Let #f(x) = (x²-3x+3)/(x-1)#
Differentiate using the quotient rule and simplify to get:
#f'(x) = (x^2-2x)/(x-1)^2 = (x(x-2))/(x-1)^2#
#f'(x) = 0# at #0# and #2# and #f'(x)# is not defined at #1#.
The domain of #f# includes all Real numbers except #1#,
so the critical numbers for #f# are: #0# and #2#
Test the critical number #0#
If #x# is a little less that #0#, then #x# and #x-2# are negative, while #(x-1)^2# is positive, so #f'(x) > 0#
When #x# is a little greater than #0#, the sign of #x# is positive, but the signs of the other factor remain the same. So #f'(x)# changes to a negative value.
This tells us that #f(0)# (which is #-3#) is a local maximum.
Test the critical number #2#
For#x# a little less than #2#, #f'(x) < 0# and
For #x# a little greater than #2#, the sign changes to #f'(x) > 0#
Therefore #f(2)# (which is #(4-6+3)/1 = 1#) is a local minimum.
