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

1 Answer
Apr 6, 2018

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

Explanation:

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

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)