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

The extreme f(x)=x^2(x+3)^4 is Minimum(-3,0),Maximum(-1,16),Minimum(0,0)

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

extreme points

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

Minimum (- 3, 0), Maximum (- 1, 16), Minimum (0, 0)

Solution steps

f^{'} (x) = 6 x (x + 3)^{3} (x + 1)

Find intervals:

Decreasing

: - \infty < x < - 3,

Increasing

: - 3 < x < - 1,

Decreasing

: - 1 < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - 3

into

x^{2} (x + 3)^{4} : 0

Minimum (- 3, 0)

Plug

x = - 1

into

x^{2} (x + 3)^{4} : 16

Maximum (- 1, 16)

Plug

x = 0

into

x^{2} (x + 3)^{4} : 0

Minimum (0, 0)

Minimum (- 3, 0), Maximum (- 1, 16), Minimum (0, 0)

f (y) = 6 - 3 x - 2 y

y^{2} - 120 y

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

y = ln (x + 2) + \frac{4}{x + 2}

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

What is the extreme f(x)=x^2(x+3)^4 ?
The extreme f(x)=x^2(x+3)^4 is Minimum(-3,0),Maximum(-1,16),Minimum(0,0)