What is the extreme f(x)=(3x^2-8x+4)/(x^2) ?

The extreme f(x)=(3x^2-8x+4)/(x^2) is Minimum(1,-1)

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extreme f(x)=(3x^2-8x+4)/(x^2)

extreme

Minimum (1, - 1)

Solution steps

f^{'} (x) = \frac{8 ( x - 1 )}{x ^{3}}

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = 1

into

\frac{3 x ^{2} - 8 x + 4}{x ^{2}} : - 1

Minimum (1, - 1)

What is the extreme f(x)=(3x^2-8x+4)/(x^2) ?
The extreme f(x)=(3x^2-8x+4)/(x^2) is Minimum(1,-1)