# What are the points of inflection of f(x)=x/lnx ?

May 22, 2017

$x \in \left\{1 , {e}^{2}\right\}$

#### Explanation:

Find where $f ' ' \left(x\right) = 0$ OR $f ' ' \left(x\right)$ switches from negative to positive instantaneously (like with an asymptote).

First, we find points where $f ' ' \left(x\right) = 0$.

$f ' \left(x\right) = \frac{\ln x - x \left(\frac{1}{x}\right)}{\ln x} ^ 2 = \frac{1}{\ln x} - \frac{1}{\ln x} ^ 2$

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

Now set this equal to 0.

$0 = - \frac{1}{x {\left(\ln x\right)}^{2}} + \frac{2}{x {\left(\ln x\right)}^{3}}$

$0 = - \ln x + 2$

$\ln x = 2$

$x = {e}^{2}$

This shows that the only solution to $f ' ' \left(x\right) = 0$ is $x = {e}^{2}$. Additionally, in order to be a point of inflection, the graph of $f ' ' \left(x\right)$ must cross the x-axis from positive to negative or negative to positive AT $x = {e}^{2}$ (which can be proven by showing that $\frac{d}{\mathrm{dx}} \left(\frac{x}{\ln} x\right) \ne 0$ when $x = {e}^{2}$, but I will not include this proof unless it is requested).

So, $x = {e}^{2}$ is a point of inflection.

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We also need to look for points of instantaneous change. In this case, we have one: the asymptote at $x = 1$.

At $x = 1$, $\ln \left(x\right) = 0$, and it switches from negative to positive by definition. This means that $- \frac{1}{x {\left(\ln x\right)}^{2}} + \frac{2}{x {\left(\ln x\right)}^{3}}$ will also jump from negative to positive, since the denominator is switching signs. Even though $f ' ' \left(x\right)$ never equals 0, it switches signs at $x = 1$, so $x = 1$ is a point of inflection.

Therefore the two points of inflection for $f \left(x\right) = \frac{x}{\ln} x$ are $x = 1$ and $x = {e}^{2}$