What is the extreme f(x)=sqrt(x)ln(7x),(0,infinity) ?

The extreme f(x)=sqrt(x)ln(7x),(0,infinity) is Minimum(1/(7e^2),-2/(sqrt(7)e))

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extreme f(x)=sqrt(x)ln(7x),(0,infinity)

extreme points

f (x) = x ln (7 x), (0, \infty)

Minimum (\frac{1}{7 e ^{2}}, - \frac{2}{7 e})

Solution steps

Find the critical points:

x = \frac{1}{7 e ^{2}}

Domain of

x ln (7 x), (0, \infty) : x > 0

Find intervals:

Decreasing

: 0 < x < \frac{1}{7 e ^{2}},

Increasing

: \frac{1}{7 e ^{2}} < x < \infty

Plug the extreme point

x = \frac{1}{7 e ^{2}}

into

x ln (7 x) \Rightarrow y = - \frac{2}{7 e}

Minimum (\frac{1}{7 e ^{2}}, - \frac{2}{7 e})

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

f (x, y) = x^{2} + 4 y^{2} - 2 x + 8 y + 4

2 x^{2} + 8 x + 12

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

(x - 9) ln (x - 9)

What is the extreme f(x)=sqrt(x)ln(7x),(0,infinity) ?
The extreme f(x)=sqrt(x)ln(7x),(0,infinity) is Minimum(1/(7e^2),-2/(sqrt(7)e))