Question

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

Answer 1

None

Explanation:

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