For quartic $f(x)$, why can't the number of distinct real solutions of $f(x)=1$, $f(x)=2$, $f(x)=3$, $f(x)=4$ be $1$, $3$, $2$, $4$, respectively? 
$f(x)=ax^4+bx^3+cx^2+dx+e$

$a, b, c, d, e \in\mathbb{R}$

Suppose $f(x)=1$ has p distinct real solutions, $f(x)=2$ has q distinct real solutions, $f(x)=3$ has r distinct real solutions, $f(x)=4$ has s distinct real solutions.
Why is it not possible for $p=1, q=3, r=2, s=4$?

How can you tell that it's not possible for $p=1, q=3, r=2$, and $s=4$?  I am stuck on this problem.
 A: Since $f(x) - 1$ has only one root and all coeficients are in $\mathbb{R}$, we have only two cases to consider:
Case 1. $f(x) - 1 = a(x - \alpha)^4$
In this case, the function obtained is convex (if $a > 0$) and concave otherwise. In particular, it does not have any horizontal line that crosses it $3$ times.
Case 2. $f(x) - 1 = a(x - \alpha)^2 g(x)$ with $g(x)$ without real roots.
Here, we have to use the other data. Since there is a horizontal line that crosses the function $4$ times, we know it must have $3$ local maxima or minima. Also, since the $4$ roots happen at line $y=4$, we know that $a > 0$ and also $g(x) > 0$, that is, the function is always above line $y=1$ (hitting it at point $x = \alpha$). This already provides a very precise picture of the shape of function $f$. If you start with line $y=0$ and begin moving it upwards, the number of roots changes in this sequence:
$$0 \to 1 (\text{at } y=1) \to 2 (\text{immediately after})\to 3 (\text{local min})\to 4 \to 3 (\text{local max}) \to 2 (\text{immediately after}).$$
Notice that this sequence does not have $1 \to 3 \to 2 \to 4$ as a subsequence.
A: Consider $f$ a polynomial of degree $4$, then its derivative is a polynomial of degree $3$, so it has at most three critical points. There are only two cases.
First case: localmin-localmax-localmin (or max-min-max). The plot you have to imagine is this one. If you count the roots from the global minimum upwards you find $1 \rightarrow 2 \rightarrow 3 \rightarrow 4 \rightarrow 3 \rightarrow 2$. But you can't find a subsequence $1\rightarrow 3 \rightarrow 2 \rightarrow 4$ in the roots number so it isn't possible. (And you can find $1\rightarrow 4 \rightarrow 3 \rightarrow 2$ as a subsequence). There are some details missing (what if the two min are at the same height?), but the core idea is here.
Second case: inflection point-min-inflection point (or flex-max-flex). The argument is the same as the first case.
A: If one doesn't count multiplicities of roots
$$
f(x)=36x^4-76x^3+42x^2+1
$$
has one point at which $f(x)=1$, two points at which $f(x)=2$, three points at which $f(x)=3$, and four points at which $f(x)=4$.

