For what values of c is there a straight line that intersects a four-polynomial's graph in four distinct points? 

I can't understand the part of the solution: $\textbf{If and only if}$ 
It is trivial that if a four-degree polynomial has four real solutions, then $f''(x)=0$ has two real solutions. But the converse has a counterexample($f''(x)=0$ has two real solutions can mean that $f'(x)=0$ has three extreme values. But that doesn't mean $f(x)=0$ has four real solution)
Is there a proof about the opposite side? or is the solution wrong?
*I've searched similar question in the web, and they all say that "if and only if" is trivial

You are correct that having three points where $f'(x)=0$ does not guarantee four roots, but having some line intersect four times is less restrictive than having the line $y=0$ intersect four times.  Assume the $x^4$ coefficient is positive, otherwise invert things. If there are two minima you can draw a line between the minima.  The local maximum must be above that line, so you can move it up a bit if necessary and cut the graph in four points.  There must be an inflection point between the local maximum and each local minimum because the curvature changes from down to up somewhere in the segment.  This is a rough proof that two inflections means a line cuts four times.  
If a line cuts four times, subtract the line from the polynomial to get a quartic with four roots.  There must be minimum/maximum/minimum and there must be inflections between them.
