# How do you find the inflection point(s) of the following equation (1+ln(x))^3?

Mar 29, 2015

I will assume that you know, in general how to find inflection points.

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

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

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

$= \frac{3}{x} ^ 2 \left(1 + \ln x\right) \left[- \left(1 + \ln x\right) \textcolor{red}{+} 2\right]$

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

$f ' ' \left(x\right)$ does not exist for $x \le 0$ and neither does $f \left(x\right)$, so the question of concavity there cannot arise.

Solve: $f ' ' \left(x\right) = 0$

$\frac{3}{x} ^ 2 \left(1 - {\ln}^{2} x\right) = 0$
at $\ln x = \pm 1$, so, $x = e$ or $x = \frac{1}{e}$

Because $\frac{3}{x} ^ 2 > 0$ for all $x$,

the sign of $f ' '$ depends only on the sign of $1 - {\ln}^{2} x$

On $\left(0 , \frac{1}{e}\right)$, we have $\ln x < - 1$ so ${\ln}^{2} x > 1$ and $1 - {\ln}^{2} x < 0$ $f ' ' \left(x\right) < 0$

On $\left(\frac{1}{e} , e\right)$, we have $- 1 < \ln x < 1$ so $0 < {\ln}^{2} x < 1$ and $1 - {\ln}^{2} x > 0$ so $f ' ' \left(x\right) > 0$

The concavity changes at $\left(\frac{1}{e} , f \left(\frac{1}{e}\right)\right) = \left(\frac{1}{e} , 0\right)$

On $\left(e , \infty\right)$, we have $\ln x > 1$ so ${\ln}^{2} x > 1$ and $1 - {\ln}^{2} x < 0$ $f ' ' \left(x\right) < 0$

The concavity changes at $\left(e , f \left(e\right)\right) = \left(e , 8\right)$

The inflection points are $\left(\frac{1}{e} , 0\right)$ and $\left(e , 8\right)$