Sketch y=T(x), you will find that it gas a maximum at x=0, and a minimum at x=6,

and its derivative is negative on (0,6). So |T'(x)|=-T'(x) on the interval.

So now you are looking for the maximum of -T'(x). So differentiate it and set

that derivative to zero solve for x, and substitute back into -T'(x) to find the

maximum. So you need to find the solutions of:

T''(x)=0.

RonL