What is the extreme f(x)=(x-6)*(\sqrt[3]{x+3}) ?

The extreme f(x)=(x-6)*(\sqrt[3]{x+3}) is Minimum(-3/4 ,-(27\sqrt[3]{9})/(4\sqrt[3]{4)})

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

extreme f(x)=(x-6)*(\sqrt[3]{x+3})

extreme points

Solution steps

Find intervals:

Decreasing

: - \infty < x < - 3,

Decreasing

: - 3 < x < - \frac{3}{4},

Increasing

: - \frac{3}{4} < x < \infty

Plug

x = - \frac{3}{4}

into

f (x) = 9 - x^{2}, - 1 \leq x \leq 3

y = x^{\frac{5}{3}} - 5 x^{\frac{2}{3}}

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

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

ln (x + 2) (x - 1)^{2}

What is the extreme f(x)=(x-6)*(\sqrt[3]{x+3}) ?
The extreme f(x)=(x-6)*(\sqrt[3]{x+3}) is Minimum(-3/4 ,-(27\sqrt[3]{9})/(4\sqrt[3]{4)})