There are a number of problems with this question as asked. The most glaring is that two of the possible "answers" cannot be answers at all. It is not possible in the context of this problem to have the input variable as part of your intervals, so the last two possibilities cannot be the answer.

Assuming that there is no typo in the problem, the correct solution is as follows:

`C(x)=3x^3-24x^2+9x` is the given cost function. We are asked to find the interval(s) where this function is decreasing.

From algebra we know that this function can have either no turning points or exactly two turning points. (In calculus these are local extrema). Since the leading coefficient is positive, the function is either always increasing, or has one interval where it is decreasing.

From calculus we know a function is decreasing if the first derivative is less than zero. So we compute the first derivative:

`C'(x)=9x^2-48x+9` . Setting the first derivative equal to zero we find the critical points:

`9x^2-48x+9=0==>x~~.1946"or"x~~5.1387`

From the first derivative test we see that the function increases on `(-oo,.1946)` , decreases on `(.1946,5.1387)` , and increases on `(5.1387,oo)`

**Thus the function is decreasing on `(.1946,5.1387)` **