What is the extreme f(x)=\sqrt[3]{x^3+3x^2} ?

The extreme f(x)=\sqrt[3]{x^3+3x^2} is Maximum(-2,\sqrt[3]{4}),Minimum(0,0)

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

extreme f(x)=\sqrt[3]{x^3+3x^2}

extreme

Maximum (- 2, 34), Minimum (0, 0)

Solution steps

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

Find intervals:

Increasing

: - \infty < x < - 3,

Increasing

: - 3 < x < - 2,

Decreasing

: - 2 < x < 0,

Increasing

: 0 < x < \infty

Plug

x = - 2

into

3 x^{3} + 3 x^{2} : 34

Maximum (- 2, 34)

Plug

x = 0

into

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

Minimum (0, 0)

Maximum (- 2, 34), Minimum (0, 0)

What is the extreme f(x)=\sqrt[3]{x^3+3x^2} ?
The extreme f(x)=\sqrt[3]{x^3+3x^2} is Maximum(-2,\sqrt[3]{4}),Minimum(0,0)