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

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

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

extreme points

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

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

Solution steps

f^{'} (x) = 4 x (x^{2} - 1)

Find intervals:

Decreasing

: - \infty < x < - 1,

Increasing

: - 1 < x < 0,

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = - 1

into

(x^{2} - 1)^{2} : 0

Minimum (- 1, 0)

Plug

x = 0

into

(x^{2} - 1)^{2} : 1

Maximum (0, 1)

Plug

x = 1

into

(x^{2} - 1)^{2} : 0

Minimum (1, 0)

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

f (x) = x y

f (x) = x^{3} - 6 x^{2} + 8

f (x, y) = y^{3} + 3 x^{2} y - 6 x^{2} - 6 y^{2} + 2

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

f (x, y) = e^{x + y}

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