What is the extreme f(x)=395x^2-3160x^3 ?

The extreme f(x)=395x^2-3160x^3 is Minimum(0,0),Maximum(1/12 , 395/432)

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extreme f(x)=395x^2-3160x^3

extreme

Minimum (0, 0), Maximum (\frac{1}{12}, \frac{395}{432})

Solution steps

f^{'} (x) = 790 x - 9480 x^{2}

Find intervals:

Decreasing

: - \infty < x < 0,

Increasing

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

Decreasing

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

Plug

x = 0

into

395 x^{2} - 3160 x^{3} : 0

Minimum (0, 0)

Plug

x = \frac{1}{12}

into

395 x^{2} - 3160 x^{3} : \frac{395}{432}

Maximum (\frac{1}{12}, \frac{395}{432})

Minimum (0, 0), Maximum (\frac{1}{12}, \frac{395}{432})

What is the extreme f(x)=395x^2-3160x^3 ?
The extreme f(x)=395x^2-3160x^3 is Minimum(0,0),Maximum(1/12 , 395/432)