
Re: Help Needed in math
Posted:
May 2, 2014 12:21 AM


On Thu, 1 May 2014, quasi wrote: > >>>> > >>>>f(x) = (5x + 3) / x^2, x <> 0 > >>>What's f? > >Did you define f as f(x) = (5x  3)/x^2? > > No, he wrote > 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). > >>>>The global minimum of f(x) = 25/12. > >>>>The largest integer less than this is 3.
f'(x) = 5/x^2  2(5x + 3)/x^3 = 0; 5x  6 = 0; x = 6/5
The extremum point, x = 6/5 when f(x) = 25/12, is a minimum.
> Not when you use the correct f'(x). > > >What you're doing is unfathomable. > Unfathomable to you perhaps. > 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?

