What is the extreme f(t)=sqrt(64-x^2),-8<= x<= 8 ?

The extreme f(t)=sqrt(64-x^2),-8<= x<= 8 is Minimum(-8,0),Maximum(0,8),Minimum(8,0)

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extreme f(t)=sqrt(64-x^2),-8<= x<= 8

extreme points

f (t) = 64 - x^{2}, - 8 \leq x \leq 8

Minimum (- 8, 0), Maximum (0, 8), Minimum (8, 0)

Solution steps

Find the critical points:

x = - 8, x = 0, x = 8

Domain of

64 - x^{2}, - 8 \leq x \leq 8 : - 8 \leq x \leq 8

Find intervals:

Increasing

: - 8 < x < 0,

Decreasing

: 0 < x < 8

Plug the extreme point

x = - 8

into

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

Minimum (- 8, 0)

Plug the extreme point

x = 0

into

64 - x^{2} \Rightarrow y = 8

Maximum (0, 8)

Plug the extreme point

x = 8

into

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

Minimum (8, 0)

Minimum (- 8, 0), Maximum (0, 8), Minimum (8, 0)

f (x, y) = 4 x^{2} - 8 x y + 2 y^{2}

f (x) = - 3 x^{2} + 24 x - 200

2 - x^{4} + 2 x^{2} - y^{2}

- x^{2} + 162 ln (x)

f (x) = x^{4} - 8 x^{3} + 18 x^{2} - 16 x + 40

What is the extreme f(t)=sqrt(64-x^2),-8<= x<= 8 ?
The extreme f(t)=sqrt(64-x^2),-8<= x<= 8 is Minimum(-8,0),Maximum(0,8),Minimum(8,0)