determine any inflection points for the function f(x)=ln(x)+32/x^3. separate multiple answers with commas…

Explanation:

Step1: Find the first - derivative

The function is $f(x)=\ln(x)+\frac{32}{x^{3}}$, where $x > 0$. The derivative of $\ln(x)$ is $\frac{1}{x}$ and the derivative of $\frac{32}{x^{3}}=32x^{-3}$ is $- 96x^{-4}=-\frac{96}{x^{4}}$. So, $f^\prime(x)=\frac{1}{x}-\frac{96}{x^{4}}=\frac{x^{3}-96}{x^{4}}$.

Step2: Find the second - derivative

Differentiate $f^\prime(x)$ with respect to $x$. Using the quotient rule $\left(\frac{u}{v}\right)^\prime=\frac{u^\prime v - uv^\prime}{v^{2}}$, where $u = x^{3}-96$, $u^\prime = 3x^{2}$, $v=x^{4}$, and $v^\prime = 4x^{3}$. Then $f^{\prime\prime}(x)=\frac{3x^{2}\cdot x^{4}-(x^{3}-96)\cdot4x^{3}}{x^{8}}=\frac{3x^{6}-4x^{6}+384x^{3}}{x^{8}}=\frac{-x^{6}+384x^{3}}{x^{8}}=\frac{-x^{3}(x^{3} - 384)}{x^{8}}=\frac{384 - x^{3}}{x^{5}}$.

Step3: Set the second - derivative equal to zero

Set $f^{\prime\prime}(x)=0$, so $\frac{384 - x^{3}}{x^{5}}=0$. Since the denominator $x^{5}\neq0$ for the domain $x>0$, we solve the numerator $384 - x^{3}=0$. Then $x^{3}=384$, and $x = \sqrt[3]{384}=4\sqrt[3]{6}$.

Step4: Check the concavity change

We can use test points. For $0<x<4\sqrt[3]{6}$, let $x = 1$, then $f^{\prime\prime}(1)=\frac{384 - 1}{1}=383>0$, the function is concave up. For $x>4\sqrt[3]{6}$, let $x = 8$, then $f^{\prime\prime}(8)=\frac{384 - 512}{8^{5}}=\frac{-128}{8^{5}}<0$, the function is concave down. So $x = 4\sqrt[3]{6}$ is an inflection point.

Answer:

$4\sqrt[3]{6}$