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

The extreme f(x)=3x^3-3x^2-4 is Maximum(0,-4),Minimum(2/3 ,-40/9)

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

extreme points

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

Maximum (0, - 4), Minimum (\frac{2}{3}, - \frac{40}{9})

Solution steps

f^{'} (x) = 9 x^{2} - 6 x

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

: 0 < x < \frac{2}{3},

Increasing

: \frac{2}{3} < x < \infty

Plug

x = 0

into

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

Maximum (0, - 4)

Plug

x = \frac{2}{3}

into

3 x^{3} - 3 x^{2} - 4 : - \frac{40}{9}

Minimum (\frac{2}{3}, - \frac{40}{9})

Maximum (0, - 4), Minimum (\frac{2}{3}, - \frac{40}{9})

What is the extreme f(x)=3x^3-3x^2-4 ?
The extreme f(x)=3x^3-3x^2-4 is Maximum(0,-4),Minimum(2/3 ,-40/9)