What is the extreme f(x)=sqrt(x)ln(8x) ?

The extreme f(x)=sqrt(x)ln(8x) is Minimum(1/(8e^2),-1/(sqrt(2)e))

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extreme f(x)=sqrt(x)ln(8x)

extreme points

f (x) = x ln (8 x)

Minimum (\frac{1}{8 e ^{2}}, - \frac{1}{2 e})

Solution steps

f^{'} (x) = \frac{ln ( 8 x )}{2 x} + \frac{1}{x}

Find intervals:

Decreasing

: 0 < x < \frac{1}{8 e ^{2}},

Increasing

: \frac{1}{8 e ^{2}} < x < \infty

Plug

x = \frac{1}{8 e ^{2}}

into

x ln (8 x) : - \frac{1}{2 e}

Minimum (\frac{1}{8 e ^{2}}, - \frac{1}{2 e})

x 7 - x

f (x) = 2 cos (2 x) - 2

2 x e^{- x^{2}}

R (x, y) = - 5 x^{2} - 8 y^{2} - 2 x y + 42 x + 102 y

f (x) = ln (x) + x^{2}

What is the extreme f(x)=sqrt(x)ln(8x) ?
The extreme f(x)=sqrt(x)ln(8x) is Minimum(1/(8e^2),-1/(sqrt(2)e))