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

The extreme f(x)=(2x^2)/((x-1)^2) is Minimum(0,0)

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

extreme

Minimum (0, 0)

Solution steps

f^{'} (x) = - \frac{4 x}{( x - 1 ) ^{3}}

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 1,

Decreasing

: 1 < x < \infty

Plug

x = 0

into

\frac{2 x ^{2}}{( x - 1 ) ^{2}} : 0

Minimum (0, 0)

What is the extreme f(x)=(2x^2)/((x-1)^2) ?
The extreme f(x)=(2x^2)/((x-1)^2) is Minimum(0,0)