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

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

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

extreme points

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

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

Solution steps

Find the critical points:

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

Domain of

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

Find intervals:

Decreasing

: - 2 < x < 0,

Increasing

: 0 < x < 3

Plug the extreme point

x = - 2

into

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

Maximum (- 2, - 5)

Plug the extreme point

x = 0

into

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

Minimum (0, - 3)

Plug the extreme point

x = 3

into

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

Maximum (3, 0)

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

f (y) = x^{3} - 9 x^{2} + 7 x - 6

sin (3 x), - \frac{π}{4} \leq x \leq \frac{π}{3}

\frac{x ^{2}}{( 1 - x )}

f (x) = 2 x^{3} - 39 x^{2} + 240 x - 9

f (x) = 3 x^{2} - 5

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