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

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

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

extreme points

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

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

Solution steps

Find the critical points:

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

Domain of

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

Find intervals:

Increasing

: - 1 < x < 0,

Decreasing

: 0 < x < 2

Plug the extreme point

x = - 1

into

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

Minimum (- 1, 3)

Plug the extreme point

x = 0

into

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

Maximum (0, 2)

Plug the extreme point

x = 2

into

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

Minimum (2, 0)

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

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