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

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

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

extreme points

y = 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)

110 \cdot 3. 1^{x}

\frac{8}{3} x - 3

f (x) = x + 2 sin (x)

f (x) = 4 x^{2} (5^{x})

f (x) = x^{3} - 3 x^{2} + 8

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