What is the extreme f(x)=sqrt(64-x^2) ?

The extreme f(x)=sqrt(64-x^2) is Minimum(-8,0),Maximum(0,8),Minimum(8,0)

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extreme f(x)=sqrt(64-x^2)

extreme points

f (x) = 64 - x^{2}

Minimum (- 8, 0), Maximum (0, 8), Minimum (8, 0)

Solution steps

f^{'} (x) = - \frac{x}{64 - x ^{2}}

Find intervals:

Increasing

: - 8 < x < 0,

Decreasing

: 0 < x < 8

Plug

x = - 8

into

64 - x^{2} : 0

Minimum (- 8, 0)

Plug

x = 0

into

64 - x^{2} : 8

Maximum (0, 8)

Plug

x = 8

into

64 - x^{2} : 0

Minimum (8, 0)

Minimum (- 8, 0), Maximum (0, 8), Minimum (8, 0)

f (x, y) = x y - 4 y - 16 x + 64

y = cos (x)

2 x^{2} + 24 x - 6

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

f (x) = 2 x^{3} + 3 x^{2} - 12 x + 6

What is the extreme f(x)=sqrt(64-x^2) ?
The extreme f(x)=sqrt(64-x^2) is Minimum(-8,0),Maximum(0,8),Minimum(8,0)