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

Question

Christopher Agresta · Accepted Answer

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!

Luke Alvarez · Answer

undefined To find the critical numbers for $ f(x) = x - 2\ln(x) $, first, find the derivative of the function, then set it equal to zero and solve for $ x $.

1. Find the derivative of $ f(x) $ using the product rule: 
$$ f'(x) = 1 - \frac{2}{x} $$

2. Set $ f'(x) $ equal to zero and solve for $ x $:
$$ 1 - \frac{2}{x} = 0 $$

3. Solve for $ x $:
$$ 1 = \frac{2}{x} $$
$$ x = 2 $$

So, the critical number is $ x = 2 $.

To determine whether it's a maximum or minimum, you can use the first or second derivative test.

Since $ f''(x) = \frac{2}{x^2} $, plug $ x = 2 $ into $ f''(x) $. 
$$ f''(2) = \frac{2}{2^2} = \frac{1}{2} > 0 $$

Since $ f''(2) > 0 $, $ f(x) $ has a local minimum at $ x = 2 $.