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

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

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

extreme points

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

Maximum (0, - 4), Minimum (2, - 8)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < 2,

Increasing

: 2 < x < \infty

Plug

x = 0

into

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

Maximum (0, - 4)

Plug

x = 2

into

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

Minimum (2, - 8)

Maximum (0, - 4), Minimum (2, - 8)

4 x^{2} y + 2 x y^{2} - 12 x y - 5

K (r, s) = 4 r - s

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

f (x) = 3 x^{4} + 4 x^{3} - 12 x^{2} + 7

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

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