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

The extreme f(x)=x^3+2x^2-4x is Maximum(-2,8),Minimum(2/3 ,-40/27)

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

extreme points

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

Maximum (- 2, 8), Minimum (\frac{2}{3}, - \frac{40}{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

Maximum (- 2, 8)

Plug

x = \frac{2}{3}

into

x^{3} + 2 x^{2} - 4 x : - \frac{40}{27}

Minimum (\frac{2}{3}, - \frac{40}{27})

Maximum (- 2, 8), Minimum (\frac{2}{3}, - \frac{40}{27})

- x^{2} + 1

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

tan (2 x)

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

f (x) = \frac{4}{x} + 2

What is the extreme f(x)=x^3+2x^2-4x ?
The extreme f(x)=x^3+2x^2-4x is Maximum(-2,8),Minimum(2/3 ,-40/27)