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

The extreme f(x)=3x^3-3x^2-3x+7 is Maximum(-1/3 , 68/9),Minimum(1,4)

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

extreme points

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

Maximum (- \frac{1}{3}, \frac{68}{9}), Minimum (1, 4)

Solution steps

f^{'} (x) = 9 x^{2} - 6 x - 3

Find intervals:

Increasing

: - \infty < x < - \frac{1}{3},

Decreasing

: - \frac{1}{3} < x < 1,

Increasing

: 1 < x < \infty

Plug

x = - \frac{1}{3}

into

3 x^{3} - 3 x^{2} - 3 x + 7 : \frac{68}{9}

Maximum (- \frac{1}{3}, \frac{68}{9})

Plug

x = 1

into

3 x^{3} - 3 x^{2} - 3 x + 7 : 4

Minimum (1, 4)

Maximum (- \frac{1}{3}, \frac{68}{9}), Minimum (1, 4)

What is the extreme f(x)=3x^3-3x^2-3x+7 ?
The extreme f(x)=3x^3-3x^2-3x+7 is Maximum(-1/3 , 68/9),Minimum(1,4)