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.