What is the extreme f(x)=7x^3-7x^2+7 ?

The extreme f(x)=7x^3-7x^2+7 is Maximum(0,7),Minimum(2/3 , 161/27)

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extreme f(x)=7x^3-7x^2+7

extreme points

f (x) = 7 x^{3} - 7 x^{2} + 7

Maximum (0, 7), Minimum (\frac{2}{3}, \frac{161}{27})

Solution steps

f^{'} (x) = 21 x^{2} - 14 x

Find intervals:

Increasing

: - \infty < x < 0,

Decreasing

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

Increasing

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

Plug

x = 0

into

7 x^{3} - 7 x^{2} + 7 : 7

Maximum (0, 7)

Plug

x = \frac{2}{3}

into

7 x^{3} - 7 x^{2} + 7 : \frac{161}{27}

Minimum (\frac{2}{3}, \frac{161}{27})

Maximum (0, 7), Minimum (\frac{2}{3}, \frac{161}{27})

What is the extreme f(x)=7x^3-7x^2+7 ?
The extreme f(x)=7x^3-7x^2+7 is Maximum(0,7),Minimum(2/3 , 161/27)