Find critical numbers for f(x)= x(x-2)^(-3) .explain why x= 2 is not one?

1 Answer
Sep 4, 2015

#-1# is the only critical number. #2# is not a critical number because it is not in the domain of #f#.

Explanation:

#f(x) = x(x-2)^(-3) = x/(x-2)^3 #

#f'(x) = ((x-2)^3 - 3x(x-2)^2)/(x-2)^6 = (x-2-3x)/(x-2)^4#

#f'(x) = (-2-2x)/(x-2)^4#

#f'(x) = 0# at #x=-1# and #-1# is in the domain of #f#, so #-1# is a critical number for #f#.

#f'(x)# does not exist at #x=2# but #2# is not in the domain of #f#, so #2# is not a critical number for #f#.

The only critical number of #f# is #-1#.

Recall that: if a function #f# has a local extremum at #c# (if #f(c)# is a local extremum), then #f'(c) = 0# or #f'(c)# does not exist.

A critical number for #f# is a point at which #f# might have a local extremum.
That is: it is a point where #f# is defined (a point in the domain of #f#) at which #f'(c) = 0# or #f'(c)# does not exist.

If the denominator of #f# is zero at #x=a#, then #a# is not in the domain of #f#, so #a# is not a critical number for #f#.

If the denominator or the derivative is #0# at #a#, then we have to ask whether #a# is in the domain of #f#. If it is, then #a# is a critical number. If #a# is not in the domain of #f#, then #a# is not a critical number.

Example: #g(x) = x^(2/3) = root(3)x^2# has domain, all real numbers.

#g'(x) = 2/(3root(3)x)# is not defined at #x=0#.

#0# is in the domain of #f#, so #0# is a critical number for #f#.