What is the extreme f(x)=320x^2-2560x^3 ?

The extreme f(x)=320x^2-2560x^3 is Minimum(0,0),Maximum(1/12 , 20/27)

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extreme f(x)=320x^2-2560x^3

extreme points

f (x) = 320 x^{2} - 2560 x^{3}

Minimum (0, 0), Maximum (\frac{1}{12}, \frac{20}{27})

Solution steps

f^{'} (x) = 640 x - 7680 x^{2}

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \frac{1}{12},

Decreasing

: \frac{1}{12} < x < \infty

Plug

x = 0

into

320 x^{2} - 2560 x^{3} : 0

Minimum (0, 0)

Plug

x = \frac{1}{12}

into

320 x^{2} - 2560 x^{3} : \frac{20}{27}

Maximum (\frac{1}{12}, \frac{20}{27})

Minimum (0, 0), Maximum (\frac{1}{12}, \frac{20}{27})

f (x, y) = x^{2} - 4 x + y^{2} - 6 y - 2

f (x) = 4 x^{3} - x^{2} + 9

x^{2} - 6 x + 12

f (x) = \cdot f (y) - f (x, y) (x y) = 3 x + 3 y + 6

f (x) = 3 - sin (3 x)

What is the extreme f(x)=320x^2-2560x^3 ?
The extreme f(x)=320x^2-2560x^3 is Minimum(0,0),Maximum(1/12 , 20/27)