Question

Find an example of a function that has a local minimum at \(x=c\), \(f''(c)>0\) but \(f\) never concave up in any neighborhood containing \(c\)

Answer 1

How can a function have a local minimum at x=c but not be concave up in that area... that doesn't make sense to me. Doesn't a minimum have to be concave up by definition? I don't think you can count an inflection point as a minimum.

Answer 2

Trust me it exist! :D

Answer 3

The definition of local minimum is that there is a neighborhood around c such that to the left of c the function is decreasing and to the right of c the function is increasing

Answer 4

...which is concave up.

Answer 5

not necessarily

Answer 6

I disagree. That to me is the definition of concave up.

Answer 7

If a function is concave up then it is true that there is a neighborhood around c such that f decreasing to the left of c and increasing to the right of c. But not the other way around.

Answer 8

you mean like this? http://tutorial.math.lamar.edu/Classes/CalcI/ShapeofGraphPtII_files/image011.gif

Answer 9

from http://tutorial.math.lamar.edu/Classes/CalcI/ShapeofGraphPtII.aspx

Answer 10

In that picture f'(c) does not exist and here f''(c)=0

Answer 11

Maybe I should make it more clear what I meant I mean by never concave up in any neighborhood. What I wanted to say is that for any open interval \(I\) containing c, there is always an \(x\in I\) such that \(f''(x)<0\)

Answer 12

Hi @myininaya :)

Answer 13

don't run shamil ;3