What is the extreme f(x)=x^4-4x^3-8x^2 ?

The extreme f(x)=x^4-4x^3-8x^2 is Minimum(-1,-3),Maximum(0,0),Minimum(4,-128)

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extreme f(x)=x^4-4x^3-8x^2

extreme points

f (x) = x^{4} - 4 x^{3} - 8 x^{2}

Minimum (- 1, - 3), Maximum (0, 0), Minimum (4, - 128)

Solution steps

f^{'} (x) = 4 x^{3} - 12 x^{2} - 16 x

Find intervals:

Decreasing

: - \infty < x < - 1,

Increasing

: - 1 < x < 0,

Decreasing

: 0 < x < 4,

Increasing

: 4 < x < \infty

Plug

x = - 1

into

x^{4} - 4 x^{3} - 8 x^{2} : - 3

Minimum (- 1, - 3)

Plug

x = 0

into

x^{4} - 4 x^{3} - 8 x^{2} : 0

Maximum (0, 0)

Plug

x = 4

into

x^{4} - 4 x^{3} - 8 x^{2} : - 128

Minimum (4, - 128)

Minimum (- 1, - 3), Maximum (0, 0), Minimum (4, - 128)

f (x, y) = x^{3} ln (y^{2} + 1) - 2 y

f (x) = (\frac{x}{5})^{1} \cdot (\frac{y}{5})^{x - 1}

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

f (x) = x^{3} - 3 x^{2} + 10

f (x) = (x^{2} - 3) e^{x}

What is the extreme f(x)=x^4-4x^3-8x^2 ?
The extreme f(x)=x^4-4x^3-8x^2 is Minimum(-1,-3),Maximum(0,0),Minimum(4,-128)