What is the extreme f(x)=2x^4-4x^2+1 ?

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

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extreme f(x)=2x^4-4x^2+1

extreme points

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

Minimum (- 1, - 1), Maximum (0, 1), 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} + 1 : - 1

Minimum (- 1, - 1)

Plug

x = 0

into

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

Maximum (0, 1)

Plug

x = 1

into

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

Minimum (1, - 1)

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

f (x) = 6 x^{5} - 160 x^{3} + 19

f (x) = 2 x^{3} - 9 x^{2} - 24 x + 1

f (x) = \frac{8 x ^{3}}{3} - 5 x^{2} - 3 x + 4

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

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

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