Investigating Turning Points: What if Every Derivative = 0?

[Owen-]

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.

 

[Owen-]

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

 

[physeven]

I can't see the picture :S

 

[HallsofIvy]

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.

 

[Owen-]

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?

 

[HallsofIvy]

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!

 

[Owen-]

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.

 

[Office_Shredder]

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