On $ (-\infty , 0]$ .

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

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

Let us look for points where the first derivative vanishes:

$\displaystyle x \in (-\infty, 0); \; f'(x)=0 \Longleftrightarrow \begin{cases}x=0  x = \pm \sqrt{2} \end{cases}.$    

Actually, only $ -\sqrt{2}$ is in the domain where we work now. Moreover we have:

$\displaystyle \begin{matrix}x<-\sqrt{2} \Longrightarrow f'(x)<0  -\sqrt{2} < x < 0 \Longleftrightarrow f'(x)>0 \end{matrix}$    

Thus, $ f$ decreases on $ (-\infty, -\sqrt{2})$ and increases on $ (-\sqrt{2},0)$ .

What about differentiability on the left 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 left at 0 at its first left-derivative at 0 is equal to 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 $ .

Noah Dana-Picard 2007-12-28