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Increasing Function values Watch

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    State the set of values of x for which 27 + 9x ? 3x^2 ? x^3 is an increasing function.

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    I first find dy/dx which is:
    9 - 6x - 3x^2 > 0

    Divide it by 3 to get:
    3 - 2x - x^2 > 0

    Bring over everything to right to factorise easily
    x^2 + 2x - 3 >0

    Which factorises into:
    (x-1) and ( x+3)

    I sketch out the graph, and see where its greater than zero , but I get x >1 and x <-3 , but the answer is -3 < x < 1 :confused:
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    (Original post by Gelato)
    Divide it by 3 to get:
    3 - 2x - x^2 > 0

    Bring over everything to right to factorise easily
    x^2 + 2x - 3 >0
    Here's your error. You didn't change the sign of the inequality -- the second one should be a <.
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    (Original post by nuodai)
    Here's your error. You didn't change the sign of the inequality -- the second one should be a <.
    Why does the sign swap round ? I'm not dividing by anything negative ?

    I done it now as you said and got it right thanks , im just not sure how it comes about
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    (Original post by Gelato)
    Why does the sign swap round ? I'm not dividing by anything negative ?

    I done it now as you said and got it right thanks , im just not sure how it comes about
    You took everything to the other side of the inequality. You had:
    3 - 2x - x² > 0

    And so taking things term-by-term we get:

    3 - 2x > x²
    3 > x² + 2x
    0 > x² + 2x - 3

    Which is the same as writing x² + 2x - 3 < 0.

    Alternatively, you can look at it as multiplying through by -1 (which is a negative number), so:

    3 - 2x - x² > 0
    -1×(3 - 2x - x²) < -1×0
    x²+2x-3 < 0
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    Thanks for the help, +rep
 
 
 
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