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

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

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extreme f(x)=|2-x^2|

extreme points

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

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

Solution steps

Find the critical points:

x = - 2, x = 0, x = 2

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < 0,

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug the extreme point

x = - 2

into

2 - x^{2} \Rightarrow y = 0

Minimum (- 2, 0)

Plug the extreme point

x = 0

into

2 - x^{2} \Rightarrow y = 2

Maximum (0, 2)

Plug the extreme point

x = 2

into

2 - x^{2} \Rightarrow y = 0

Minimum (2, 0)

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

x^{4} - 32 x^{2}

f (x, y) = 16 + (x - 2)^{2} - 4 (y + 1)^{2}

f (x, y) = x e^{- 2 x^{2} - 2 y^{2}}

f (x) = 3 x^{4} - 4 x^{3} - 8, - 1 \leq x \leq 2

4 x^{4} + 4 y^{4} - 2 x y

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