What is the extreme f(x)=x^3-6x^2+8 ?

The extreme f(x)=x^3-6x^2+8 is Maximum(0,8),Minimum(4,-24)

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extreme f(x)=x^3-6x^2+8

extreme points

f (x) = x^{3} - 6 x^{2} + 8

Maximum (0, 8), Minimum (4, - 24)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 4,

Increasing

: 4 < x < \infty

Plug

x = 0

into

x^{3} - 6 x^{2} + 8 : 8

Maximum (0, 8)

Plug

x = 4

into

x^{3} - 6 x^{2} + 8 : - 24

Minimum (4, - 24)

Maximum (0, 8), Minimum (4, - 24)

f (x, y) = y^{3} + 3 x^{2} y - 6 x^{2} - 6 y^{2} + 2

2 x^{3} + 3 x^{2} - 12 x

f (x, y) = e^{x + y}

f (x) = 2 x^{3} + 3 x^{2} - 120 x

f (x) = - 4 x^{3} + 3 x^{2} + 18 x

What is the extreme f(x)=x^3-6x^2+8 ?
The extreme f(x)=x^3-6x^2+8 is Maximum(0,8),Minimum(4,-24)