What is the extreme f(x)=-4x^6ln(3x) ?

The extreme f(x)=-4x^6ln(3x) is Maximum(1/(3\sqrt[6]{e)}, 2/(2187e))

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extreme f(x)=-4x^6ln(3x)

extreme points

f (x) = - 4 x^{6} ln (3 x)

Maximum (\frac{1}{3 6 e}, \frac{2}{2187 e})

Solution steps

f^{'} (x) = - 4 (6 x^{5} ln (3 x) + x^{5})

Find intervals:

Increasing

: 0 < x < \frac{1}{3 6 e},

Decreasing

: \frac{1}{3 6 e} < x < \infty

Plug

x = \frac{1}{3 6 e}

into

- 4 x^{6} ln (3 x) : \frac{2}{2187 e}

Maximum (\frac{1}{3 6 e}, \frac{2}{2187 e})

f (x) = - x^{2} + 100 x + 8

f (x) = - x^{3} + 4 x^{2} + 16 x - 7

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

f (x) = - 7 x^{3} + 21 x + 4

x^{3} - 2 x^{2} - 4 x + 4

What is the extreme f(x)=-4x^6ln(3x) ?
The extreme f(x)=-4x^6ln(3x) is Maximum(1/(3\sqrt[6]{e)}, 2/(2187e))