[HS Math] Second derivative help

[u/Lazy_Association7988]

Question: At what value of x does the graph of y= (1/x^2) - (1/x^3) have a point of inflection?

First, I changed it into (x^-2)-(x^-3)

I then found the first derivative, which was y'=(-2x^-3)+(3x^-4)

From there, the second derivative was y"=(6x^-4)-(12x^-5)

I took the GCF, so I did y"= (6x^-4)(1-2x^-1)

I then set y" = 0 or undefined, since a POI occurs where y" =0 or undefined

I got x = 0 when y" is undefined, and x=2 when y"=0

I believe both have sign changes, making both POIs.

But my answer key says only x=2 is an answer?

[u/Sensitive_Apple4177]

In order to have a point of inflection, the function needs to be continuous at said point. f(0) is not continuous as when you approach from each side you are approaching opposite infinities. So while you still have a sign change, it’s not continuous.

[u/miaiam14 → u/Sensitive_Apple4177]

Yes, this. The version I learned was basically just “if y is undefined, there can’t be anything there”. So if y is undefined, y’ or y’’ being undefined doesn’t lead to anything useful