What is the extreme y=x^4-6x^2=x^2(x^2-6) ?

The extreme y=x^4-6x^2=x^2(x^2-6) is Minimum(-sqrt(3),-9),Maximum(0,0),Minimum(sqrt(3),-9)

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extreme y=x^4-6x^2=x^2(x^2-6)

extreme points

y = x^{4} - 6 x^{2} = x^{2} (x^{2} - 6)

Minimum (- 3, - 9), Maximum (0, 0), Minimum (3, - 9)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 3,

Increasing

: - 3 < x < 0,

Decreasing

: 0 < x < 3,

Increasing

: 3 < x < \infty

Plug

x = - 3

into

x^{2} (x^{2} - 6) : - 9

Minimum (- 3, - 9)

Plug

x = 0

into

x^{2} (x^{2} - 6) : 0

Maximum (0, 0)

Plug

x = 3

into

x^{2} (x^{2} - 6) : - 9

Minimum (3, - 9)

Minimum (- 3, - 9), Maximum (0, 0), Minimum (3, - 9)

f (x) = 5 + 3 x - 3 x^{2}

x^{4} - 16

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

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

f (x) = - x^{\frac{2}{3}} (x - 1)

What is the extreme y=x^4-6x^2=x^2(x^2-6) ?
The extreme y=x^4-6x^2=x^2(x^2-6) is Minimum(-sqrt(3),-9),Maximum(0,0),Minimum(sqrt(3),-9)