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

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

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

extreme points

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

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

Solution steps

f^{'} (x) = 4 x (x - 2) (x - 1)

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < 1,

Decreasing

: 1 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = 0

into

x^{2} \cdot (2 - x)^{2} : 0

Minimum (0, 0)

Plug

x = 1

into

x^{2} \cdot (2 - x)^{2} : 1

Maximum (1, 1)

Plug

x = 2

into

x^{2} \cdot (2 - x)^{2} : 0

Minimum (2, 0)

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

f (x, y) = 4 x^{2} - 8 x + 5 y^{2} + 6

- (x^{2} - y^{2}) e^{- x^{2} - y^{2}}

6 3 cos (x) + 6 sin^{2} (x)

\frac{( 4 x - 12 )}{( x - 2 ) ^{2}}

f (x, y) = x + 3 y

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