How many extremums does a quartic have? Is it true that quartics only have either 1 extremum or 3 extremums? So they can't have 2 extremums? 
 A: If it had only $2$ extremum, the derivative must have $2$ real zeroes $x_1, x_2$  with multiplicity one. But then there exists a $c \neq x_1, x_2$ such that 
$f'(x) = a(x-x_1)(x-x_2)(x-c)$. 
which is another zero of multiplicity one, contradiction. 
A: Here's an explanation that involves some counting and the fundamental theorem or algebra:
If $p$ is your quartic with real coefficients say, the critical points of $p$ are found by solving $p'=0$. Now $p'$ is a cubic with leading coefficient non-zero (else $p$ wouldn't have been a quartic!) so by the fundamental theorem of algebra $p'$ has $3$ roots counting multiplicities in $\Bbb C$. At least one of these is real. 
Now simply look at how you can decompose the number $3$ into smaller integers: 1+1+1, three distinct critical points, $2+1$ two distinct critical points, one repeated, $3$ one distinct critical point repeated three times.
If you'd like examples of each case let me know.
A: If it had two local extrema one would be a minimum and one would be a maximum and on the other side of the maximum it would go towards $-\infty$ and on the other side of the minimum it would go towards $+\infty$ and then it would have an odd degree because the largest exponent monomial dominates for large $|x|$ and an even degree like $x^4$ has would have to go same ways for - as for +.
A: Because the leading term has an even power, a quartic polynomial starts decreasing and ends increasing, or conversely. So there is no room for an even number of extrema.
The derivative, which is a cubic polynomial, can have two roots, but one of them must be double because the number of roots must be odd. In this case, two extrema of the quartic are merged in a so-called stationary inflection point (both $p'$ and $p''$ are zero).

