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# Increasing Function values Watch

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1. 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
2. (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 <.
3. (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
4. (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
5. Thanks for the help, +rep

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Updated: December 18, 2010
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