
Re: Help Needed in math
Posted:
May 1, 2014 6:06 AM


On Thursday, May 1, 2014 5:59:46 AM UTC+2, William Elliot wrote: > > > > > > > Find the largest values of the integer a for which ax^25x3 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? > > > a < f(x) = (5x + 3) / x^2, x <> 0 > > What's f?
It's the sixth letter in the alphabet. I don't understand the question.
> > > 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.

