What is the extreme f(x)=340x^2-2040x^3 ?

The extreme f(x)=340x^2-2040x^3 is Minimum(0,0),Maximum(1/9 , 340/243)

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extreme f(x)=340x^2-2040x^3

extreme

Minimum (0, 0), Maximum (\frac{1}{9}, \frac{340}{243})

Solution steps

f^{'} (x) = 680 x - 6120 x^{2}

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

: 0 < x < \frac{1}{9},

Decreasing

: \frac{1}{9} < x < \infty

Plug

x = 0

into

340 x^{2} - 2040 x^{3} : 0

Minimum (0, 0)

Plug

x = \frac{1}{9}

into

340 x^{2} - 2040 x^{3} : \frac{340}{243}

Maximum (\frac{1}{9}, \frac{340}{243})

Minimum (0, 0), Maximum (\frac{1}{9}, \frac{340}{243})

What is the extreme f(x)=340x^2-2040x^3 ?
The extreme f(x)=340x^2-2040x^3 is Minimum(0,0),Maximum(1/9 , 340/243)