Idon't see how a 4 digit code can be formed given that the antiderivative is a cubic. can anyone give me insight?

ignore the scribble, .

I believe what this question is saying (albeit poorly) is that you have 4 possible ways to encode what happens on a given interval between critical points, and not that the sequence you would submit as an answer consists of 4 digits.

For a basic example, let's consider . We have



so that is the only critical point, and



so that is the only inflection point.

for all , while for , and vice versa.

We have two intervals to work with, and . So in this case, the answer would be 2 over and 1 over .

That's a guess at any rate. Also, I don't know how helpful this is 4 days after the fact...

(m+11n)^2=(m+11n)(m+11n)=m^2+11mn+11mn+121n^2=m^2+22mn×121n^2 for example; use foil: (a+b)(a+b)first. a^2outside. abinside. ablast. b^2