# How do you find all points of inflection given y=((x-3)/(x+1))^2?

Dec 1, 2016

The only inflection point is $\overline{x} = 5$

#### Explanation:

The necessary condition for $y \left(x\right)$ to have an inflection point in $\overline{x}$ is that:

$y ' ' \left(\overline{x}\right) = 0$

Calculate the second derivative:

$y ' \left(x\right) = 2 \left(\frac{x - 3}{x + 1}\right) \left(\frac{x + 1 - x + 3}{x + 1} ^ 2\right) = 8 \frac{x - 3}{x + 1} ^ 3$

$y ' ' \left(x\right) = 8 \frac{\left({\left(x + 1\right)}^{3} - 3 \left(x - 3\right) {\left(x + 1\right)}^{2}\right)}{x + 1} ^ 6 =$

$= 8 \frac{x + 1 - 3 x + 9}{x + 1} ^ 4 = - 16 \frac{x - 5}{x + 1} ^ 4$

So, the only candidate inflection point is:

$\overline{x} = 5$

As in the neighborhood of $\overline{x} = 5$ $y ' ' \left(x\right)$ changes sign, also the sufficient condition is met and this is effectively an inflection point.

graph{((x-3)/(x+1))^2 [-21, 19, -10.48, 9.52]}