Is the function concave up or down if $f(x)= (lnx)^2$ ?

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Answer 1 · 2015-04-13T17:55:56

It is concave up on the interval $(0,e)$ and concave down on $(e, oo)$ .

$f(x)=(lnx)^2$ .

$f'(x) = 2(lnx)*1/x = (2lnx)/x$ .

$f''(x) = ((2/x)*x - 2lnx *1)/x^2 = (2(1-lnx))/x^2$ .

$1-lnx = 0$ where $lnx = 1$ which is at $x=e$ .

All other factors of $f''(x)$ are always positive, so the sign of $f''(x)$ is the same as the sign of $1-lnx$ .

That is:
$f''(x)$ is positive if $x < e$ (so $lnx < 1$ ) and
it is negative for $x > e$ (where $lnx > 1$ ).

Is the function concave up or down if f(x)= (lnx)^2f(x)= (lnx)^2?