If $f(x) = (\ln x)^2$, at which of the following $x$ values is the function both increasing and concave down?
i) $x = 0$
ii) $x = e$
iii) $x = e^2$
iv)$x=1/e$
Hi. So in order to get a negative concave, f′′(x)=−(2(1−lnx))/x2 would have to be <0. However, the domain of the function, f(x)=(lnx)2 is x>0 because lnx does not exist for values of x<=0. Which would mean that for every x value, the concave is positive, therefore concave's up. Is this correct?
Note that: \begin{align} f(x)&=[\ln(x)]^{2} \\ f'(x)&=\frac{2\ln(x)}{x} \\ f''(x)&=\frac{2-2\ln(x)}{x^{2}}\end{align} Your function is increasing if $f'(x)>0$ which, in this case, is true if $x>1$. This rules out $x=\frac{1}{e}$ and, as you noted in a comment, $x=0$ is already ruled out because it is not even in the domain of $f$ (so $f$ cannot be increasing there). Now, you need $f''(x)<0$ for concave down which is equivalent to \begin{equation} 2-2\ln(x)<0 \iff 2<2\ln(x) \iff 1<\ln(x)\iff e<x \end{equation} and this rules our $x=e$. So the only remaining choice is $x=e^{2}$.