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

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Answer 1 · 2016-12-01T12:24:30

The only inflection point is $bar x = 5$

The necessary condition for $y(x)$ to have an inflection point in $bar x$ is that:

$y''(bar x) =0$

Calculate the second derivative:

$y'(x) = 2 ((x-3)/(x+1)) ((x+1-x+3)/ (x+1)^2) = 8 (x-3)/(x+1)^3$

$y''(x) = 8 ( ( (x+1)^3 - 3(x-3)(x+1)^2))/(x+1)^6 =$

$= 8 (x+1-3x+9)/(x+1)^4=-16(x-5)/(x+1)^4$

So, the only candidate inflection point is:

$bar x = 5$

As in the neighborhood of $bar x = 5$ $y''(x)$ 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]}

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