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.