What is the extreme f(x)=sqrt(9-x^2),-1<= x<= 3 ?

The extreme f(x)=sqrt(9-x^2),-1<= x<= 3 is Minimum(-1,2sqrt(2)),Maximum(0,3),Minimum(3,0)

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extreme f(x)=sqrt(9-x^2),-1<= x<= 3

extreme points

f (x) = 9 - x^{2}, - 1 \leq x \leq 3

Minimum (- 1, 22), Maximum (0, 3), Minimum (3, 0)

Solution steps

Find the critical points:

x = - 1, x = 0, x = 3

Domain of

9 - x^{2}, - 1 \leq x \leq 3 : - 1 \leq x \leq 3

Find intervals:

Increasing

: - 1 < x < 0,

Decreasing

: 0 < x < 3

Plug the extreme point

x = - 1

into

9 - x^{2} \Rightarrow y = 22

Minimum (- 1, 22)

Plug the extreme point

x = 0

into

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

Maximum (0, 3)

Plug the extreme point

x = 3

into

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

Minimum (3, 0)

Minimum (- 1, 22), Maximum (0, 3), Minimum (3, 0)

y = x^{\frac{5}{3}} - 5 x^{\frac{2}{3}}

f (x) = - (x - 1)^{2} (x - 5)

f (x, y) = 3 x^{3} - 9 x + 9 x y^{2}

ln (x + 2) (x - 1)^{2}

f (x, y) = 4 x + 6 y - x^{2} - y^{2} + 9

What is the extreme f(x)=sqrt(9-x^2),-1<= x<= 3 ?
The extreme f(x)=sqrt(9-x^2),-1<= x<= 3 is Minimum(-1,2sqrt(2)),Maximum(0,3),Minimum(3,0)