<s>[INST] Find the inflection points of $f(x)={1 +\ln^2 x \over x}$ 
Given $$f(x)={1+\ln^2x\over x}, x>0$$
• Find the inflection points of $C_f.$

Personal work:
In order to find the inflection points of $C_f$ we first need to find the second derivative and then at which "$x$", $f''(x)=0.$
$${d\over dx}({1+\ln^2 x \over x})={2lnx-1-\ln^2x\over x^2}$$
$${d\over dx}({2lnx-1-\ln^2x\over x^2})=\cdots={2\ln^2-6\ln x+4 \over x^3}$$
or
$${d\over dx}({d\over dx}({1+\ln^2 x \over x}))={2\ln^2-6\ln x+4 \over x^3}$$
I'm struggling on finding the sign of $f'(x)$ and the points that $f''(x)=0.$
 [/INST] We have $$f''(x)=\frac{\left(\frac{2}{x}-2\ln(x)\cdot \frac{1}{x}\right)x^2-\left(2\ln(x)-1-\ln(x)^2\right)\cdot 2x}{x^4} \\ = \frac{\left(2-2\ln(x)\right)+\left(-2\ln(x)+1+\ln(x)^2\right)\cdot 2}{x^3} \\ = \frac{2-2\ln(x) -4\ln(x)+2+2\ln(x)^2}{x^3} \\ = \frac{2\ln(x)^2 -6\ln(x)+4}{x^3}$$
$f''(x)=0 \Longleftrightarrow 2\ln(x)^2 -6\ln(x)+4=0$
Now, leting $\ln(x) = u$, we have: $$2u^2 -6u+4=0 \implies u=1,2 \\ \implies x=e,e^2$$ which are your possible points of inflection.
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