What is the extreme f(x)=((x^2+x+1))/(x^2) ?

The extreme f(x)=((x^2+x+1))/(x^2) is Minimum(-2, 3/4)

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extreme f(x)=((x^2+x+1))/(x^2)

extreme

Minimum (- 2, \frac{3}{4})

Solution steps

f^{'} (x) = \frac{- x - 2}{x ^{3}}

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < 0,

Decreasing

: 0 < x < \infty

Plug

x = - 2

into

\frac{x ^{2} + x + 1}{x ^{2}} : \frac{3}{4}

Minimum (- 2, \frac{3}{4})

What is the extreme f(x)=((x^2+x+1))/(x^2) ?
The extreme f(x)=((x^2+x+1))/(x^2) is Minimum(-2, 3/4)