What is the extreme f(x)=sqrt(16-x^2),-4<= x<= 4 ?

The extreme f(x)=sqrt(16-x^2),-4<= x<= 4 is Minimum(-4,0),Maximum(0,4),Minimum(4,0)

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extreme f(x)=sqrt(16-x^2),-4<= x<= 4

extreme

Minimum (- 4, 0), Maximum (0, 4), Minimum (4, 0)

Solution steps

Find the critical points:

x = - 4, x = 0, x = 4

Domain of

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

Find intervals:

Increasing

: - 4 < x < 0,

Decreasing

: 0 < x < 4

Plug the extreme point

x = - 4

into

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

Minimum (- 4, 0)

Plug the extreme point

x = 0

into

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

Maximum (0, 4)

Plug the extreme point

x = 4

into

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

Minimum (4, 0)

Minimum (- 4, 0), Maximum (0, 4), Minimum (4, 0)

What is the extreme f(x)=sqrt(16-x^2),-4<= x<= 4 ?
The extreme f(x)=sqrt(16-x^2),-4<= x<= 4 is Minimum(-4,0),Maximum(0,4),Minimum(4,0)