Step 1 :Given the function \(F(x)=2-2 x^{3}\), we first find the derivative of the function, which is \(F'(x)=-6x^{2}\).
Step 2 :We then set the derivative equal to zero to find the critical points: \(-6x^{2}=0\), which gives us \(x=0\).
Step 3 :We then use the second derivative test to determine whether these points are relative minima, maxima, or neither. The second derivative of the function is \(F''(x)=-12x\). At \(x=0\), the second derivative is also 0, which means it's not a relative extrema.
Step 4 :Therefore, there are no relative minimum points and there are no relative maximum points.
Step 5 :To find the intervals over which the function is increasing or decreasing, we again use the derivative. The function is increasing where the derivative is positive and decreasing where the derivative is negative. In this case, the derivative is negative for all \(x\) except 0, so the function is decreasing over the entire real line except at \(x=0\).
Step 6 :Final Answer: \(\boxed{\text{The correct choice for the relative extrema is D. There are no relative minimum points and there are no relative maximum points.}}\)
Step 7 :Final Answer: \(\boxed{\text{The correct choice for the intervals of increase and decrease is B. The function } F(x) \text{ is decreasing over the interval(s) } (-\infty, 0) \cup (0, \infty) \text{ and is not increasing anywhere.}}\)