Extrema of quartic functions $f(x)=ax^4+bx^3+cx^2+dx$ with $a>0$, $c<0$ I am looking for some general statement how the extrema (or only the maximum) of a quartic function looks like. 
My quartic polynomial is bounded from below (fourth order coefficient is positive), the second order coefficient is negative (can look like a "mexican hat" potential), and $f(0)=0 $ (no constant term).
To summarize:
$f(x)=ax^4+bx^3+cx^2+dx$ with $a>0, c<0,(e=0)$
What I want to know: Is there any possibility that the function value at the local maximum is negative? If yes, can we say something about relations of the coefficients?
What I have tried so far: If the maximum becomes negative, we are left with two real roots. There is a condition for this to happen, which I can use to derive some relations 
(see http://en.wikipedia.org/wiki/Quartic_function#Nature_of_the_roots, which becomes much simpler in my case since $e=0$). The problem is, that these conditions are necessary but not sufficient. It does not exclude the case where the function looks effectively quadratic or only has a saddle point. I still need the maximum. Unfortunately, I wasn't able to go on from here, since the equations become messy for me.
Any hint on literature or general theorems are appreciated. 
 A: There is no such polynomial. If there is a negative local maximum at a point $x_0$, then
we can write $f(x_0)=-p$, $f'(x_0)=0$ and $f''(x_0)=-2q$ with $p,q \geq 0$. Using 
Taylor expansion,
$$
\begin{array}{lcl}
f(x) &=& f(x_0)+f'(x_0)(x-x_0)+\frac{f''(x_0)}{2}(x-x_0)^2+\frac{f^{(3)}(x_0)}{6}(x-x_0)^2+\frac{f^{(4)}(x_0)}{4}(x-x_0)^4 \\ 
&=& -p-q(x-x_0)^2+w(x-x_0)^3+a(x-x_0)^4 \ (\text{where} \ w=\frac{f^{(3)}(x_0)}{6})\\
&=& ax^4+(w - 4ax_0)x^3 + (6ax_0^2 - 3wx_0 -q)x^2 + (-4ax_0^3 + 3wx_0^2+qx_0)x + (ax_0^4 -wx_0^3 - qx_0^2 - p)
\end{array}
$$
We see then that $f(0)=ax_0^4 -wx_0^3 - qx_0^2 - p$ and $c=6ax_0^2 - 3wx_0 -q$. Now, consider
$$
t_1=3ax_0^4,\ t_2=3p, \ t_3=2qx_0^2, \ t_4=-cx_0^2, \ t_5=3f(0) 
$$
Then, by hypothesis, all the $t_i$ are nonnegative. But
$$
t_1+t_2+t_3+t_4+t_5+t_6=
3ax_0^4+3p+2qx_0^2+(-6ax_0^4 + 3wx_0^3 +qx_0^2) +(3ax_0^4 -3wx_0^3 - 3qx_0^2 - 3p)=0.
$$
So all the $t_i$ must in fact be zero. In particular $t_1=0$, whence  $x_0=0$. And also $p=0$ since $t_2=0$. Then the local maximum $f(x_0)$
is zero and is not negative.
