How do you find the critical points of #f(x) = x - 3ln(x)#?

1 Answer
Jim H
Apr 1, 2015

I use the definition: a critical point for a function, #f# is a point (number, value), #c# in the domain of #f# at which either #f'(c)=0# or #f'(c)# does not exist.

#f(x) = x-3lnx#

#f'(x)=1-3/x=(x-3)/x#

For this function, #f'(x)# does not exist for #x=0#, but the domain of #f# is #(0,oo)#, so #0# is not a critical point for #f#.

#f'(x)=0# at #x=3# which is in the domain of #f#, so #3# is a critical number for #f#.

(I understand that there are some who would make the critical point #(9, 3-3ln3)#.)

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