What is the extreme f(x)= 1/9*((x^3)/(x+2)) ?

The extreme f(x)= 1/9*((x^3)/(x+2)) is Minimum(-3,3),Saddle(0,0)

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extreme f(x)= 1/9*((x^3)/(x+2))

extreme points

f (x) = \frac{1}{9} \cdot (\frac{x ^{3}}{x + 2})

Minimum (- 3, 3), Saddle (0, 0)

Solution steps

f^{'} (x) = \frac{2 x ^{3} + 6 x ^{2}}{9 ( x + 2 ) ^{2}}

Find intervals:

Decreasing

: - \infty < x < - 3,

Increasing

: - 3 < x < - 2,

Increasing

: - 2 < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - 3

into

\frac{1}{9} \cdot \frac{x ^{3}}{x + 2} : 3

Minimum (- 3, 3)

Plug

x = 0

into

\frac{1}{9} \cdot \frac{x ^{3}}{x + 2} : 0

Saddle (0, 0)

Minimum (- 3, 3), Saddle (0, 0)

f (x) = x^{3} - 5 x^{2} + 3 x + 10, - 2 \leq x \leq 4

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

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

ln (x^{4} + 27)

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

What is the extreme f(x)= 1/9*((x^3)/(x+2)) ?
The extreme f(x)= 1/9*((x^3)/(x+2)) is Minimum(-3,3),Saddle(0,0)