The interval is the largest interval (containing x=0) on which:
We know that this interval is symmetric about so to find the upper end point of the interval you are looking for the solution of:
on (the -ve sign is because close to ).
This has to be solved numerically, when we find the interval is:
An interval can be found by using one of the forms of the remainder of the Taylor series. Which in this case can be slightly modified to give the result that on the interval the error is less than:
so if we set and solve for we will have found such an interval. Doing this we find , and so an interval solving the problem is: