What is the extreme f(x)=sqrt(x)ln(8x),(0,infinity) ?

The extreme f(x)=sqrt(x)ln(8x),(0,infinity) is Minimum(1/(8e^2),-1/(sqrt(2)e))

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extreme f(x)=sqrt(x)ln(8x),(0,infinity)

extreme points

f (x) = x ln (8 x), (0, \infty)

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

Solution steps

Find the critical points:

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

Domain of

x ln (8 x), (0, \infty) : x > 0

Find intervals:

Decreasing

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

Increasing

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

Plug the extreme point

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

into

x ln (8 x) \Rightarrow y = - \frac{1}{2 e}

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

f (x, y) = 3 x^{3} - 12 x y + y^{3}

f (x) = 19 + 2 x - x^{2}, 0 \leq x \leq 5

f (x) = 2 x^{3} + 3 x^{2} - 2 x - 0

f (x) = 2 x^{3} + 3 x^{2} - 2 x - 1

f (x) = \frac{4 x ^{2} + 12}{x - 1}

What is the extreme f(x)=sqrt(x)ln(8x),(0,infinity) ?
The extreme f(x)=sqrt(x)ln(8x),(0,infinity) is Minimum(1/(8e^2),-1/(sqrt(2)e))