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

The extreme f(x)=x^3-2x^2-4x+4 is Maximum(-2/3 , 148/27),Minimum(2,-4)

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

extreme points

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

Maximum (- \frac{2}{3}, \frac{148}{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 + 4 : \frac{148}{27}

Maximum (- \frac{2}{3}, \frac{148}{27})

Plug

x = 2

into

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

Minimum (2, - 4)

Maximum (- \frac{2}{3}, \frac{148}{27}), Minimum (2, - 4)

(x + 1)^{5} - 5 x - 2

f (x) = \frac{2 x - 1}{x ^{2}}

f (x, y) = (2 x - x^{2}) \cdot (2 y - y^{2})

4 x - x^{2}

f (x) = - 5 x^{3} + 15 x + 8

What is the extreme f(x)=x^3-2x^2-4x+4 ?
The extreme f(x)=x^3-2x^2-4x+4 is Maximum(-2/3 , 148/27),Minimum(2,-4)