How do you determine where the function is increasing or decreasing, and determine where relative maxima and minima occur for #f(x) = x ln x#?

1 Answer

#f(x)# is decreasing on #(0,1/e)# and increasing on #(1/e, oo)#. Therefore, this point is a relative minimum. There are no relative maxima.

Explanation:

In order to find the relative extrema (maxima or minima) for a function, we must find its derivative. This is true because a function is increasing when its derivative is positive and decreasing when its derivative is negative.

However, before we do this, a step students often forget is to consider the domain of the function. Relative extrema cannot exist at points where the function itself does not exist! In this case, #lnx# does not exist when #x<=0#. Therefore, the domain of #f(x)# is #(0,oo)#.

We now find the derivative using the product rule, since #f(x)# is a product of two functions:

#f(x)=xlnx#
#f'(x)=x*(1/x) +lnx*1#
#f'(x)=1+lnx#

We now consider the "critical points" (points in the domain of #f(x)# where #f'(x)=0# or does not exist). These would be places where #f'# could change signs, causing #f# to change directions.

In the domain of #(0,oo)#:
there are no points where #f'# does not exist.
We now look for points in the domain where #f'(x)=0#.
#f'(x)=0#
#1+lnx=0#
#lnx=-1#
#x=e^(-1)=1/e=0#
We now apply the first derivative test. If we select a number in the domain smaller than #x1/e=0.36787944 . . . # such as #x=0.1#,
#f'(0.1) =1+ln(0.1)=-1.30258 . . .#.
If we select a number in the domain larger than #x=1/e# such as #x=5#,
#f'(5) =1+ln(5)=2.68943 . . .#.

Since the sign of #f'(x)# changes from negative to positive at #x=1/e#, the direction of #f(x)# changes from decreasing to increasing at this point, making it a relative minimum by the first derivative test. In fact, since it is the only critical number, it is also the absolute minimum.