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

Question

Cameron Alexander · Accepted Answer

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]}

Adam Adkins · Answer

undefined To find the points of inflection for $ y = \left(\frac{x-3}{x+1}\right)^2 $, follow these steps:

1. Find the second derivative of the function.
2. Set the second derivative equal to zero and solve for $ x $.
3. Determine the $ y $-coordinate of each point of inflection by substituting the $ x $-values into the original function.

Let's go through these steps:

1. First derivative:
   $ y' = 2\left(\frac{x-3}{x+1}\right) \left(\frac{(x+1) - (x-3)}{(x+1)^2}\right) $

2. Second derivative:
   $ y'' = 2\left(\frac{x-3}{x+1}\right) \left(\frac{(x+1) - (x-3)}{(x+1)^2}\right) + 2\left(\frac{x-3}{x+1}\right)^2 \left(\frac{-1}{(x+1)^2}\right) $

3. Set $ y'' $ equal to zero and solve for $ x $ to find critical points.
4. Once you find the critical points, determine if they are points of inflection by evaluating the concavity of the function around these points.
5. Find the $ y $-coordinates of the points of inflection by substituting the $ x $-values into the original function.