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)})

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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

(x + 2)^{\frac{2}{3}}

f (x) = x^{3} - 3 x^{2} - 24 x

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

f (x) = 9 x^{3} - \frac{1}{x ^{3}}

x e^{- 3 x}

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)})