What is the extreme y=x^4-4x^2=x^2(x^2-4) ?

The extreme y=x^4-4x^2=x^2(x^2-4) is Minimum(-sqrt(2),-4),Maximum(0,0),Minimum(sqrt(2),-4)

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extreme y=x^4-4x^2=x^2(x^2-4)

extreme points

y = x^{4} - 4 x^{2} = x^{2} (x^{2} - 4)

Minimum (- 2, - 4), Maximum (0, 0), Minimum (2, - 4)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 2,

Increasing

: - 2 < x < 0,

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = - 2

into

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

Minimum (- 2, - 4)

Plug

x = 0

into

x^{2} (x^{2} - 4) : 0

Maximum (0, 0)

Plug

x = 2

into

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

Minimum (2, - 4)

Minimum (- 2, - 4), Maximum (0, 0), Minimum (2, - 4)

f (x) = 2 x^{3} + 3 x^{2} - 12 x + 7

- x + 2 cos (x)

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

f (x) = cos (x) - 3 x

f (x) = 3 x^{2} - 24 ln (x)

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