How do you find all local maximum and minimum points given $y=(x^2-1)/x$ ?

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Answer 1 · 2018-03-15T15:35:18

None

The lokal minimum and maximum can be found by $f'(x)=0$

$y=(x^2-1)/x$
$y'=(2x)/x-(x^2-1)/x^2=1+1/x^2$

$0=1+1/x^2| -1 | *x ^ 2$

$-x^2=1|*(-1)|sqrt()$

$x=+-sqrt(-1)$

Can't take the square root of negative numbers->No roots->No maximum or minimum

How do you find all local maximum and minimum points given y=(x^2-1)/xy=(x^2-1)/x?