Optimizing application of derivative

Question

g(t) = t/t-1 on [3,5]

find the absolute minimum and maximun

Answer 1

g(t) = t/(t-1) on [3,5] (endpoints are included and must be checked for global min, max)

To find critical points (not t=1 is critical but not included in the interval)

set dg/dt to 0:

((t-1)-t)/(t-1)² = -1/(t-1)² which is negative definite from 3 to 5

Plug in endpoints 3 and 5 into g. One is max, the other the min.

Please consider a tutor. Take care.