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

The extreme f(x)= 1/3 x^3+2x^2+4x is Saddle(-2,-8/3)

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

extreme points

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

Saddle (- 2, - \frac{8}{3})

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 2,

Increasing

: - 2 < x < \infty

Plug

x = - 2

into

\frac{1}{3} x^{3} + 2 x^{2} + 4 x : - \frac{8}{3}

Saddle (- 2, - \frac{8}{3})

x^{5} - 5 x^{3}

y = \frac{- 2 x}{9 + x ^{2}}

f (x) = - x^{2} - 12 x - 27

f (x) = x 9 - x^{2}, - 3 \leq x \leq 3

f (x) = \frac{x ^{2} + 3}{x ^{3} - 9}

What is the extreme f(x)= 1/3 x^3+2x^2+4x ?
The extreme f(x)= 1/3 x^3+2x^2+4x is Saddle(-2,-8/3)