What is the extreme f(x)=sqrt(4-x^2),-2<= x<= 1 ?

The extreme f(x)=sqrt(4-x^2),-2<= x<= 1 is Minimum(-2,0),Maximum(0,2),Minimum(1,sqrt(3))

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extreme f(x)=sqrt(4-x^2),-2<= x<= 1

extreme points

f (x) = 4 - x^{2}, - 2 \leq x \leq 1

Minimum (- 2, 0), Maximum (0, 2), Minimum (1, 3)

Solution steps

Find the critical points:

x = - 2, x = 0, x = 1

Domain of

4 - x^{2}, - 2 \leq x \leq 1 : - 2 \leq x \leq 1

Find intervals:

Increasing

: - 2 < x < 0,

Decreasing

: 0 < x < 1

Plug the extreme point

x = - 2

into

4 - x^{2} \Rightarrow y = 0

Minimum (- 2, 0)

Plug the extreme point

x = 0

into

4 - x^{2} \Rightarrow y = 2

Maximum (0, 2)

Plug the extreme point

x = 1

into

4 - x^{2} \Rightarrow y = 3

Minimum (1, 3)

Minimum (- 2, 0), Maximum (0, 2), Minimum (1, 3)

x y

f (x) = 2 x y

f

f (x, y) = - 2 x^{3} - 2 y^{3} + 6 x y + 10

f (x) = x^{\frac{1}{3}} + 1

What is the extreme f(x)=sqrt(4-x^2),-2<= x<= 1 ?
The extreme f(x)=sqrt(4-x^2),-2<= x<= 1 is Minimum(-2,0),Maximum(0,2),Minimum(1,sqrt(3))