What is the extreme f(x)=x^2ln(x/6) ?

The extreme f(x)=x^2ln(x/6) is Minimum(6/(sqrt(e)),-(18)/e)

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extreme f(x)=x^2ln(x/6)

extreme points

f (x) = x^{2} ln (\frac{x}{6})

Minimum (\frac{6}{e}, - \frac{18}{e})

Solution steps

f^{'} (x) = 2 x ln (\frac{x}{6}) + x

Find intervals:

Decreasing

: 0 < x < \frac{6}{e},

Increasing

: \frac{6}{e} < x < \infty

Plug

x = \frac{6}{e}

into

x^{2} ln (\frac{x}{6}) : - \frac{18}{e}

Minimum (\frac{6}{e}, - \frac{18}{e})

f (x) = x + 3 - x

e^{3 x} cos (2 x)

\frac{x + 4}{20}

2 x + 3 y = 12

(- 3, 2), (2, - 5)

What is the extreme f(x)=x^2ln(x/6) ?
The extreme f(x)=x^2ln(x/6) is Minimum(6/(sqrt(e)),-(18)/e)