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

The extreme f(x)=6x^2-3x^3 is Minimum(0,0),Maximum(4/3 , 32/9)

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

extreme points

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

Minimum (0, 0), Maximum (\frac{4}{3}, \frac{32}{9})

Solution steps

f^{'} (x) = 12 x - 9 x^{2}

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \frac{4}{3},

Decreasing

: \frac{4}{3} < x < \infty

Plug

x = 0

into

6 x^{2} - 3 x^{3} : 0

Minimum (0, 0)

Plug

x = \frac{4}{3}

into

6 x^{2} - 3 x^{3} : \frac{32}{9}

Maximum (\frac{4}{3}, \frac{32}{9})

Minimum (0, 0), Maximum (\frac{4}{3}, \frac{32}{9})

f (t) = 2 x

f (x, y) = 12 x y^{2} - 16 x + 25 x y^{3} - 18 x^{2} y^{3}

f (x, y) = \frac{1}{3} x^{3} + \frac{1}{3} y^{3} - x y + 4

f (x) = y^{2} - x^{2}

f (x) = \frac{1}{x - 2} - 3

What is the extreme f(x)=6x^2-3x^3 ?
The extreme f(x)=6x^2-3x^3 is Minimum(0,0),Maximum(4/3 , 32/9)