# What are the points of inflection, if any, of f(x) =(x+4)/(x-2)^2?

Dec 9, 2016

$f \left(x\right)$ has an inflection point for $x = - 16$.

#### Explanation:

A necessary condition for the function to have an inflection point is that the second derivative vanishes.

Calculate the second derivative:

$f \left(x\right) = \frac{x + 4}{{\left(x - 2\right)}^{2}}$

$f ' \left(x\right) = \frac{{\left(x - 2\right)}^{2} - 2 \left(x - 2\right) \left(x + 4\right)}{{\left(x - 2\right)}^{4}} = \frac{\left(x - 2\right) - 2 \left(x + 4\right)}{{\left(x - 2\right)}^{3}} = \frac{x - 2 - 2 x - 8}{{\left(x - 2\right)}^{3}} = - \frac{x + 10}{{\left(x - 2\right)}^{3}}$

$f ' ' \left(x\right) = \frac{- {\left(x - 2\right)}^{3} + 3 \left(x + 10\right) {\left(x - 2\right)}^{2}}{{\left(x - 2\right)}^{6}} = \frac{- \left(x - 2\right) + 3 \left(x + 10\right)}{{\left(x - 2\right)}^{4}} = \frac{- x + 2 + 3 x + 30}{{\left(x - 2\right)}^{4}} = 2 \frac{x + 16}{{\left(x - 2\right)}^{4}}$

So $f ' ' \left(x\right) = 0$ for $x = - 16$.

To be sure this is effectively an inflection point we have to check that $f ' ' \left(x\right)$ changes sign around $x = - 16$. We can see that the denominator is always positive, and the numerator is of first grade in x, so it definitely changes sign around the root.

We can conclude that $x = - 16$ is an inflection point.