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

The extreme f(x)=(2x^2-2)^2 is Minimum(-1,0),Maximum(0,4),Minimum(1,0)

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

extreme

Minimum (- 1, 0), Maximum (0, 4), Minimum (1, 0)

Solution steps

f^{'} (x) = 8 x (2 x^{2} - 2)

Find intervals:

Decreasing

: - \infty < x < - 1,

Increasing

: - 1 < x < 0,

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = - 1

into

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

Minimum (- 1, 0)

Plug

x = 0

into

(2 x^{2} - 2)^{2} : 4

Maximum (0, 4)

Plug

x = 1

into

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

Minimum (1, 0)

Minimum (- 1, 0), Maximum (0, 4), Minimum (1, 0)

What is the extreme f(x)=(2x^2-2)^2 ?
The extreme f(x)=(2x^2-2)^2 is Minimum(-1,0),Maximum(0,4),Minimum(1,0)