a small rough sketch of what it's supposed to look like - that'll help.
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    f(x)=12x-x^3
    f is derivable on R as sum of two functions derivable on R [u(x)=12x and v(x)=-x^3]
    Therefore, for all x of R, f'(x)=12-3x²
    f'(x)=0 <=> 12=3x² <=> x²=4 <=> x=-2 or x=2
    For x<2 and x>2, f'(x)<0 so f decreasing
    For -2<x<2, f'(x)>0 so f increasing
 
 
 
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Updated: January 14, 2010

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