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:

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

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

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

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

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

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

It follows that

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

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:

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

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

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

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 $f$ :

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

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

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 $f(x) = \frac 14 x^4 + x\vert x\vert$ .