What is the extreme f(x)=sqrt(-x^2+4) ?

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

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extreme f(x)=sqrt(-x^2+4)

extreme points

f (x) = - x^{2} + 4

Minimum (- 2, 0), Maximum (0, 2), Minimum (2, 0)

Solution steps

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

Find intervals:

Increasing

: - 2 < x < 0,

Decreasing

: 0 < x < 2

Plug

x = - 2

into

- x^{2} + 4 : 0

Minimum (- 2, 0)

Plug

x = 0

into

- x^{2} + 4 : 2

Maximum (0, 2)

Plug

x = 2

into

- x^{2} + 4 : 0

Minimum (2, 0)

Minimum (- 2, 0), Maximum (0, 2), Minimum (2, 0)

f (x) = 2.4

f (x) = 2 - x

P (x, y) = x^{2} + 5 x y + 4 y^{2} + 2 x + 5 y + 1

f (x) = 2 x^{3} - 24 x, - 4 \leq x \leq 4

\frac{1}{4} x^{4} + \frac{1}{2} x^{3} - 3 x^{2} + 8

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