What is the extreme g(x)=x^3-2x^2-4x+8 ?

The extreme g(x)=x^3-2x^2-4x+8 is Maximum(-2/3 , 256/27),Minimum(2,0)

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extreme g(x)=x^3-2x^2-4x+8

extreme points

g (x) = x^{3} - 2 x^{2} - 4 x + 8

Maximum (- \frac{2}{3}, \frac{256}{27}), Minimum (2, 0)

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 + 8 : \frac{256}{27}

Maximum (- \frac{2}{3}, \frac{256}{27})

Plug

x = 2

into

x^{3} - 2 x^{2} - 4 x + 8 : 0

Minimum (2, 0)

Maximum (- \frac{2}{3}, \frac{256}{27}), Minimum (2, 0)

f (x) = x^{3} + 6 x^{2} - 63 x + 5, - 8 \leq x \leq 0

f (x) = 3 x^{2} - 6 x + 24

f (x) = (9 x^{2} + 4 x + 5) (9 y^{2} + 6 y + 6)

f (x) = \frac{4860}{x} + 12 x + 709407

f (x) = x^{3} + 6 x^{2} - 63 x + 12, - 5 \leq x \leq 4

What is the extreme g(x)=x^3-2x^2-4x+8 ?
The extreme g(x)=x^3-2x^2-4x+8 is Maximum(-2/3 , 256/27),Minimum(2,0)