# I think this is my first proof but need help constructing it

1. Aug 29, 2011

### DrummingAtom

I should say that I've never really try to "prove" many math theorems from my own point of view. Usually, I will just read through a proof and try to grasp the main concept.

I stumbled upon this problem:

Show that y = x3 - 3x + 6 only has one real root.

It got me thinking that this can be generalized to a whole set of x3 functions will have only one real root as long as both of the min/max values are in the positive y of the graph.

y' = 3x2 - 3

Min/max values: x = -1, 1

If I plug these values into the function I get:

y(-1) = 8
y(1) = 4

Which are both positive.

I then took the 2nd derivative to show that the concavity of both min/max values can only possibly go through the x-axis once. Which happens to be when x < -1.

I'm lost on if that information was enough to "show that" for that particular problem. Furthermore, how can I begin generalizing this to a set of x3 equations?

Thanks for any help.

2. Aug 29, 2011

### jambaugh

Nice work, the question begged is how rigorous do you need to be.

What are you assuming about the cubic function which makes your proof work? This is what you need to be explicit about. For example you are assuming continuity. (How do you know cubics are continuous?)
I believe you are invoking also the mean value theorem and the limiting behavior of the cubic as x->infinity and x-> - infinity.

Given the limits you can invoke the limit definition, i.e. there exist sufficiently large x and sufficiently small x such that f(x) is negative for one and positive for the other.

That's the sort of stuff you need to fill in to increase the rigor of your argument.