What is the extreme f(x)=sqrt(36-x^2) ?

The extreme f(x)=sqrt(36-x^2) is Minimum(-6,0),Maximum(0,6),Minimum(6,0)

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extreme f(x)=sqrt(36-x^2)

extreme points

f (x) = 36 - x^{2}

Minimum (- 6, 0), Maximum (0, 6), Minimum (6, 0)

Solution steps

f^{'} (x) = - \frac{x}{36 - x ^{2}}

Find intervals:

Increasing

: - 6 < x < 0,

Decreasing

: 0 < x < 6

Plug

x = - 6

into

36 - x^{2} : 0

Minimum (- 6, 0)

Plug

x = 0

into

36 - x^{2} : 6

Maximum (0, 6)

Plug

x = 6

into

36 - x^{2} : 0

Minimum (6, 0)

Minimum (- 6, 0), Maximum (0, 6), Minimum (6, 0)

(- 1, - 4), (6, - 5)

f (x) = x^{3} - 9

f (x) = (- 5 x^{2} + 5) e^{x}

x^{3} - 2 x^{2} - 7 x - 4

8 x

What is the extreme f(x)=sqrt(36-x^2) ?
The extreme f(x)=sqrt(36-x^2) is Minimum(-6,0),Maximum(0,6),Minimum(6,0)