
Re: Help Needed in math
Posted:
May 1, 2014 9:45 PM


> > > > > > > > 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. The function f. Did you define f as f(x) = (5x  3)/x^2? > > > If a < f(x) for all x, then a has to be less than the global minimum of > > > f(x). That's obvious > > > The global minimum of f(x) = 25/12. The largest integer less than this > > > is 3. f'(x) = 5/x^2  (5x + 3)/2x^3 = 0; 5x/2  3/2 = 0; x = 3/5. The extremum point, x = 3/5 when f(x) = 0, is an inflection point. What you're doing is unfathomable.

