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

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

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

extreme points

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

Minimum (- 3, 0)

Solution steps

f^{'} (x) = \frac{12 ( x + 3 )}{( x + 1 ) ^{4}}

Find intervals:

Decreasing

: - \infty < x < - 3,

Increasing

: - 3 < x < - 1,

Increasing

: - 1 < x < \infty

Plug

x = - 3

into

\frac{( x - 3 ) ( x + 3 ) ^{2}}{2 ( x + 1 ) ^{3}} : 0

Minimum (- 3, 0)

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

f (x) = (5 - x) (x + 1)^{2}

f (x, y) = 4 x^{2} - 12 x + y^{2} + 2 y - 10

x^{4} - 8 x^{2} + 1

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

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