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

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

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extreme f(x)=sqrt(9-x^2)

extreme points

f (x) = 9 - x^{2}

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

Solution steps

f^{'} (x) = - \frac{x}{9 - x ^{2}}

Find intervals:

Increasing

: - 3 < x < 0,

Decreasing

: 0 < x < 3

Plug

x = - 3

into

9 - x^{2} : 0

Minimum (- 3, 0)

Plug

x = 0

into

9 - x^{2} : 3

Maximum (0, 3)

Plug

x = 3

into

9 - x^{2} : 0

Minimum (3, 0)

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

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