Posted by **Rochelle F** on Wednesday, April 11, 2012 at 8:45pm.

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)

- calculus -
**Steve**, Thursday, April 12, 2012 at 12:20am
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)

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