What is the extreme f(x)=sqrt(x)ln(9x),(0,infinity) ?

The extreme f(x)=sqrt(x)ln(9x),(0,infinity) is Minimum(1/(9e^2),-2/(3e))

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extreme f(x)=sqrt(x)ln(9x),(0,infinity)

extreme

Minimum (\frac{1}{9 e ^{2}}, - \frac{2}{3 e})

Solution steps

Find the critical points:

x = \frac{1}{9 e ^{2}}

Domain of

x ln (9 x), (0, \infty) : x > 0

Find intervals:

Decreasing

: 0 < x < \frac{1}{9 e ^{2}},

Increasing

: \frac{1}{9 e ^{2}} < x < \infty

Plug the extreme point

x = \frac{1}{9 e ^{2}}

into

x ln (9 x) \Rightarrow y = - \frac{2}{3 e}

Minimum (\frac{1}{9 e ^{2}}, - \frac{2}{3 e})

What is the extreme f(x)=sqrt(x)ln(9x),(0,infinity) ?
The extreme f(x)=sqrt(x)ln(9x),(0,infinity) is Minimum(1/(9e^2),-2/(3e))