What is the extreme f(x)=-x^7ln(3x),(0,4) ?

The extreme f(x)=-x^7ln(3x),(0,4) is Maximum(1/(3\sqrt[7]{e)}, 1/(15309e))

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extreme f(x)=-x^7ln(3x),(0,4)

extreme points

f (x) = - x^{7} ln (3 x), (0, 4)

Solution steps

Find the critical points:

Domain of

- x^{7} ln (3 x), (0, 4) : 0 < x < 4

Find intervals:

Increasing

Decreasing

Plug the extreme point

into

- x^{7} ln (3 x) \Rightarrow y = \frac{1}{15309 e}

\frac{x ^{2}}{- x ^{2} + 4}

y = x^{x} I n (x^{5} + x)

f (x) = 2 x^{2} - 4 x + y^{2} - 6 y + 3

\frac{x + 2}{x ^{2} - x - 6}

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

What is the extreme f(x)=-x^7ln(3x),(0,4) ?
The extreme f(x)=-x^7ln(3x),(0,4) is Maximum(1/(3\sqrt[7]{e)}, 1/(15309e))