What are the values and types of the critical points, if any, of #f(x)=xlnx#?

1 Answer
VNVDVI
Apr 6, 2018

Local minimum: #(1/e, -1/e)#

Explanation:

Take the first derivative, noting that the domain of the original function is #(0, oo).#

#f'(x)=x/x+lnx#

#f'(x)=1+lnx#

The domain of the first derivative is also #(0, oo),# so there won't be any critical points where the first derivative does not exist.

Set to zero and solve for #x.#

#1+lnx=0#

#lnx=-1#

#e^lnx=e^-1#

Recalling that #e^lnx=x:#

#x=1/e#

This is not a very complex function, so we can take the second derivative and apply the Second Derivative Test, which tells us that if #x=a# is a critical value and #f''(a)>0,# then it is a minimum, and if #f''(a)<0,# then it is a maximum.

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

#f''(1/e)=1/(1/e)=e>0#

Thus, we have a local minimum at #x=1/e.# Find the #y-#coordinate:

#f(1/e)=1/eln(1/e)=1/e(-lne)=-1/e#

Local minimum: #(1/e, -1/e)#

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