Originally Posted by

**Hweengee** i didn't realise that there were 6 possible combinations of x and y such that dy/dx=0. thanks for the help so far, so if there are 3 solutions to the quintic would it be right to say that implies 3 horizontal tangents? making for a total of 9?

yes, it would be right to say that

you would have 3 nw solutions corresponding to a third x-value. making 9 pairs of solutions in all

or would it be best to use a CAS to solve this problem as this is a question that's supposed to be solved using MAPLE.

you can use Maple to find the solutions to the quintic and thus find the specific points. if you are required to do so, then do it

if my memory of Maple serves me correctly,

Code:

f:=y-> y^5 - y + (1/16);
fsolve(f(y)= 0);

should work