# How do you find the inflection point of the function f(x) = x9ln(x)?

Oct 24, 2015

Assuming that this is $f \left(x\right) = {x}^{9} \ln x$, see the explanation section, below

#### Explanation:

$f \left(x\right) = {x}^{9} \ln x$,

Note that: $\text{dom} \left(f\right) = \left(0 , \infty\right)$

$f ' \left(x\right) = 9 {x}^{8} \ln x + {x}^{8}$

$f ' ' \left(x\right) - 72 {x}^{7} \ln x + 9 {x}^{7} + 8 {x}^{7} = {x}^{7} \left(72 \ln x + 17\right)$

$f ' ' \left(x\right)$ is undefined at $x = 0$, but there cannot be an inflection point at $x = 0$ $\text{ }$(Because $\text{dom} \left(f\right) = \left(0 , \infty\right)$)

$f ' ' \left(x\right) = 0$ at $\ln x = - \frac{17}{72}$, so $x = {e}^{- \frac{17}{72}}$

On the interval $\left(0 , {e}^{- \frac{17}{72}}\right)$,

${x}^{7}$ is positive and $\left(72 \ln x + 17\right)$ is negative.

(Recall that $\ln x$ is increasing, so
$x < {e}^{- \frac{17}{72}} \Rightarrow \ln x < \ln \left({e}^{- \frac{17}{72}}\right) = - \frac{17}{72}$

$\Rightarrow 72 \ln x < - 17$

$\Rightarrow 72 \ln x + 17 < 0$)

So, on the interval $\left(0 , {e}^{- \frac{17}{72}}\right)$, we have $f ' ' \left(x\right)$ is negative.

By similar reasoning, on the interval $\left({e}^{- \frac{17}{72}} , \infty\right)$, we have $f ' ' \left(x\right)$ is positive.

Therefore, the concavity changes at $x = {e}^{- \frac{17}{72}}$, which is in the domain of $f$ and finally, we conclude that there is one inflection point.

The inflection point is $\left({e}^{- \frac{17}{72}} , f \left({e}^{- \frac{17}{72}}\right)\right) = \left({e}^{- \frac{17}{72}} , - \frac{17}{72} \left({e}^{- \frac{17}{8}}\right)\right)$

(Do the arithmetic if someone insists.)