Investigating Turning Points: What if Every Derivative = 0?

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Discussion Overview

The discussion revolves around the classification of turning points in functions where every derivative at a certain point is zero. Participants explore the implications of this scenario using the function \( f(x) = e^{-\frac{1}{x^2}} \) and question the mathematical methods for proving the nature of such turning points, particularly at \( x = 0 \).

Discussion Character

  • Exploratory
  • Technical explanation
  • Debate/contested

Main Points Raised

  • Owen raises a question about how to classify a turning point when every derivative at that point is zero, specifically referencing the function \( f(x) = e^{-\frac{1}{x^2}} \).
  • One participant clarifies that \( x = 0 \) is a minimum point since the function is positive for all \( x \neq 0 \).
  • Owen questions whether there is a mathematical proof for the minimum classification, expressing concern about the possibility of other local minima nearby.
  • Another participant argues that \( x = 0 \) is a global minimum based on the definition of minima, stating that the existence of other minima does not negate this classification.
  • Participants discuss the limitations of calculus in finding extrema and suggest that original definitions may provide alternative insights.

Areas of Agreement / Disagreement

Participants express differing views on the sufficiency of the arguments presented for classifying the turning point at \( x = 0 \). While some assert it is a minimum based on the function's behavior, others raise concerns about the potential for local minima, indicating that the discussion remains unresolved.

Contextual Notes

There is an implicit assumption that the function's behavior near \( x = 0 \) is well understood, but the discussion does not delve into the details of other potential minima or the broader implications of the function's shape.

Owen-
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Tried a couple of turning points questions recently, because its going to come up on my course again soon and I was rusty.

Only thing i don't get is: what happens if every derivative is 0? How do you classify the turning point, for example:

f(x) =e^-(1/x)

you can differentiate once and set = to 0 to find a turning point at x=0

However every order after that will =0 when x=0

What do?

Thanks,
Owen.
 
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Sorry i didnt mean to say

f(x) =e^-(1/x)

I meant this equation:

[PLAIN]http://www4a.wolframalpha.com/Calculate/MSP/MSP470119d4a2b6bf4c85da00004f4eb393i4c22719?MSPStoreType=image/gif&s=57&w=32&h=31
 
Last edited by a moderator:
I can't see the picture :S
 
Your picture did not load properly. If you meant
y= e^{-\frac{1}{x^2}}
if x is not 0, y(0)= 0, then, yes, that function has every derivative 0 at x= 0. But it is easy to see that if x is not 0, then y> 0 so x= 0 is obviously a minimum point.
 
Ah thanks! That is the equation i was talking about - don't know what hapened to the picture, sorry - Yea i know you can tell if you look at it, but I am just wondering - is there no mathematical way to prove its a minimum?
 
What I said before, " But it is easy to see that if x is not 0, then y> 0 so x= 0 is obviously a minimum point." is a perfectly good "mathematical" way!
 
I see your point, but what if its only a local minimum? Say there was another minimum close to it that was lower. Its not really mathematical if you have to assume that the curve only has one minimum point.
 
0 is a minimum if and only if y(0)\leq y(x) for all values of x. HallsOfIvy perfectly described why this is true. The existence of other possible local minima is irrelevant, because 0 still has to be a global minimum.

While calculus is convenient for finding and classifying extrema, keep in mind the original definitions because there are other ways of finding that information
 

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