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

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

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

extreme points

f (x) = x ln (x)

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

Solution steps

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

Find intervals:

Decreasing

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

Increasing

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

Plug

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

into

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

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

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

f (t) = e^{a t}

f (x) = x^{5} - 5 x^{3} - 20 x - 2

f (x) = 3 x^{4} - 8 x^{3} + 6 x^{2}

f (x) = x^{3} - 6 x^{2} + 9 x - 8

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