How do you find the critical numbers for #f(x) = x-2ln(x)# to determine the maximum and minimum?

1 Answer
Aug 13, 2017

The first derivative is given by

#f'(x) = 1 - 2/x#

Which has critical points at #0# and when #f'(x) = 0#.

#0 = 1- 2/x -> 2/x = 1 -> x = 2#

However, #x = 0# is not really a critical point because the initial function is undefined there. Recall that #ln(0) = O/#. Now let's see if #x = 2# is a maximum or a minimum. At #x = 1#, the function is decreasing because #f'(1) < 0#. Hence, #x = 2# will be an absolute minimum.

Hopefully this helps!