An inflection point occurs at points of the domain at which the function changes concavity. The second derivative at an inflection point may be zero but it also may be undefined. If we look at the second derivative for your function
2 & x < 0 \\
0 & 0 \le x \le 3 \\
-2 & x>3
then $F''(x) > 0$ when $x < 0$ and $F''(x)<0$ when $x>3$. We know that the function $F(x)$ is concave up when $F''(x) > 0$ and concave down when $F''(x) < 0$. Therefore, it is clear that $F(x)$ is concave up on $(-\infty,0)$ and concave down on $(3,\infty)$.
But, what about the interval $[0,3]$? Since $F(x)\equiv 0$ when $0\le x \le 3$, the second derivative is also zero. As $F(x)$ changes concavity at the end points we know that there must be at least one inflection point. Therefore, we must decide whether there is one inflection point or two inflection points.
Suppose that we assume that there are two inflection points. Then, we would need to show that the function changes concavity twice - there would be two intervals where it was concave up and one interval where it is concave down. Alternatively, the function could be concave down on two intervals and then concave up on one interval.
Your function is concave up on one interval and then concave down on another interval. In the interval between these two intervals, the function is neither concave up nor concave down. Hence, your function only changes concavity once.
So, $F(x)$ has one inflection point. If we plot the function on Desmos, then we see that this point of inflection occurs in the middle of the interval $[0,3]$.
Interestingly, the function $f(x)=x^3$ exhibits similar behavior. This function is concave up on $(0,\infty)$ and concave down on $(-\infty,0)$. It has one inflection point at $(0,0)$.