What is the extreme f(x)=2x^4-4x^2+3 ?

The extreme f(x)=2x^4-4x^2+3 is Minimum(-1,1),Maximum(0,3),Minimum(1,1)

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extreme f(x)=2x^4-4x^2+3

extreme points

f (x) = 2 x^{4} - 4 x^{2} + 3

Minimum (- 1, 1), Maximum (0, 3), Minimum (1, 1)

Solution steps

f^{'} (x) = 8 x^{3} - 8 x

Find intervals:

Decreasing

: - \infty < x < - 1,

Increasing

: - 1 < x < 0,

Decreasing

: 0 < x < 1,

Increasing

: 1 < x < \infty

Plug

x = - 1

into

2 x^{4} - 4 x^{2} + 3 : 1

Minimum (- 1, 1)

Plug

x = 0

into

2 x^{4} - 4 x^{2} + 3 : 3

Maximum (0, 3)

Plug

x = 1

into

2 x^{4} - 4 x^{2} + 3 : 1

Minimum (1, 1)

Minimum (- 1, 1), Maximum (0, 3), Minimum (1, 1)

What is the extreme f(x)=2x^4-4x^2+3 ?
The extreme f(x)=2x^4-4x^2+3 is Minimum(-1,1),Maximum(0,3),Minimum(1,1)