determine any inflection points for the function f(x) = ln(x)+\\frac{32}{x^{3}}. separate multiple answers…

Explanation:

Step1: Find the first - derivative

The function is $f(x)=\ln(x)+\frac{32}{x^{3}}=\ln(x) + 32x^{-3}$, $x>0$. Using the sum - rule of differentiation and the power - rule $(\ln(x))^\prime=\frac{1}{x}$ and $(x^{n})^\prime=nx^{n - 1}$, we have $f^\prime(x)=\frac{1}{x}-96x^{-4}=\frac{1}{x}-\frac{96}{x^{4}}=\frac{x^{3}-96}{x^{4}}$.

Step2: Find the second - derivative

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

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 $384 - x^{3}=0$. Rearranging gives $x^{3}=384$, and then $x=\sqrt[3]{384}=4\sqrt[3]{6}$.

Step4: Check the concavity change

We can use test points. For $0<x<4\sqrt[3]{6}$, let's take $x = 1$. Then $f^{\prime\prime}(1)=\frac{384 - 1}{1}=383>0$, so the function is concave up. For $x>4\sqrt[3]{6}$, let's take $x = 10$. Then $f^{\prime\prime}(10)=\frac{384 - 1000}{10^{5}}=\frac{- 616}{10^{5}}<0$, so the function is concave down.

Answer:

$4\sqrt[3]{6}$