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

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

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extreme y=sqrt(4-x^2)

extreme points

y = 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)

2 cos (x) + sin (2 x)

f (x) = x^{3} - 18 x^{2} + 81 x + 45

2 x^{3} - 6 x^{2} - 18 x

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

f (x, y) = 2 x^{2} + 3 x y + 4 y^{2} - 7 x - 11 y

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