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

Question

2315 views around the world

You can reuse this answer
Creative Commons License

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]}