How do you find the critical points for $x^{4} (ln x)$ $x^4(lnx)$ ?

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Answer 1 · 2018-06-03T07:18:30

$P_min(e^(-1/4),-1/(4e))$

we get
$f'(x)=x^3(4ln(x)+1)$
and since we have $x>0$
we get
$f'(x)=0$ if $x=e^(-1/4)$
so $f''(x)=3x^2(4ln(x)+1)+x^3+4/x$

$f''(x)=3x^2(4ln(x)+1)+4x^2$

and

$f''(e^(-1/4))=4(e^(-1/4))^2>0$
and
$f(e^(-1/4))=-1/4*e^(-1)$

How do you find the critical points for x^4(lnx) x4(lnx)x^4(lnx) ?