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.