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Topic: Help Needed in math
Replies: 25   Last Post: May 2, 2014 6:01 PM

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Karl-Olav Nyberg

Posts: 352
Registered: 12/6/04
Re: Help Needed in math
Posted: May 1, 2014 6:06 AM
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On Thursday, May 1, 2014 5:59:46 AM UTC+2, William Elliot wrote:
> > > > > > > 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?


It's the sixth letter in the alphabet. I don't understand the question.

>
> > 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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