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

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William Elliot

Posts: 1,673
Registered: 1/8/12
Re: Help Needed in math
Posted: May 1, 2014 9:45 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?
> It's the sixth letter in the alphabet. I don't understand the question.
The function f. Did you define f as f(x) = (5x - 3)/x^2?
> > > If a < f(x) for all x, then a has to be less than the global minimum of
> > > f(x).

That's obvious
> > > The global minimum of f(x) = -25/12. The largest integer less than this
> > > is -3.

f'(x) = 5/x^2 - (5x + 3)/2x^3 = 0; 5x/2 - 3/2 = 0; x = 3/5.
The extremum point, x = 3/5 when f(x) = 0, is an inflection point.
What you're doing is unfathomable.



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