How to determine how many local minima / maxima a polynomial function has? 
The graph of the quintic function $a x^5 + b x^4 + c x^3 + d x^2 + e x + f$ is sketched, where $a, b, c, d, e, f \in \mathbb R$ are given and ܽ$a \neq 0$.
Which one of the following is not possible?
A) The graph has two local minima and two local maxima.
B) The graph has one local minimum and two local maxima.
C) The graph has one local minimum and one local maximum.
D) The graph has no local minima or local maxima.

I was trying to solve this problem, but I'm completely lost. I know that for a quadratic you can find the local minimum/maximum using derivatives.
 A: Since the polynomial has odd degree, it tends towards $\infty$ in one direction and $-\infty$ in the other.  If you try drawing a few possible polynomial graphs like that, you'll see that the number of local maxima and local minima must be the same (and possibly zero of each, as it would be for the graph $y=x^5$.  Therefore, option B is the only one that cannot happen.
A: You can differentiate $f(x)$ to find the maxima/minima in the same way as for a quadratic - in this case, you will obtain a polynomial of order 4 for $f'(x)$. This will have 4 roots, however we are only interested in the real roots of this equation to find stationary points. As the number of complex roots must be even, we know that there can only be 0,2 or 4 real roots, and so B is not possible as the total number of maxima and minima cannot be odd - we cannot have a quartic equation with 3 real roots.
A: It is 5th degree polynomial hence derivative will be 4th degree polynomial.
 4th degree polynomial can have 4 real roots $$ let f'(x) = 0 \;is \,at \;x =\alpha ,\,\beta ,\,\gamma, \,\delta$$
If all distinct they will be alternately points of local Maxima.or minimum
Always local Maxima and minimum will exist alternately even if f'(x)=0 and $x=\alpha = \beta $ is point of local Maxima then $ x=\gamma $ will be point of local minimum 
Since odd degree polynomial has Range as R Thus Maxima and minimum will exist in pair. either $ 1. $No  Maxima nor minimum 
$2.  $1 minimum and 1 maximum 
$ 3. $2 maximum and 2 minimum 
$4$  there will exist complex root win pairs.
Hence (B)  is correct answer.
