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

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

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

extreme points

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

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

Solution steps

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

Find intervals:

Increasing

: - 2 < x < 0,

Decreasing

: 0 < x < 2

Plug

x = - 2

into

- 2 x^{2} + 4 : 0

Minimum (- 2, 0)

Plug

x = 0

into

- 2 x^{2} + 4 : 2

Maximum (0, 2)

Plug

x = 2

into

- 2 x^{2} + 4 : 0

Minimum (2, 0)

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

\frac{1 - 2 x}{x ^{2} - x}

x^{4} + x^{3} - 8 x^{2} - 12 x

f (x) = e^{x} (x - 9)

3 x^{2} + 2 x [- 2.1]

f (x) = x^{3} - 48 x y + 64 y^{3}

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