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

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

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

extreme points

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

Minimum (0, 0), Maximum (1, 1), Minimum (2, 0)

Solution steps

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

Find intervals:

Increasing

: 0 < x < 1,

Decreasing

: 1 < x < 2

Plug

x = 0

into

2 x - x^{2} : 0

Minimum (0, 0)

Plug

x = 1

into

2 x - x^{2} : 1

Maximum (1, 1)

Plug

x = 2

into

2 x - x^{2} : 0

Minimum (2, 0)

Minimum (0, 0), Maximum (1, 1), Minimum (2, 0)

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

f (x) = 3 x^{2} - 4 x + 5

f (x) = 2 x - 3 x^{\frac{2}{3}}, - 1 \leq x \leq 3

x (1 - x^{2})^{2}

f (x, y) = x^{2} + x y + y^{2} - 10 y + 33

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