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

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snmpprotocol@gmail.com

Posts: 28
Registered: 4/16/14
Re: Help Needed in math
Posted: May 2, 2014 12:50 PM
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On Friday, May 2, 2014 12:34:17 AM UTC-7, William Elliot wrote:
> On Fri, 2 May 2014, quasi wrote:
>

> > William Elliot wrote:
>
> > >quasi wrote:
>
> > >>...
>
> > >>Nevertheless, konyberg's argument is correct.
>
> > >His answer is correct. I don't see his reasoning
>
> > >How does f(x) relate to the missing problem?
>
> >
>
> > This is all very elementary.
>
> > Let g(x) = a*x^2 - 5*x - 3, for some fixed a in R.
>
> >
>
> > The problem was to find the largest integer value of a such that
>
> > g(x) < 0 for all x in R
>
>
>
> g(x) = 0 iff x = (5 +- sqr(25 + 12a))/2a. (WLOG a /= 0.)
>
> Thus
>
> 25 + 12a < 0; a < -25/12.
>
> With that,
>
> ax^2 - 5x - 3 < -25x^2 / 12 - 5x - 3
>
> = -(25x^2 + 60 + 36)/12 = -(5x + 3)^2 <= 0
>
> This is all elementary, simple alegbra stuff.
>
>
>

> > Note that g(0) = -3 < 0, regardless of the value of a.
>
> >
>
> > Let f(x) = (5x + 3)/(x^2), x != 0.
>
> > For x != 0,
>
> > g(x) < 0 <=> a < f(x)
>
> >
>
> > Hence
>
> > g(x) < 0 for all x in R
>
> >
>
> > <=> a < f(x) for all x in R with x != 0
>
> >
>
> > Such a value of a exists iff f is bounded below.
>
> >
>
> > As konyberg showed, f has a global minimum value of -25/12,
>
> > hence
>
>
>
> Uses basic calculus.
>
>
>

> > g(x) < 0 for all x in R <=> a < -25/12
>
> >
>
> > It follows that the largest integer value of a such that
>
> > g(x) < 0 for all x in R
>
> > is -3.
>
>
>
> Ok, I get it, as complicated as it is.




There is an easier way to look at this (even though it uses similar principles).

Axis of symmetry for ax^2 + bx +c = 0 is x = -b/2a = 5/2a

Value of function at axis symmetry (gives either maxima when a < 0 (curve is concave down) and minima when a > 0 (curve is concave up)):
a(5/2a)^2 -5(5/2a) - 3 = 0 --> a = -25/12

So when a = -25/12 the maxima has value 0 and so largest integer for maxima to be less than 0 is -3.

snmp



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