What is the extreme f(x)=|x^2-1| ?

The extreme f(x)=|x^2-1| is Minimum(-1,0),Maximum(0,1),Minimum(1,0)

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extreme f(x)=|x^2-1|

extreme points

f (x) = x^{2} - 1

Minimum (- 1, 0), Maximum (0, 1), Minimum (1, 0)

Solution steps

Find the critical points:

x = - 1, x = 0, x = 1

Find intervals:

Decreasing

: - \infty < x < - 1,

Increasing

: - 1 < x < 0,

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < \infty

Plug the extreme point

x = - 1

into

x^{2} - 1 \Rightarrow y = 0

Minimum (- 1, 0)

Plug the extreme point

x = 0

into

x^{2} - 1 \Rightarrow y = 1

Maximum (0, 1)

Plug the extreme point

x = 1

into

x^{2} - 1 \Rightarrow y = 0

Minimum (1, 0)

Minimum (- 1, 0), Maximum (0, 1), Minimum (1, 0)

f (x) = sin (4 x)

f (x) = x + 2 y

f (x) = - x^{3} + 3 x - 7

f (x, y) = x^{2} + 4 y^{2} - 6 x + 16 y

f (x, y) = 8 x y - x^{3} - 4 y^{2}

What is the extreme f(x)=|x^2-1| ?
The extreme f(x)=|x^2-1| is Minimum(-1,0),Maximum(0,1),Minimum(1,0)