Question

determine any inflection points for the function $f(x)=ln(x)+\frac{9}{x^{2}}$. separate multiple answers with commas. enter dne if no such points exist. inflection points: attempt 1: 2 attempts remaining. submit answer next item

Explanation:

Step1: Find the first - derivative

The function is $f(x)=\ln(x)+\frac{9}{x^{2}}$, $x > 0$. The derivative of $\ln(x)$ is $\frac{1}{x}$ and the derivative of $\frac{9}{x^{2}}=9x^{-2}$ is $- 18x^{-3}=-\frac{18}{x^{3}}$. So, $f^\prime(x)=\frac{1}{x}-\frac{18}{x^{3}}=\frac{x^{2}-18}{x^{3}}$.

Step2: Find the second - derivative

Using the quotient rule, if $y = \frac{u}{v}$, then $y^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$. Here, $u = x^{2}-18$, $u^\prime = 2x$, $v=x^{3}$, $v^\prime = 3x^{2}$. So, $f^{\prime\prime}(x)=\frac{2x\cdot x^{3}-(x^{2}-18)\cdot3x^{2}}{x^{6}}=\frac{2x^{4}-3x^{4}+54x^{2}}{x^{6}}=\frac{-x^{4}+54x^{2}}{x^{6}}=\frac{-x^{2}(x^{2} - 54)}{x^{6}}=\frac{-(x^{2}-54)}{x^{4}}$, $x>0$.

Step3: Set the second - derivative equal to zero

Set $f^{\prime\prime}(x)=0$. Then $\frac{-(x^{2}-54)}{x^{4}} = 0$. Since $x^{4}>0$ for $x>0$, we solve $x^{2}-54 = 0$. So, $x^{2}=54$, and $x=\sqrt{54}=3\sqrt{6}$ (we ignore $x = - 3\sqrt{6}$ because the domain of $y = \ln(x)+\frac{9}{x^{2}}$ is $x>0$).

Answer:

$3\sqrt{6}$