Find the open interval on which the function is increasing and decreasing and find the local and absolute extreme value????? f(x)= xLnx

2 Answers
Dec 12, 2017

Increasing on #(1/e,oo)# decreasing on #(0,1/e)# Minimum is #f(1/e) = -1/e#

Explanation:

The derivative of #f# is
#f'(x) = lnx+1#

#f'# might change signs where it is #0# or undefined.

This #f'# is defined for all #x# in the domain of #f# and

#f'(x) = 0# at #lnx+1=0#

so #lnx = -1# and #x=e^-1 = 1/e#

On #(0,1/e)#, test using a number in the interval
(I like #e^-2# since #ln(e^-2) = -2# so I can quickly see that #f'(3^-2) = -2+1 < 0#))

#f'(x) < 0# so #f# is decreasing.

On #(1/e,oo)#, #f'(x) > 0# so #f# is increasing. (Test using a number greater than #1/e#, like #1# or #e#)

Therefore #f(1/e)# is a local minimum.
Because there are no other critical numbers, #f(1/e)# is also the absolute minimum.

Dec 12, 2017

The function #f(x) = xlnx# is defined for #x in (0,+oo)#.

Evaluate the derivative:

#d/dx xlnx = x (d/dx lnx) + (d/dx x) lnx = x*1/x +1*lnx = 1+lnx#

Therefore the function is decreasing for:

#1+lnx < 0 => x < 1/e#

and increasing for #x > 1/e#. #f(x)# has correspondingly a local minimum for #x=1/e# with the value #f(1/e) = -1/e#

Note that:

#lim_(x->oo) xf(x) = +oo#

and using l'Hospital's rule:

#lim_(x->0^+) xlnx = lim_(x->0^+) lnx/(1/x) = lim_(x->0^+) (d/dx lnx)/(d/dx 1/x) = lim_(x->0^+) (1/x)/(-1/x^2) = lim_(x->0^+) -x = 0#

So #f(1/e) = -1/e# is also the absolute minimum, and the function does not have an absolute maximum because it is not upper bounded.

graph{xlnx [-10, 10, -5, 5]}