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

Minimum (- \frac{3}{4}, - \frac{27 3 9}{4 3 4})

Solution steps

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

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 - 6) \cdot 3 x + 3 : - \frac{27 3 9}{4 3 4}

Minimum (- \frac{3}{4}, - \frac{27 3 9}{4 3 4})

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)})