Let

for each . Let denote the minimum of for . Determine for each . The minimum of the function may occur at the endpoints 0 and 1 in some cases.

Since Apostol specifically warns us to watch out for the minimum solution being on the endpoints of we calculate the value at 0 and 1,

Now, we take the derivative

Setting this equal to 0, we find that if has an extremum in the interior of then it occurs at

Then, evaluating at we have

But this is greater than or equal to 0 for all values of , so it cannot be the minimum (since we know , so it is at least greater than ). Therefore, the minimum must occur on the end points. From our evaluation of and we have

Beautiful solution: finding the positive extrema as the absolute value and comparing it to the zero endpoint.