What is the extreme f(x)=x^2ln(x/8) ?

The extreme f(x)=x^2ln(x/8) is Minimum(8/(sqrt(e)),-(32)/e)

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extreme f(x)=x^2ln(x/8)

extreme points

f (x) = x^{2} ln (\frac{x}{8})

Minimum (\frac{8}{e}, - \frac{32}{e})

Solution steps

f^{'} (x) = 2 x ln (\frac{x}{8}) + x

Find intervals:

Decreasing

: 0 < x < \frac{8}{e},

Increasing

: \frac{8}{e} < x < \infty

Plug

x = \frac{8}{e}

into

x^{2} ln (\frac{x}{8}) : - \frac{32}{e}

Minimum (\frac{8}{e}, - \frac{32}{e})

x^{2} - 1

f (x) = \frac{1}{2} x + 16

f (x) = \frac{4 x + 7}{9 - x ^{2}}

f (x) = - \frac{2}{3} (x - \frac{1}{2}) - 3

f (x) = lo g_{10} (x + 4)

What is the extreme f(x)=x^2ln(x/8) ?
The extreme f(x)=x^2ln(x/8) is Minimum(8/(sqrt(e)),-(32)/e)