What is the extreme f(x)=x^3-2x^2-4x+12 ?

The extreme f(x)=x^3-2x^2-4x+12 is Maximum(-2/3 , 364/27),Minimum(2,4)

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extreme f(x)=x^3-2x^2-4x+12

extreme points

f (x) = x^{3} - 2 x^{2} - 4 x + 12

Maximum (- \frac{2}{3}, \frac{364}{27}), Minimum (2, 4)

Solution steps

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

Find intervals:

Increasing

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

Decreasing

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

Increasing

: 2 < x < \infty

Plug

x = - \frac{2}{3}

into

x^{3} - 2 x^{2} - 4 x + 12 : \frac{364}{27}

Maximum (- \frac{2}{3}, \frac{364}{27})

Plug

x = 2

into

x^{3} - 2 x^{2} - 4 x + 12 : 4

Minimum (2, 4)

Maximum (- \frac{2}{3}, \frac{364}{27}), Minimum (2, 4)

f (x) = - 5 cos (3 x) - 2

f (x) = x^{3} - 9 x^{2} + 24 x - 12

f (x) = \frac{x ^{3}}{3} - 6 x^{2} + 32 x

y^{2} - x^{2}

f (x) = 4 x - 5 x^{\frac{4}{5}}

What is the extreme f(x)=x^3-2x^2-4x+12 ?
The extreme f(x)=x^3-2x^2-4x+12 is Maximum(-2/3 , 364/27),Minimum(2,4)