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

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

Popular >

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

extreme points

f (x) = 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) = \frac{1}{x} - \frac{64}{y} + x y

f (x) = x^{2} + 8 x + 13

f (x) = 15 x^{4} - 4 x^{3}

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

f (x) = 2 + x - x^{2}

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