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

The extreme f(x)=x^3+2x^2-4x+8 is Maximum(-2,16),Minimum(2/3 , 176/27)

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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, 16), Minimum (\frac{2}{3}, \frac{176}{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 : 16

Maximum (- 2, 16)

Plug

x = \frac{2}{3}

into

x^{3} + 2 x^{2} - 4 x + 8 : \frac{176}{27}

Minimum (\frac{2}{3}, \frac{176}{27})

Maximum (- 2, 16), Minimum (\frac{2}{3}, \frac{176}{27})

What is the extreme f(x)=x^3+2x^2-4x+8 ?
The extreme f(x)=x^3+2x^2-4x+8 is Maximum(-2,16),Minimum(2/3 , 176/27)