What is the extreme f(x)=sqrt(16-x^2) ?

The extreme f(x)=sqrt(16-x^2) is Minimum(-4,0),Maximum(0,4),Minimum(4,0)

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extreme f(x)=sqrt(16-x^2)

extreme points

f (x) = 16 - x^{2}

Minimum (- 4, 0), Maximum (0, 4), Minimum (4, 0)

Solution steps

f^{'} (x) = - \frac{x}{16 - x ^{2}}

Find intervals:

Increasing

: - 4 < x < 0,

Decreasing

: 0 < x < 4

Plug

x = - 4

into

16 - x^{2} : 0

Minimum (- 4, 0)

Plug

x = 0

into

16 - x^{2} : 4

Maximum (0, 4)

Plug

x = 4

into

16 - x^{2} : 0

Minimum (4, 0)

Minimum (- 4, 0), Maximum (0, 4), Minimum (4, 0)

2 x - 2 y = 8

\frac{x + 3}{4 x ^{2} + 3 x - 2}

f (x) = 0.09 x + 17 + \frac{350}{x}

\frac{( 7 + \frac{1}{x} )}{( \frac{1}{x} )}

(- 5, - 5), (- 9, - 2)

What is the extreme f(x)=sqrt(16-x^2) ?
The extreme f(x)=sqrt(16-x^2) is Minimum(-4,0),Maximum(0,4),Minimum(4,0)