What is the extreme f(x)=sqrt(81-x^4) ?

The extreme f(x)=sqrt(81-x^4) is Minimum(-3,0),Maximum(0,9),Minimum(3,0)

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extreme f(x)=sqrt(81-x^4)

extreme points

f (x) = 81 - x^{4}

Minimum (- 3, 0), Maximum (0, 9), Minimum (3, 0)

Solution steps

f^{'} (x) = - \frac{2 x ^{3}}{81 - x ^{4}}

Find intervals:

Increasing

: - 3 < x < 0,

Decreasing

: 0 < x < 3

Plug

x = - 3

into

81 - x^{4} : 0

Minimum (- 3, 0)

Plug

x = 0

into

81 - x^{4} : 9

Maximum (0, 9)

Plug

x = 3

into

81 - x^{4} : 0

Minimum (3, 0)

Minimum (- 3, 0), Maximum (0, 9), Minimum (3, 0)

\frac{x}{x - 5}

- x^{3} + 6 x^{2}

3 x^{2} - 6 x + 12

y = x^{3} + 10 x

f (x) = \frac{1}{x ^{2} - 8 x - 9}

What is the extreme f(x)=sqrt(81-x^4) ?
The extreme f(x)=sqrt(81-x^4) is Minimum(-3,0),Maximum(0,9),Minimum(3,0)