Question

if
$f(x) = {(ln \: x)}^{2}$
at which of the following x values is the function both decreasing and concave up?
$x = 0 \\ x = e \\ x = {e}^{2} \\ x = {e}^{3} \\ x = \frac{1}{e}$
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Answer 1

Answer:

$x = \frac{1}{e}$

Step-by-step explanation:

$f(x) = {(lnx)}^{2}$

$f1st(x) = \frac{2 lnx }{x} = 0$

$x < 1 \: \: f1 < 0 \: \: de creasing$

$f2nd(x) = \frac{2 - 2 lnx}{ {x}^{2} } = 0$

$x < e \: \: f2 > 0 \: \: concave \: up$

So x<1 is the condition which contains x=1/e