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

The extreme f(x)=x^3(x+4) is Minimum(-3,-27),Saddle(0,0)

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

extreme points

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

Minimum (- 3, - 27), Saddle (0, 0)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - 3,

Increasing

: - 3 < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - 3

into

x^{3} (x + 4) : - 27

Minimum (- 3, - 27)

Plug

x = 0

into

x^{3} (x + 4) : 0

Saddle (0, 0)

Minimum (- 3, - 27), Saddle (0, 0)

f (x, z) = x^{3} + z^{3} - 3 x - 3 z

f (x) = \frac{1}{3} x^{3} - x^{2} - 3 x + 5

f (x, y) = x^{3} + y^{3} - 27 x y

f (x) = x^{3} + x^{2} - 5 x - 5

f (x, y) = 5 x^{2} + 3 x y + e^{x y}

What is the extreme f(x)=x^3(x+4) ?
The extreme f(x)=x^3(x+4) is Minimum(-3,-27),Saddle(0,0)