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

The extreme f(x)=x^4-4x^3-8x^2+8 is Minimum(-1,5),Maximum(0,8),Minimum(4,-120)

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

extreme points

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

Minimum (- 1, 5), Maximum (0, 8), Minimum (4, - 120)

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} + 8 : 5

Minimum (- 1, 5)

Plug

x = 0

into

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

Maximum (0, 8)

Plug

x = 4

into

x^{4} - 4 x^{3} - 8 x^{2} + 8 : - 120

Minimum (4, - 120)

Minimum (- 1, 5), Maximum (0, 8), Minimum (4, - 120)

f (x, y) = 6 - x - 2 y

2 x^{3} + 3 x^{2} - 36 x

f (x, y) = x^{2} + x y + y^{2} - 31 y + 320

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

What is the extreme f(x)=x^4-4x^3-8x^2+8 ?
The extreme f(x)=x^4-4x^3-8x^2+8 is Minimum(-1,5),Maximum(0,8),Minimum(4,-120)