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

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

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

extreme points

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

Minimum (2, 0)

Solution steps

f^{'} (x) = - \frac{1}{x - x ^{2} + 2 x}

Find intervals:

Decreasing

: 0 < x < 2

Plug

x = 2

into

\frac{2 x - x ^{2}}{x} : 0

Minimum (2, 0)

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