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

The extreme sqrt(9-x^2),-3<= x<= 2 is Minimum(-3,0),Maximum(0,3),Minimum(2,sqrt(5))

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

extreme points

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

Minimum (- 3, 0), Maximum (0, 3), Minimum (2, 5)

Solution steps

Find the critical points:

x = - 3, x = 0, x = 2

Domain of

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

Find intervals:

Increasing

: - 3 < x < 0,

Decreasing

: 0 < x < 2

Plug the extreme point

x = - 3

into

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

Minimum (- 3, 0)

Plug the extreme point

x = 0

into

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

Maximum (0, 3)

Plug the extreme point

x = 2

into

9 - x^{2} \Rightarrow y = 5

Minimum (2, 5)

Minimum (- 3, 0), Maximum (0, 3), Minimum (2, 5)

f (x) = 1 + \frac{x}{x - 2}

y = \frac{x ^{2} + x - 1}{x - 1}

f (x) = \frac{x ^{6}}{10} - \frac{2 x ^{5}}{3}

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

f (x) = 2 x + 3

What is the extreme sqrt(9-x^2),-3<= x<= 2 ?
The extreme sqrt(9-x^2),-3<= x<= 2 is Minimum(-3,0),Maximum(0,3),Minimum(2,sqrt(5))