No, you don't have to show that- you only have to show that f(x)> 0 for .
From , you have so that , for all positive . Now take (or any number less than f''(c))..
So to start off with, fix . Then since f is there exists such thatSuppose that is everywhere, and suppose that satisfies and .
(a) Prove that there exists such that
I'm not sure where I have to go from here. I know I have to show something like:
It might help to draw yourself a picture: geometrically, what is happening is that by stating that f'' is continuous, and that f''(c) > 0, we are saying f'' stays positive in some small interval containing c (symmetric about c).
To show this fact, the hypothesis that f'(c) = 0 is not needed (but other parts of the problem may require it).
The purpose epsilon and delta serve in all this, is to SPECIFY "how small is small enough" (if f is a quadratic polynomial with positive leading coefficient, for example, ANY delta would do).
What HallsOfIvy's post shows is that we are free to choose ANY epsilon we like-so in particular, we can choose an epsilon $0 < \epsilon < f''(c)$.
This, then spits out an appropriate delta-the same one were are given by continuity (the epsilon in a delta/epsilon limit definition tells us "how close the values of the function need to be", and delta tells us "how close the input values need to be to get there").
The reason I say it like this, is it is somewhat easy to get lost in the formal machinery of inequalities, and forget what it is you are actually doing.