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
.
Figure 1:
The graph of
.
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Noah Dana-Picard
2007-12-28