# What are the points of inflection, if any, of f(x)=x^4/(x^3+6 ?

Dec 12, 2016

$\left(\sqrt{12} , \frac{2 \sqrt{12}}{3}\right)$

#### Explanation:

$f ' ' \left(x\right) = - \frac{36 {x}^{2} \left({x}^{3} - 12\right)}{{x}^{3} + 6} ^ 3$

An inflection point of $f$ is a point on the graph of $f$ at which the concavity changes.

$f ' '$ changes signs at $x = \sqrt{12}$ and at $- \sqrt{6}$

$- \sqrt{6}$ is not in the domain of $f$, so there is no point on the graph pf $f$ with $x = - \sqrt{6}$.

$\sqrt{12}$ is in the domain of $f$, and $f \left(\sqrt{12}\right) = \frac{2 \sqrt{12}}{3}$
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