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Re: Help Needed in math
Posted:
Apr 30, 2014 11:59 PM
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> > > > > > Find the largest values of the integer a for which ax^2-5x-3 is
> > > > > > negative for all values of x.
> > > > > set a = (5x + 3) / x^2 and find the minima of this function. This
> > > > > will, with your constraints, give the answer: a = -3
> > > > 5/x + 3/x^2; -5/x^2 - 6/x^3 = 0; -5x + -6 = 0; x = -6/5
> > > > a = (-6 + 3)/(36/25) = -25/12
> > > > Why does the minimum work?
> a < f(x) = (5x + 3) / x^2, x <> 0
What's f?
> If a < f(x) for all x, then a has to be less than the global minimum of f(x).
> The global minimum of f(x) = -25/12. The largest integer less than this is -3.
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