What is the extreme f(x)=sqrt(49-x^2) ?

The extreme f(x)=sqrt(49-x^2) is Minimum(-7,0),Maximum(0,7),Minimum(7,0)

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extreme f(x)=sqrt(49-x^2)

extreme points

f (x) = 49 - x^{2}

Minimum (- 7, 0), Maximum (0, 7), Minimum (7, 0)

Solution steps

f^{'} (x) = - \frac{x}{49 - x ^{2}}

Find intervals:

Increasing

: - 7 < x < 0,

Decreasing

: 0 < x < 7

Plug

x = - 7

into

49 - x^{2} : 0

Minimum (- 7, 0)

Plug

x = 0

into

49 - x^{2} : 7

Maximum (0, 7)

Plug

x = 7

into

49 - x^{2} : 0

Minimum (7, 0)

Minimum (- 7, 0), Maximum (0, 7), Minimum (7, 0)

f (x) = \frac{20 x ^{2}}{x ^{4} + 100}

f (x) = \frac{2 x ^{3} + 3}{x ^{3} - 1}

\frac{4 x + 20}{- x ^{2} - 5 x}

f (x) = \frac{2 a ^{3}}{17}

\frac{9 x}{16 - x ^{2}}

What is the extreme f(x)=sqrt(49-x^2) ?
The extreme f(x)=sqrt(49-x^2) is Minimum(-7,0),Maximum(0,7),Minimum(7,0)