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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: 1,704
Registered: 1/8/12
Re: Help Needed in math
Posted: May 2, 2014 3:34 AM
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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.



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