What is the extreme f(x)=sqrt(4-x^2) ?

The extreme f(x)=sqrt(4-x^2) is Minimum(-2,0),Maximum(0,2),Minimum(2,0)

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extreme f(x)=sqrt(4-x^2)

extreme points

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

Minimum (- 2, 0), Maximum (0, 2), Minimum (2, 0)

Solution steps

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

Find intervals:

Increasing

: - 2 < x < 0,

Decreasing

: 0 < x < 2

Plug

x = - 2

into

4 - x^{2} : 0

Minimum (- 2, 0)

Plug

x = 0

into

4 - x^{2} : 2

Maximum (0, 2)

Plug

x = 2

into

4 - x^{2} : 0

Minimum (2, 0)

Minimum (- 2, 0), Maximum (0, 2), Minimum (2, 0)

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

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

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

f (x, y) = 3 x y - x^{2} y - x y^{2}

f (x) = x^{4} - 12 x^{3} - 29

What is the extreme f(x)=sqrt(4-x^2) ?
The extreme f(x)=sqrt(4-x^2) is Minimum(-2,0),Maximum(0,2),Minimum(2,0)