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

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

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

extreme points

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

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

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 2,

Decreasing

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

Increasing

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

Plug

x = - 2

into

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

Maximum (- 2, 0)

Plug

x = \frac{2}{3}

into

x^{3} + 2 x^{2} - 4 x - 8 : - \frac{256}{27}

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

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

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f (x) = \frac{( x + 7 )}{x - 3}

\frac{x - 3}{7}

f (x) = \frac{x + 14}{x - 11}

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