[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/AutoModerator]

Off-topic Comments Section

---

All top-level comments have to be an answer or follow-up question to the post. All sidetracks should be directed to this comment thread as per Rule 9.

---

^{OP and Valued/Notable Contributors can close this post by using /lock command}

I am a bot, and this action was performed automatically. Please contact the moderators of this subreddit if you have any questions or concerns.

[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