What is the extreme f(x)=x^3-2x^2-4x+1 ?

The extreme f(x)=x^3-2x^2-4x+1 is Maximum(-2/3 , 67/27),Minimum(2,-7)

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extreme f(x)=x^3-2x^2-4x+1

extreme points

f (x) = x^{3} - 2 x^{2} - 4 x + 1

Maximum (- \frac{2}{3}, \frac{67}{27}), Minimum (2, - 7)

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 + 1 : \frac{67}{27}

Maximum (- \frac{2}{3}, \frac{67}{27})

Plug

x = 2

into

x^{3} - 2 x^{2} - 4 x + 1 : - 7

Minimum (2, - 7)

Maximum (- \frac{2}{3}, \frac{67}{27}), Minimum (2, - 7)

f (x, y) = 3 x^{2} + 2 y^{2} - 6 x - 4 y + 16

f (x) = 2 x^{5} - 4 x^{2} + 5 x - 3

2^{x}

y = (4 - x) 4^{x}

- 8 t^{2} + 36 t + 99

What is the extreme f(x)=x^3-2x^2-4x+1 ?
The extreme f(x)=x^3-2x^2-4x+1 is Maximum(-2/3 , 67/27),Minimum(2,-7)