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

The extreme f(x)=x(ln(x))^5 is Minimum(1/(e^5),-(3125)/(e^5)),Saddle(1,0)

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extreme f(x)=x(ln(x))^5

extreme points

f (x) = x (ln (x))^{5}

Minimum (\frac{1}{e ^{5}}, - \frac{3125}{e ^{5}}), Saddle (1, 0)

Solution steps

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

Find intervals:

Decreasing

: 0 < x < \frac{1}{e ^{5}},

Increasing

: \frac{1}{e ^{5}} < x < 1,

Increasing

: 1 < x < \infty

Plug

x = \frac{1}{e ^{5}}

into

x ln^{5} (x) : - \frac{3125}{e ^{5}}

Minimum (\frac{1}{e ^{5}}, - \frac{3125}{e ^{5}})

Plug

x = 1

into

x ln^{5} (x) : 0

Saddle (1, 0)

Minimum (\frac{1}{e ^{5}}, - \frac{3125}{e ^{5}}), Saddle (1, 0)

f (x, y) = \frac{e ^{(- x^{2} + y^{2})} ( x + y ^{2} ) + 1}{2}

f (x) = x^{3} - 4500 x^{2} + 6 \cdot 1 0^{6} x

f (x, y) = e^{- (x^{2} + y^{2} - 6 y)}

f (x) = - x^{3} (x - 3)

f (x) = - x^{3} + 3 x + 5

What is the extreme f(x)=x(ln(x))^5 ?
The extreme f(x)=x(ln(x))^5 is Minimum(1/(e^5),-(3125)/(e^5)),Saddle(1,0)