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

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

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

extreme points

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

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

f (x) = 102 t + 0.5 t^{2} - t^{3}

f (x) = x^{3} - 8 x^{2} - 12 x + 5

f (x) = x^{3} - 8 x^{2} - 12 x + 9

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