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

Jun 3, 2018

${P}_{\min} \left({e}^{- \frac{1}{4}} , - \frac{1}{4 e}\right)$

Explanation:

we get
$f ' \left(x\right) = {x}^{3} \left(4 \ln \left(x\right) + 1\right)$
and since we have $x > 0$
we get
$f ' \left(x\right) = 0$ if $x = {e}^{- \frac{1}{4}}$
so $f ' ' \left(x\right) = 3 {x}^{2} \left(4 \ln \left(x\right) + 1\right) + {x}^{3} + \frac{4}{x}$

$f ' ' \left(x\right) = 3 {x}^{2} \left(4 \ln \left(x\right) + 1\right) + 4 {x}^{2}$

and

$f ' ' \left({e}^{- \frac{1}{4}}\right) = 4 {\left({e}^{- \frac{1}{4}}\right)}^{2} > 0$
and
$f \left({e}^{- \frac{1}{4}}\right) = - \frac{1}{4} \cdot {e}^{- 1}$