What is the extreme f(x)=-4x^5ln(2x) ?

The extreme f(x)=-4x^5ln(2x) is Maximum(1/(2\sqrt[5]{e)}, 1/(40e))

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extreme f(x)=-4x^5ln(2x)

extreme points

f (x) = - 4 x^{5} ln (2 x)

Solution steps

f^{'} (x) = - 4 (5 x^{4} ln (2 x) + x^{4})

Find intervals:

Increasing

Decreasing

Plug

into

- 4 x^{5} ln (2 x) : \frac{1}{40 e}

f (x) = \frac{x ^{3}}{x} - 3 x^{2} - 7 x

y = x^{3} + 3 x^{2} + 3 x + 1

f (x) = \frac{8 x - 3}{x}

1 + \frac{7}{x} - \frac{5}{x ^{2}}

f (x) = 4 + (4 + 3 x)^{\frac{2}{3}}

What is the extreme f(x)=-4x^5ln(2x) ?
The extreme f(x)=-4x^5ln(2x) is Maximum(1/(2\sqrt[5]{e)}, 1/(40e))