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

The extreme f(x)=x^3(x+2) is Minimum(-3/2 ,-27/16),Saddle(0,0)

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

extreme points

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

Minimum (- \frac{3}{2}, - \frac{27}{16}), Saddle (0, 0)

Solution steps

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

Find intervals:

Decreasing

: - \infty < x < - \frac{3}{2},

Increasing

: - \frac{3}{2} < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - \frac{3}{2}

into

x^{3} (x + 2) : - \frac{27}{16}

Minimum (- \frac{3}{2}, - \frac{27}{16})

Plug

x = 0

into

x^{3} (x + 2) : 0

Saddle (0, 0)

Minimum (- \frac{3}{2}, - \frac{27}{16}), Saddle (0, 0)

f (x) = 2 sin (x) - cos (2 x)

y = x^{3} - 3 x

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

w (x, y) = 3 x + 7 y

f (x) = - 2 x^{3} + 6 x^{2} - 3

What is the extreme f(x)=x^3(x+2) ?
The extreme f(x)=x^3(x+2) is Minimum(-3/2 ,-27/16),Saddle(0,0)