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    A function f is defined by f(x)=4-3x-x^3

    a) Find the derivative f'(x)

    Which I found to be f'(x)=-3x^2-3 After rearranging.

    b) Use your answer from (a) to deduce whether the function is an increasing or decreasing function.

    Do I make x<o and go from there or is this wrong?

    Thanks for the help.
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    doesn't it give you a value of x?
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    (Original post by Jfranny)
    doesn't it give you a value of x?
    This as what I though.. :confused:
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    f(x)=4-3x-x^3

    f'(x)=-3x^2-3

    f'(x) is always going to be negative, so the gradient of f(x) is always negative - so it is a decreasing function
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    (Original post by markioe)
    For all values of x>1 its a positive gradient (it should say x>1 somewhere in the question) thus the function is increasing as x increases.
    Are you reading it right? The derivative is -3x^2 - 3, so it's decreasing at all points.
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    (Original post by starofale)
    f(x)=4-3x-x^3

    f'(x)=-3x^2-3

    f'(x) is always going to be negative, so the gradient of f(x) is always negative - so it is a decreasing function
    Thanks I'll be sure to rep you! :yes:
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    (Original post by starofale)
    f(x)=4-3x-x^3

    f'(x)=-3x^2-3

    f'(x) is always going to be negative, so the gradient of f(x) is always negative - so it is a decreasing function
    Could you point me in the right way for this question?

    Find the set of values of x for which the function h, where h(x)=2-(x+3)^2 is increasing.
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    Oh misread, indeed it is decreasing.
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    In this case f'(x) can only be negative...any x value will do.
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    (Original post by econ1)
    Could you point me in the right way for this question?

    Find the set of values of x for which the function h, where h(x)=2-(x+3)^2 is increasing.
    h(x)=2-(x+3)^2
    h(x) will be decreasing when h'(x) is negative. Find h'(x) and solve for h'(x)<0
 
 
 
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Updated: January 18, 2010
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