How do use the first derivative test to determine the local extrema #y= (x²-3x+3)/ (x-1) #?

1 Answer
Jul 30, 2015

Answer:

There is a local maximum of #-3# at #x=0#
and a local minimum of #1# at #x=2#.

Explanation:

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.