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

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

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extreme f(x)=|x^2-4|

extreme

Minimum (- 2, 0), Maximum (0, 4), 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

x^{2} - 4 \Rightarrow y = 0

Minimum (- 2, 0)

Plug the extreme point

x = 0

into

x^{2} - 4 \Rightarrow y = 4

Maximum (0, 4)

Plug the extreme point

x = 2

into

x^{2} - 4 \Rightarrow y = 0

Minimum (2, 0)

Minimum (- 2, 0), Maximum (0, 4), Minimum (2, 0)

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