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

 Messages: [ Previous | Next ]
 Karl-Olav Nyberg Posts: 1,575 Registered: 12/6/04
Re: Help Needed in math
Posted: Apr 30, 2014 7:03 AM

On Wednesday, April 30, 2014 11:50:32 AM UTC+2, William Elliot wrote:
> On Wed, 30 Apr 2014, konyberg wrote:
>

> > On Wednesday, April 30, 2014 5:02:27 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?
>
> > Are you asking me why it works?
>
> Yes.

a < f(x) = (5x + 3) / x^2, x <> 0

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.

KON

Date Subject Author
4/28/14 punisher
4/28/14 quasi
4/29/14 Peter Percival
4/29/14 Karl-Olav Nyberg
4/28/14 Peter Percival
4/29/14 William Elliot
4/29/14 Peter Percival
4/29/14 Karl-Olav Nyberg
4/29/14 Karl-Olav Nyberg
4/29/14 Karl-Olav Nyberg
4/29/14 Karl-Olav Nyberg
4/29/14 William Elliot
4/30/14 Karl-Olav Nyberg
4/30/14 William Elliot
4/30/14 Karl-Olav Nyberg
4/30/14 William Elliot
5/1/14 Karl-Olav Nyberg
5/1/14 William Elliot
5/1/14 quasi
5/2/14 William Elliot
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5/2/14 Karl-Olav Nyberg
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