On $ [0,+\infty )$ .

Here $ f(x)=\frac 14 x^4 +x^2$ . On the current interval, $ f$ is a polynomial functrion, therefore it is differentiable; we have:

$\displaystyle \forall x \in (0,+\infty), \; f'(x)=x^3+2x=x(x^2+2).$    

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

$\displaystyle \forall x \in (0, +\infty), \; f'(x)>0$    

hence, $ f$ is an increasing function on $ (0,+\infty)$ .

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

$\displaystyle x>0 \Longrightarrow \frac {f(x)-f(0)}{x-0}=\frac{\frac 14 x^4+x^2}{x} =\frac 14 x^3+x.$    

It follows that

$\displaystyle \underset{x \underset{>}{\rightarrow}0}{\lim} \frac {f(x)-f(0)}{x-0} =\underset{x \underset{>}{\rightarrow}0}{\lim} (\frac 14 x^3+x)=0$    

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

As $ f$ 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 $ f$ is differentiable at 0 and $ f'(0)=0$ .

We have one limit to compute here:

$\displaystyle \underset{x \rightarrow +\infty}{\lim} f(x)= \underset{x \rightarrow + \infty}{\lim} \frac 14 x^4 = + \infty.$    

Does the graph of $ f$ have an oblique asymptote here?

$\displaystyle \underset{x \rightarrow + \infty}{\lim} \frac {f(x)}{x}= \underset{x \rightarrow + \infty}{\lim} \frac 14 x^3 = + \infty.$    

therefore the graph of $ f$ has no oblique asymptote, but a parabolic branch for $ x$ in neighborhood of $ +\infty$ .

Here is the variation table of $ f$ :

\begin{figure}\centering
\mbox{
\epsfig{file=VariationTable-01.eps,height=3cm}
}
\end{figure}

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

$\displaystyle \begin{matrix}\forall x \in (-\infty, 0), \; f''(x)=\frac 34 x^2-1 \forall x \in (0,+\infty), \; f''(x)=\frac 34 x^2+1 \end{matrix}$    

The second derivative does not vanish on $ (0,+\infty))$ , but it vanishes at $ -\sqrt{4/3}$ , and there its sign changes: $ f''(x)>0$ for $ x<-\sqrt{4/3}$ and $ f''(x)<0$ for $ x>-\sqrt{4/3}$ . Therefore $ f$ has an inflexion point at $ -\sqrt{4/3}$ .

Figure 1: The graph of $ f(x) = \frac 14 x^4 + x\vert x\vert$ .
\begin{figure}\mbox{\epsfig{file=graph1.eps,height=4cm}}\end{figure}

Noah Dana-Picard 2007-12-28