What is the extreme f(x)=x^2ln(x/4) ?

The extreme f(x)=x^2ln(x/4) is Minimum(4/(sqrt(e)),-8/e)

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extreme f(x)=x^2ln(x/4)

extreme points

f (x) = x^{2} ln (\frac{x}{4})

Minimum (\frac{4}{e}, - \frac{8}{e})

Solution steps

f^{'} (x) = 2 x ln (\frac{x}{4}) + x

Find intervals:

Decreasing

: 0 < x < \frac{4}{e},

Increasing

: \frac{4}{e} < x < \infty

Plug

x = \frac{4}{e}

into

x^{2} ln (\frac{x}{4}) : - \frac{8}{e}

Minimum (\frac{4}{e}, - \frac{8}{e})

\frac{x}{x ^{3} + 8}

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

x + 2 y = 10

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

y = 2 x + 4

What is the extreme f(x)=x^2ln(x/4) ?
The extreme f(x)=x^2ln(x/4) is Minimum(4/(sqrt(e)),-8/e)