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September 3, 2014

September 3, 2014

Posted by **Anonymous** on Friday, November 9, 2012 at 10:51am.

- Calculus -
**Steve**, Friday, November 9, 2012 at 11:28amyou know that

f'(x) has a root at x=2.

It is negative for -3/2 < x < 2

It is positive for 2<x<5

Nothing is said about any domain outside (-3/2,5)

But, since you want a cubic f(x), f'(x) must be positive for x < -3/2, so it has another root at x = -3/2

f'(x) = a(2x+3)(x-2) = a(2x^2 - x - 6)

now we can get somewhere

f(x) = a(2/3 x^3 - 1/2 x^2 - 6x) + c

or, changing a and c (which are arbitrary) and clearing fractions,

f(x) = a(4x^3 - 3x^2 - 36x) + c

Now, we want f(x) as described above, so a = 1 and

f(x) = 4x^3 - 3x^2 - 36x - 15

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