Respuesta :

LammettHash
You're minimizing the function [tex]f(x)=x+\dfrac1x[/tex] where [tex]x>0[/tex].

Take the derivative, set to zero and find the critical points:

[tex]f'(x)=1-\dfrac1{x^2}=0\implies1=\dfrac1{x^2}[/tex]

This has two roots, [tex]x=\pm1[/tex], but we ignore the negative root.

Check the signs of the derivative to the left and right of [tex]x=1[/tex] and you'll see that [tex]f'(x)<0[/tex] for [tex]x<1[/tex], and [tex]f'(x)>0[/tex] for [tex]x>1[/tex], which means a minimum occurs at [tex]x=1[/tex].

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