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

The extreme f(x)=4x^3ln(4x) is Minimum(1/(4\sqrt[3]{e)},-1/(48e))

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extreme f(x)=4x^3ln(4x)

extreme points

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

Solution steps

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

Find intervals:

Decreasing

Increasing

Plug

into

4 x^{3} ln (4 x) : - \frac{1}{48 e}

f (x) = 6 x + ln (x)

f (x, y) = x \cdot y^{2} - 6 \cdot x^{2} - 3 x + 2 \cdot x \cdot y + 9

U (x, y) = ln (1 + x y)

f (x) = \frac{- 5 x}{x ^{2} + 5}

f (x, y) = x^{3} + y^{2} - 6 x y + 9 x + 5 y + 2

What is the extreme f(x)=4x^3ln(4x) ?
The extreme f(x)=4x^3ln(4x) is Minimum(1/(4\sqrt[3]{e)},-1/(48e))