f(x)=
f ( [tex] \frac{x+1}{2} [/tex] )= [tex] x^{2} + \frac{1}{ x^{2} } [/tex]
----> f([tex] \frac{x+1}{2} [/tex])(x)= [tex] x^{2} + \frac{1}{ x^{2} } [/tex] (x)
f (x+1) +(-1)= [ ( [tex] x^{2} + \frac{1}{ x^{2} } [/tex]) (x)] +(-1)
f (x) = [tex]( x^{3} + \frac{1}{x} )-1
[/tex]
Dato x mayor igual que dos, entonces
x= 2, 3, 4, 5, 6, 7.....
Podemos remplazar
