Question 1170207
volume is π*r^2*h
=(5-x)^2*(2x+1)*π
The domain is [0,5), so while the function is graphed outside that here, it is only valid within the domain.
{{{graph(300,300,-10,10,-10,200,(5-x)^2*(2x+1)*pi,(2x^3-19x^2+40x+25)*pi)}}}

derivative ix (5-x)^2*(2)+(2x+1)*2(5-x)(-1)
=this is 6x^2-38x+40
That equals 0 when x=1.33 and when x=5
{{{graph(300,300,-10,10,-10,200,6x^2-38x+40)}}} graph of derivative

So the polynomial is increasing on the interval [0, 4/3)] and decreasing on the interval ((4/3), 5)
Note: if you use 1.3, to the nearest tenth, then there is a closed interval for the first [0, 1.3], because 1.3 is in the interval. But the second would not be [1.3 and 5) because it is not decreasing on the interval 1.30 to 1.33.  I suspect this will not be an issue in the answer, but it is worth considering.  There are more issues with intervals even when there isn't rounding, but they are omitted here.