calculus
posted by Rochelle F .
find the absolute minimum value on (0,infinity)...f(x)= x(1/x)+(2/x^3)..
The example given is x(1/x)+(10/x^3) which is F(sqrt5)=(6/sqrt5)

f = x  1/x + 2/x^3
f' = 1 + 2/x^2  6/x^4
= (x^4 + 2x^2  6)/x^4
f'=0 when x^4 + 2x^2  6 = 0
x^2 = 1 ± √7
1  √7 < 0. so we have
x^2 = 1 + √7
x = ±√(√7  1)
we want x in (0,∞), so
f(1.28287) = 1.451
or, more exactly,
f(√(√7  1)) = √((97√7  143)/54)