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

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

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

extreme points

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

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

Solution steps

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

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

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

Plug

x = 1

into

ln^{5} (x) x : 0

Saddle (1, 0)

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

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

f (x) = \frac{ln ( 5 x )}{x}

f (x, y) = 3 x^{2} + 8 x y - 7 y^{2} - 3 x + 6 y

f (x, y) = x^{2} - 4 x - y^{2} - 6 y + 7

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

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