## On .

Here . On the current interval, is a polynomial functrion, therefore it is differentiable; we have:

Here the he first derivative does not vanish; as it is a continuous function, it has a constant sign, nmaely

hence, is an increasing function on .

Let us check whether the function is differentiable on the right at 0:

It follows that

hence, the function is differentiable on the right at 0 at its first right-derivative at 0 is equal to 0.

As is differentiable on the left and on the right at 0 and as the two one-sided derivatives are equal to 0, we conclude that is differentiable at 0 and .

We have one limit to compute here:

Does the graph of have an oblique asymptote here?

therefore the graph of has no oblique asymptote, but a parabolic branch for in neighborhood of .

Here is the variation table of :

Let us now check convexity-concavity of and look for possible inflexion points:

The second derivative does not vanish on , but it vanishes at , and there its sign changes: for and for . Therefore has an inflexion point at .

Noah Dana-Picard 2007-12-28