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

Answer 1

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

#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#.
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