y= x 1x y =x ( 1x) + 1x y x= 0 y= x^2+1x⇒ x + 1x y = 1 1x^2⇒ 0 1x^2= 1 x= ± 1 if x= 11 , y=2 if x= 1, y= 2 So, ...

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Question

Find Maxima $y = x \frac{1}{x}$

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Solution

$y = x \frac{1}{x}$
$y = x (\frac{- 1}{x}) + \frac{1}{x}$
$y x = 0$
$y = \frac{x^{2} + 1}{x} \Rightarrow x + \frac{1}{x}$
$y = 1 - \frac{1}{x^{2}} \Rightarrow 0$
$\frac{1}{x^{2}} = 1$
$x = \pm 1$
if $x = 11, y = 2$
if $x = - 1, y = - 2$
So, $(1, 2)$ is maxim of the given function.

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