What is the extreme sqrt(81-x^2),-9<= x<= 9 ?

The extreme sqrt(81-x^2),-9<= x<= 9 is Minimum(-9,0),Maximum(0,9),Minimum(9,0)

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extreme sqrt(81-x^2),-9<= x<= 9

extreme points

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

Minimum (- 9, 0), Maximum (0, 9), Minimum (9, 0)

Solution steps

Find the critical points:

x = - 9, x = 0, x = 9

Domain of

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

Find intervals:

Increasing

: - 9 < x < 0,

Decreasing

: 0 < x < 9

Plug the extreme point

x = - 9

into

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

Minimum (- 9, 0)

Plug the extreme point

x = 0

into

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

Maximum (0, 9)

Plug the extreme point

x = 9

into

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

Minimum (9, 0)

Minimum (- 9, 0), Maximum (0, 9), Minimum (9, 0)

\frac{( - 4 )}{x - 7}

f (x) = (x - 3 x^{2})^{\frac{1}{5}}, - 1 \leq x \leq 5

g (x) = x 2 - x^{2}

ln (x^{2} + 144)

f (x, y) = - 5 x^{2} - 3 y^{2} + 48 x - 4 y + 2 x y + 300

What is the extreme sqrt(81-x^2),-9<= x<= 9 ?
The extreme sqrt(81-x^2),-9<= x<= 9 is Minimum(-9,0),Maximum(0,9),Minimum(9,0)